CTAN sync 202208110300

49 files changed, 3094 insertions, 2454 deletions

diff --git a/FILES.byname b/FILES.byname
index d31bd251fb..29587b2b40 100644
--- a/FILES.byname
+++ b/FILES.byname
@@ -5267,7 +5267,7 @@
 1990/10/02 |       5440 | dviware/vutex/vaxvms.hlp
 1990/10/02 |        812 | dviware/vutex/vaxvms.ins
 1990/10/02 |     127576 | dviware/vutex/vutex.web
-2022/08/09 |     239285 | FILES.last07days
+2022/08/10 |     243093 | FILES.last07days
 2021/07/24 |    1139410 | fonts/aboensis/Aboensis.glyphs
 2021/07/24 |   18907382 | fonts/aboensis/aboensis.pdf
 2021/07/24 |     137860 | fonts/aboensis/Aboensis-Regular.otf
@@ -158407,11 +158407,11 @@
 2020/06/12 |       4382 | graphics/pgf/contrib/commutative-diagrams/tikzlibrarycommutative-diagrams.ramma.code.tex
 2020/06/12 |       7595 | graphics/pgf/contrib/commutative-diagrams/tikzlibrarycommutative-diagrams.velos.code.tex
 2020/06/12 |     569843 | graphics/pgf/contrib/commutative-diagrams.zip
-2022/07/31 |      45786 | graphics/pgf/contrib/customdice/customdice.dtx
+2022/08/10 |      51109 | graphics/pgf/contrib/customdice/customdice.dtx
 2022/07/31 |        417 | graphics/pgf/contrib/customdice/customdice.ins
-2022/07/31 |     369482 | graphics/pgf/contrib/customdice/customdice.pdf
-2022/07/31 |        534 | graphics/pgf/contrib/customdice/README.md
-2022/08/01 |     363456 | graphics/pgf/contrib/customdice.zip
+2022/08/10 |     372025 | graphics/pgf/contrib/customdice/customdice.pdf
+2022/08/10 |        872 | graphics/pgf/contrib/customdice/README.md
+2022/08/10 |     366577 | graphics/pgf/contrib/customdice.zip
 2020/04/15 |      29754 | graphics/pgf/contrib/dpcircling/DPcircling.pdf
 2020/04/15 |       9595 | graphics/pgf/contrib/dpcircling/DPcircling.sty
 2020/04/15 |       2836 | graphics/pgf/contrib/dpcircling/DPcircling.tex
@@ -208214,15 +208214,6 @@
 2017/05/04 |       1031 | macros/latex/contrib/bgteubner/usefiles.txt
 2017/05/05 |    2855695 | macros/latex/contrib/bgteubner.zip
 2009/04/29 |     122631 | macros/latex/contrib/bg.zip
-2015/12/07 |      24203 | macros/latex/contrib/bhcexam/BHCexam.dtx
-2015/10/11 |       3586 | macros/latex/contrib/bhcexam/BHCexam.ins
-2015/10/11 |       3078 | macros/latex/contrib/bhcexam/Makefile
-2015/10/11 |        641 | macros/latex/contrib/bhcexam/README
-2015/10/11 |       5723 | macros/latex/contrib/bhcexam/test1.tex
-2015/10/11 |       5694 | macros/latex/contrib/bhcexam/test2.tex
-2015/10/11 |       5661 | macros/latex/contrib/bhcexam/test3.tex
-2015/10/11 |       5695 | macros/latex/contrib/bhcexam/test4.tex
-2015/12/07 |      18556 | macros/latex/contrib/bhcexam.zip
 2022/01/10 |     332462 | macros/latex/contrib/bibarts/ba-short.pdf
 2022/01/10 |      29417 | macros/latex/contrib/bibarts/ba-short.tex
 2022/01/10 |    1044258 | macros/latex/contrib/bibarts/bibarts.pdf
@@ -226360,17 +226351,20 @@
 2022/04/26 |        323 | macros/latex/contrib/proflabo/README.md
 2022/04/26 |       9327 | macros/latex/contrib/proflabo/tex/ProfLabo.sty
 2022/04/26 |     136164 | macros/latex/contrib/proflabo.zip
+2022/08/10 |         44 | macros/latex/contrib/proflycee/doc/basique.csv
+2022/08/10 |       1034 | macros/latex/contrib/proflycee/doc/cap.csv
 2022/06/28 |       9278 | macros/latex/contrib/proflycee/doc/graphics/pl-doc-stats_a.png
 2022/06/28 |      13958 | macros/latex/contrib/proflycee/doc/graphics/pl-doc-stats_b.png
 2022/06/28 |       9989 | macros/latex/contrib/proflycee/doc/graphics/pl-doc-stats_c2.png
 2022/06/28 |      11759 | macros/latex/contrib/proflycee/doc/graphics/pl-doc-stats_c.png
 2022/06/28 |       7316 | macros/latex/contrib/proflycee/doc/graphics/pl-doc-stats_d.png
 2022/06/28 |       9599 | macros/latex/contrib/proflycee/doc/graphics/pl-doc-stats_e.png
-2022/08/01 |     678803 | macros/latex/contrib/proflycee/doc/ProfLycee-doc.pdf
-2022/08/01 |     104452 | macros/latex/contrib/proflycee/doc/ProfLycee-doc.tex
+2022/08/10 |        805 | macros/latex/contrib/proflycee/doc/perroquet.csv
+2022/08/10 |     732702 | macros/latex/contrib/proflycee/doc/ProfLycee-doc.pdf
+2022/08/10 |     116384 | macros/latex/contrib/proflycee/doc/ProfLycee-doc.tex
 2022/02/16 |        435 | macros/latex/contrib/proflycee/README.md
-2022/08/01 |      50200 | macros/latex/contrib/proflycee/tex/ProfLycee.sty
-2022/08/01 |     757998 | macros/latex/contrib/proflycee.zip
+2022/08/10 |      52218 | macros/latex/contrib/proflycee/tex/ProfLycee.sty
+2022/08/10 |     811440 | macros/latex/contrib/proflycee.zip
 1996/02/05 |       1760 | macros/latex/contrib/progkeys/demo/demo.tex
 1996/02/05 |        459 | macros/latex/contrib/progkeys/demo/prog1.tex
 1996/02/05 |        347 | macros/latex/contrib/progkeys/demo/prog2.tex
@@ -228945,12 +228939,12 @@
 2022/08/02 |     219736 | macros/latex/contrib/sillypage/sillypage.pdf
 2022/08/02 |      18636 | macros/latex/contrib/sillypage/sillywalk-map.pdf
 2022/08/02 |     317481 | macros/latex/contrib/sillypage.zip
-2022/05/08 |       1100 | macros/latex/contrib/simplebnf/LICENSE
+2022/08/10 |       1100 | macros/latex/contrib/simplebnf/LICENSE
 2022/05/08 |        510 | macros/latex/contrib/simplebnf/README.md
 2022/05/08 |     152698 | macros/latex/contrib/simplebnf/simplebnf-doc.pdf
-2022/05/08 |       3626 | macros/latex/contrib/simplebnf/simplebnf-doc.tex
-2022/05/08 |       6898 | macros/latex/contrib/simplebnf/simplebnf.sty
-2022/05/08 |     152593 | macros/latex/contrib/simplebnf.zip
+2022/08/10 |       4526 | macros/latex/contrib/simplebnf/simplebnf-doc.tex
+2022/08/10 |       7101 | macros/latex/contrib/simplebnf/simplebnf.sty
+2022/08/10 |     152771 | macros/latex/contrib/simplebnf.zip
 2013/02/28 |     223740 | macros/latex/contrib/simplecd/examples.pdf
 2012/12/20 |       1574 | macros/latex/contrib/simplecd/examples.tex
 2013/02/28 |       3060 | macros/latex/contrib/simplecd/README
@@ -236687,114 +236681,114 @@
 2015/10/04 |     263467 | macros/luatex/generic/luatexbase/luatexbase.pdf
 2015/10/04 |        450 | macros/luatex/generic/luatexbase/README.md
 2015/10/04 |     270547 | macros/luatex/generic/luatexbase.zip
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-2022/08/07 |       3238 | macros/luatex/generic/luatexja/doc/jfm-test.lua
-2022/08/07 |       4452 | macros/luatex/generic/luatexja/doc/jfm-ujisc33.lua
-2022/08/07 |     166557 | macros/luatex/generic/luatexja/doc/lltjp-geometry.pdf
-2022/08/07 |      13726 | macros/luatex/generic/luatexja/doc/lltjp-geometry.tex
-2022/08/07 |     397581 | macros/luatex/generic/luatexja/doc/ltjclasses.pdf
-2022/08/07 |      69217 | macros/luatex/generic/luatexja/doc/ltjltxdoc.pdf
-2022/08/07 |     456577 | macros/luatex/generic/luatexja/doc/ltjsclasses.pdf
-2022/08/07 |     541346 | macros/luatex/generic/luatexja/doc/luatexja.dtx
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-2022/08/07 |     390299 | macros/luatex/generic/luatexja/doc/luatexja-ruby.pdf
-2022/08/07 |      57155 | macros/luatex/generic/luatexja/doc/luatexja-ruby.tex
-2022/08/07 |       3796 | macros/luatex/generic/luatexja/README
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-2022/08/03 |      12842 | macros/latex/contrib/europecv/europasslogo.pdf
-2022/08/03 |    1210836 | systems/texlive/tlnet/archive/profcollege.r64032.tar.xz
-2022/08/03 |    1187253 | macros/latex/contrib/europecv.zip
-2022/08/03 |      10499 | macros/latex/contrib/europecv/templates/cv_template_en_bieber.tex
-2022/08/03 |      10078 | macros/latex/contrib/tuda-ci/tex/beamerthemeTUDa-mecheng.sty
-2022/08/03 |       1002 | macros/latex/contrib/europecv/examples/bulgarian-utf8.tex
-2022/08/03 |   10015344 | systems/texlive/tlnet/archive/profcollege.doc.r64032.tar.xz
diff --git a/graphics/pgf/contrib/customdice/README.md b/graphics/pgf/contrib/customdice/README.md
index a898f330ce..29012a12f8 100644
--- a/graphics/pgf/contrib/customdice/README.md
+++ b/graphics/pgf/contrib/customdice/README.md
@@ -9,3 +9,15 @@
 By [Peter Rowlett](https://github.com/prowlett/).
 
 The package is licenced under [Creative Commons Attribution-ShareAlike 4.0 International (CC BY-SA 4.0)](https://creativecommons.org/licenses/by-sa/4.0/).
+
+## Changes
+
+### [1.1] - 2022-08-10
+
+- Added option to include 7, 8 or 9 dots on a standard dice face via `\dice{}`
+- Added control over dice face border colour.
+- Fixed bug meaning extra whitespace appeared after e.g. `\Largedice{5}` that did not appear with `\Large\dice{5}`.
+
+### [1.0] - 2022-07-31
+
+- First working version of package
diff --git a/graphics/pgf/contrib/customdice/customdice.dtx b/graphics/pgf/contrib/customdice/customdice.dtx
index ac44ae9fd9..ed1540d3f9 100644
--- a/graphics/pgf/contrib/customdice/customdice.dtx
+++ b/graphics/pgf/contrib/customdice/customdice.dtx
@@ -3,7 +3,7 @@
 %
 %The package is licenced under Creative Commons Attribution-ShareAlike 4.0 International (CC BY-SA 4.0). To view a copy of this license, visit http://creativecommons.org/licenses/by-sa/4.0/
 %\fi
-%\title{customdice v1.0}
+%\title{customdice v1.1}
 %\author{Peter Rowlett}
 %\maketitle
 %    
@@ -44,6 +44,21 @@
 %
 %\centerline{\Huge \dice{1} \dice{2} \dice{3} \dice{4} \dice{5} \dice{6}}
 %
+%Apparently there are occasions where people desire dice faces as if from a six-sided cube that show 7, 8 or 9 dots. This is incorporated into \verb|\dice| also. 
+%
+%For example
+%
+%\begin{verbatim}
+%\dice{7} \dice{8} \dice{9}\end{verbatim}
+%
+%produces output like this: 
+%
+%\centerline{\Huge \dice{7} \dice{8} \dice{9}}
+%
+%Also, inputting \verb|\dice{0}| produces a blank face, like this:
+%
+%\centerline{\Huge \dice{0}}
+%
 %\subsection{Big dot dice}
 %
 %The command \verb|\bigdotdice| produces a dice face with one big dot in the centre.
@@ -386,6 +401,25 @@
 %
 %Note that calling \verb|\setdicefacesize| will reset \verb|\customdicetextscale| relative to the new \verb|\customdicefacesize|. 
 %
+%\subsection{Border colour}
+%
+%The background and foreground colours can be set by passing parameters to the commands as outlined above. The border of the dice face is set globally to \verb|darkgray|, with the assumption that this will be fine with any colour combination. It is customisable though. You can set the global dice face border colour to any colour name using \verb|\setdicefaceoutlinecol{name}| and restore the default using \verb|\defaultdicefaceoutlinecol|. 
+%
+%For example, 
+%
+%\begin{verbatim}
+%\setdicefaceoutlinecol{red}\dice{5}
+%\setdicefaceoutlinecol{orange}\dice{5}
+%\setdicefaceoutlinecol{yellow}\dice{5}
+%\setdicefaceoutlinecol{green}\dice{5}
+%\setdicefaceoutlinecol{blue}\dice{5}
+%\setdicefaceoutlinecol{violet}\dice{5}
+%\defaultdicefaceoutlinecol\dice{5}\end{verbatim}
+%
+%produces output like this: 
+%
+%\centerline{\Huge \setdicefaceoutlinecol{red}\dice{5} \setdicefaceoutlinecol{orange}\dice{5} \setdicefaceoutlinecol{yellow}\dice{5} \setdicefaceoutlinecol{green}\dice{5} \setdicefaceoutlinecol{blue}\dice{5} \setdicefaceoutlinecol{violet}\dice{5} \defaultdicefaceoutlinecol\dice{5}}
+%
 %\subsection{Arbitrary drawing on dice faces}
 %
 %The commands \verb|\dice|, \verb|\bigdotdice|, \verb|\textdice| and \verb|\textdicedot| all start an environment called \verb|customdiceenv| and do some drawing on it. \verb|customdiceenv| is really a \verb|tikzpicture| with baseline set to \verb|\customdicebaseline ex| and the dice face outline drawn from \verb|(0ex,0ex)| to \\
@@ -570,12 +604,26 @@
 {\PackageWarning{customdice}{command `hugetextdicebot' already defined}}
 \@ifundefined{Hugetextdicebot}{}
 {\PackageWarning{customdice}{command `Hugetextdicebot' already defined}}
+\@ifundefined{dicefaceoutlinecol}{}
+{\PackageWarning{customdice}{command `dicefaceoutlinecol' already defined}}
+\@ifundefined{setdicefaceoutlinecol}{}
+{\PackageWarning{customdice}{command `setdicefaceoutlinecol' already defined}}
+\@ifundefined{defaultdicefaceoutlinecol}{}
+{\PackageWarning{customdice}{command `defaultdicefaceoutlinecol' already defined}}
 \makeatother
 
-% sizes
+% colours
 \newcommand{\customdicecoldefault}{f40fc2340eb940fc8a170db0f81d7139} % using a UUID so we are extremely unlikely to clash with any user-specified colour name
 \global\let\customdicebg\customdicecoldefault
 \global\let\customdicefg\customdicecoldefault
+\newcommand{\dicefaceoutlinecol}{darkgray}
+\newcommand{\setdicefaceoutlinecol}[1]{%
+    \renewcommand{\dicefaceoutlinecol}{#1}%
+}
+\newcommand{\defaultdicefaceoutlinecol}{%
+    \renewcommand{\dicefaceoutlinecol}{darkgray}%
+}
+% sizes
 \newcommand{\customdicebaseline}{0.02}
 \newcommand{\setdicebaseline}[1]{%
     \renewcommand{\customdicebaseline}{#1}%
@@ -607,7 +655,7 @@
         \ifdefequal{\customdicebg}{\customdicecoldefault}{\global\let\customdicebg\colour}{\global\let\customdicefg\colour}%
     }%
     \begin{tikzpicture}[baseline=\customdicebaseline ex]
-    \draw[rounded corners=\customdicecornerrounding ex,darkgray,fill=\customdicebg,line width=\customdiceborderthickness ex] (0ex,0ex) rectangle (\customdicefacesize ex,\customdicefacesize ex);
+    \draw[rounded corners=\customdicecornerrounding ex,\dicefaceoutlinecol,fill=\customdicebg,line width=\customdiceborderthickness ex] (0ex,0ex) rectangle (\customdicefacesize ex,\customdicefacesize ex);
 }{\end{tikzpicture}%
     \global\let\customdicebg\customdicecoldefault%
     \global\let\customdicefg\customdicecoldefault%
@@ -643,10 +691,39 @@
             \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdiceupper ex,\customdicelower ex) {};
             \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdiceupper ex,\customdicehalfway ex) {};
             \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdiceupper ex,\customdiceupper ex) {};
-        }{}}}}}}
+        }{\ifnumequal{#2}{7}{
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicelower ex,\customdicelower ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicelower ex,\customdicehalfway ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicelower ex,\customdiceupper ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicehalfway ex,\customdicehalfway ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdiceupper ex,\customdicelower ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdiceupper ex,\customdicehalfway ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdiceupper ex,\customdiceupper ex) {};
+        }{\ifnumequal{#2}{8}{
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicelower ex,\customdicelower ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicelower ex,\customdicehalfway ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicelower ex,\customdiceupper ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicehalfway ex,\customdicelower ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicehalfway ex,\customdiceupper ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdiceupper ex,\customdicelower ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdiceupper ex,\customdicehalfway ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdiceupper ex,\customdiceupper ex) {};
+        }{\ifnumequal{#2}{9}{
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicelower ex,\customdicelower ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicelower ex,\customdicehalfway ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicelower ex,\customdiceupper ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicehalfway ex,\customdicelower ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicehalfway ex,\customdicehalfway ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdicehalfway ex,\customdiceupper ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdiceupper ex,\customdicelower ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdiceupper ex,\customdicehalfway ex) {};
+            \node[circle,fill=\customdicefg,inner sep=\customdicedotsize ex] at (\customdiceupper ex,\customdiceupper ex) {};
+        }{}}}}}}}}}
     \end{customdiceenv}%
 }
 
+
+
 % \bigdotdice draws dice with a big dot
 \newcommand{\bigdotdice}[1][white,black]{%
     \begin{customdiceenv}[#1]
@@ -683,124 +760,124 @@
 
 % for convenience, passes \sizecommand as {\size\command}
 \newcommand{\tinydice}[2][white,black]{%
-    {\tiny\dice[#1]{#2}}
+    {\tiny\dice[#1]{#2}}%
 }
 \newcommand{\scriptsizedice}[2][white,black]{%
-    {\scriptsize\dice[#1]{#2}}
+    {\scriptsize\dice[#1]{#2}}%
 }
 \newcommand{\footnotesizedice}[2][white,black]{%
-    {\footnotesize\dice[#1]{#2}}
+    {\footnotesize\dice[#1]{#2}}%
 }
 \newcommand{\smalldice}[2][white,black]{%
-    {\small\dice[#1]{#2}}
+    {\small\dice[#1]{#2}}%
 }
 \newcommand{\normalsizedice}[2][white,black]{%
-    {\normalsize\dice[#1]{#2}}
+    {\normalsize\dice[#1]{#2}}%
 }
 \newcommand{\largedice}[2][white,black]{%
-    {\large\dice[#1]{#2}}
+    {\large\dice[#1]{#2}}%
 }
 \newcommand{\Largedice}[2][white,black]{%
-    {\Large\dice[#1]{#2}}
+    {\Large\dice[#1]{#2}}%
 }
 \newcommand{\LARGEdice}[2][white,black]{%
-    {\LARGE\dice[#1]{#2}}
+    {\LARGE\dice[#1]{#2}}%
 }
 \newcommand{\hugedice}[2][white,black]{%
-    {\huge\dice[#1]{#2}}
+    {\huge\dice[#1]{#2}}%
 }
 \newcommand{\Hugedice}[2][white,black]{%
-    {\Huge\dice[#1]{#2}}
+    {\Huge\dice[#1]{#2}}%
 }
 \newcommand{\tinybigdotdice}[1][white,black]{%
-    {\tiny\bigdotdice[#1]}
+    {\tiny\bigdotdice[#1]}%
 }
 \newcommand{\scriptsizebigdotdice}[1][white,black]{%
-    {\scriptsize\bigdotdice[#1]}
+    {\scriptsize\bigdotdice[#1]}%
 }
 \newcommand{\footnotesizebigdotdice}[1][white,black]{%
-    {\footnotesize\bigdotdice[#1]}
+    {\footnotesize\bigdotdice[#1]}%
 }
 \newcommand{\smallbigdotdice}[1][white,black]{%
-    {\small\bigdotdice[#1]}
+    {\small\bigdotdice[#1]}%
 }
 \newcommand{\normalsizebigdotdice}[1][white,black]{%
-    {\normalsize\bigdotdice[#1]}
+    {\normalsize\bigdotdice[#1]}%
 }
 \newcommand{\largebigdotdice}[1][white,black]{%
-    {\large\bigdotdice[#1]}
+    {\large\bigdotdice[#1]}%
 }
 \newcommand{\LARGEbigdotdice}[1][white,black]{%
-    {\LARGE\bigdotdice[#1]}
+    {\LARGE\bigdotdice[#1]}%
 }
 \newcommand{\Largebigdotdice}[1][white,black]{%
-    {\Large\bigdotdice[#1]}
+    {\Large\bigdotdice[#1]}%
 }
 \newcommand{\hugebigdotdice}[1][white,black]{%
-    {\huge\bigdotdice[#1]}
+    {\huge\bigdotdice[#1]}%
 }
 \newcommand{\Hugebigdotdice}[1][white,black]{%
-    {\Huge\bigdotdice[#1]}
+    {\Huge\bigdotdice[#1]}%
 }
 \newcommand{\tinytextdice}[2][white,black]{%
-    {\tiny\textdice[#1]{#2}}
+    {\tiny\textdice[#1]{#2}}%
 }
 \newcommand{\scriptsizetextdice}[2][white,black]{%
-    {\scriptsize\textdice[#1]{#2}}
+    {\scriptsize\textdice[#1]{#2}}%
 }
 \newcommand{\footnotesizetextdice}[2][white,black]{%
-    {\footnotesize\textdice[#1]{#2}}
+    {\footnotesize\textdice[#1]{#2}}%
 }
 \newcommand{\smalltextdice}[2][white,black]{%
-    {\small\textdice[#1]{#2}}
+    {\small\textdice[#1]{#2}}%
 }
 \newcommand{\normalsizetextdice}[2][white,black]{%
-    {\normalsize\textdice[#1]{#2}}
+    {\normalsize\textdice[#1]{#2}}%
 }
 \newcommand{\largetextdice}[2][white,black]{%
-    {\large\textdice[#1]{#2}}
+    {\large\textdice[#1]{#2}}%
 }
 \newcommand{\Largetextdice}[2][white,black]{%
-    {\Large\textdice[#1]{#2}}
+    {\Large\textdice[#1]{#2}}%
 }
 \newcommand{\LARGEtextdice}[2][white,black]{%
-    {\LARGE\textdice[#1]{#2}}
+    {\LARGE\textdice[#1]{#2}}%
 }
 \newcommand{\hugetextdice}[2][white,black]{%
-    {\huge\textdice[#1]{#2}}
+    {\huge\textdice[#1]{#2}}%
 }
 \newcommand{\Hugetextdice}[2][white,black]{%
-    {\Huge\textdice[#1]{#2}}
+    {\Huge\textdice[#1]{#2}}%
 }
 \newcommand{\tinytextdicebot}[2][white,black]{%
-    {\tiny\textdicebot[#1]{#2}}
+    {\tiny\textdicebot[#1]{#2}}%
 }
 \newcommand{\scriptsizetextdicebot}[2][white,black]{%
-    {\scriptsize\textdicebot[#1]{#2}}
+    {\scriptsize\textdicebot[#1]{#2}}%
 }
 \newcommand{\footnotesizetextdicebot}[2][white,black]{%
-    {\footnotesize\textdicebot[#1]{#2}}
+    {\footnotesize\textdicebot[#1]{#2}}%
 }
 \newcommand{\smalltextdicebot}[2][white,black]{%
-    {\small\textdicebot[#1]{#2}}
+    {\small\textdicebot[#1]{#2}}%
 }
 \newcommand{\normalsizetextdicebot}[2][white,black]{%
-    {\normalsize\textdicebot[#1]{#2}}
+    {\normalsize\textdicebot[#1]{#2}}%
 }
 \newcommand{\largetextdicebot}[2][white,black]{%
-    {\large\textdicebot[#1]{#2}}
+    {\large\textdicebot[#1]{#2}}%
 }
 \newcommand{\Largetextdicebot}[2][white,black]{%
-    {\Large\textdicebot[#1]{#2}}
+    {\Large\textdicebot[#1]{#2}}%
 }
 \newcommand{\LARGEtextdicebot}[2][white,black]{%
-    {\LARGE\textdicebot[#1]{#2}}
+    {\LARGE\textdicebot[#1]{#2}}%
 }
 \newcommand{\hugetextdicebot}[2][white,black]{%
-    {\huge\textdicebot[#1]{#2}}
+    {\huge\textdicebot[#1]{#2}}%
 }
 \newcommand{\Hugetextdicebot}[2][white,black]{%
-    {\Huge\textdicebot[#1]{#2}}
+    {\Huge\textdicebot[#1]{#2}}%
 }
 %</customdice>
 %\fi
\ No newline at end of file
diff --git a/graphics/pgf/contrib/customdice/customdice.pdf b/graphics/pgf/contrib/customdice/customdice.pdf
index 4b7f9b0a21..efcfa4cfb3 100644
--- a/graphics/pgf/contrib/customdice/customdice.pdf
+++ b/graphics/pgf/contrib/customdice/customdice.pdf
diff --git a/macros/latex/contrib/bhcexam/BHCexam.dtx b/macros/latex/contrib/bhcexam/BHCexam.dtx
deleted file mode 100644
index 019d74cba0..0000000000
--- a/macros/latex/contrib/bhcexam/BHCexam.dtx
+++ /dev/null
@@ -1,859 +0,0 @@
-% \iffalse meta-comment
-%
-% Copyright (C) 2014 by Charles Bao <charley792@gmail.com>
-%
-% This file is part of the BHCexam package project.
-% ---------------------------------------------------
-%
-% It may be distributed under the conditions of the LaTeX Project Public
-% License, either version 1.2 of this license or (at your option) any
-% later version. The latest version of this license is in
-%    http://www.latex-project.org/lppl.txt
-% and version 1.2 or later is part of all distributions of LaTeX
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-% \fi
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-%%   Exclamation   \!     Double quote  \"     Hash (number) \#
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-%<cfg>\ProvidesFile{BHCexam.cfg}
-  [2015/10/10 v0.4 BHCexam
-%<cls>   document class]
-%<cfg>   configuration file]
-%
-%<*driver>
-   bundle source file]
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-%
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-\documentclass[a4paper]{ltxdoc}
-\usepackage{ctex}
-\usepackage{hyperref}
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- \textwidth 14 true cm
-\EnableCrossrefs
- %\DisableCrossrefs % Say \DisableCrossrefs if index is ready
-\CodelineIndex
-\RecordChanges      % Gather update information
- %\OnlyDescription  % comment out for implementation details
- %\OldMakeindex     % use if your MakeIndex is pre-v2.9
-\hypersetup{colorlinks,linkcolor=blue,citecolor=blue}
-\begin{document}  
- \DocInput{BHCexam.dtx}
-\end{document}
-%</driver>
-%
-% \fi
-%
-%
-% \changes{v0.0}{2011/07/22}{Initial version}
-% \changes{v0.1}{2011/07/23}{version 0.1}
-% \changes{v0.2}{2011/07/27}{version 0.2}
-% \changes{v0.3}{2014/02/18}{version 0.3}
-% \changes{v0.3}{2015/10/10}{version 0.4}
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-%
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-% \DoNotIndex{\normalfont,\bfseries,\slshape,\interlinepenalty}
-% \DoNotIndex{\hfil,\par,\vskip,\vspace,\quad}
-% \DoNotIndex{\centering,\raggedright}
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-% \DoNotIndex{\@@par}
-%
-%
-% \GetFileInfo{BHCexam.dtx}
-%
-%
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-%
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-% \newcommand{\BHCexam}{\texttt{BHCexam}}
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-%
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-% \def\@oarg#1{{\tt[}{\rm #1}{\tt]}}
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-%
-%
-% \title{\bf \BHCexam~宏包说明\thanks
-%   {这是鲍宏昌发布的第一个~\LaTeX~宏包。本文件版本号为~\fileversion{}，最后修改日期~\filedate{}。}}
-% \author{\it 鲍宏昌\thanks{charley792@gmail.com}}
-% \date{\small 打印日期：~\today}
-% \maketitle
-%
-% \begin{abstract}
-% \BHCexam~宏包提供了一个中学试卷排版的~\LaTeX{}~文档类。
-%
-% \BHCexam~主要文件包括~\texttt{BHCexam.cls}~文档类和配置文件
-% ~\texttt{BHCexam.cfg}。
-%
-% \BHCexam~宏包由鲍宏昌制作并负责维护。
-% \end{abstract}
-%
-% \tableofcontents
-% \newpage
-%
-% \section{简介}
-%
-% 本宏包以~\exam~为底层文档类，部分源代码来自于盖鹤麟开发的
-% ~\colin。不知道什么原因盖鹤麟自2004年就一直没有发布更新，
-% ~\colin~仍然使用CCT实现中文支持，而缺乏对~XeTeX~的支持。2011年7月，
-% 鲍宏昌在~\colin~的基础上改用~\ctex~实现中文支持，采用UTF8编码同时
-% 支持~XeLaTeX~和~pdfTeX~进行编译，并使用~\texttt{doc}~和
-% ~\texttt{docstrip}~工具编写了这个文档，增加了一些新的功能，
-% 并把新的宏包命名为~\BHCexam。
-%
-% 本宏包延续了~\colin~和~\exam~的特点，能让一个刚刚接触
-% ~\LaTeX~的初学者，也能轻松用它来排版试卷。希望~\BHCexam~能提高中学
-% 教师的工作效率，并把注意力放在试卷的内容上。
-% 
-% \BHCexam~由两个主要文件构成：文档类~\texttt{BHCexam.cls}~和配置文件
-% ~\texttt{BHCexam.cfg}~。后者定义了一些常用的参数。
-%
-% {\kaishu
-% 这两个文件可以通过用~XeLaTeX~编译~\texttt{BHCexam.ins}~文件来得到，
-% 而这份说明文档可以通过用~XeLaTeX~编译~\texttt{BHCexam.dtx}~文件来得到。
-% 编译说明文档需要~\ctex{}~宏包，为了生成正确的索引和版本记录，
-% 需要使用如下命令}
-% \begin{verbatim}
-% makeindex -s gind.ist -o BHCexam.ind BHCexam.idx
-% makeindex -s gglo.ist -o BHCexam.gls BHCexam.glo
-% \end{verbatim}
-%
-% \section{一个简单的例子}
-%
-% 用~\BHCexam~要排版一张基本的试卷其实很简单。如果你准备对试卷的排版进行
-% 更细致的设置，那么请参考~\exam~的文档。
-%
-% \subsection{\texttt{documentclass} 命令}
-% \label{sec:BasicDocumentclass}
-%
-% 要使用~\BHCexam~文档类，你的\verb"\documentclass" 命令应该是
-% \begin{verbatim}
-%  \documentclass{BHCexam}
-% \end{verbatim}
-% 如果，你想使用小四字体作为缺省字体大小，那么添加选项\verb"cs4size"
-% \begin{verbatim}
-% \documentclass[cs4size]{BHCexam}
-% \end{verbatim}
-% 更多的选项，请参考\ref{sec:Options}。
-%
-% \subsection{打印标题和考试须知}
-% \DescribeMacro{\maketitle}
-% \DescribeMacro{\notice}
-% 在试卷上打印标题和考试须知
-% \begin{verbatim}
-%   \maketitle
-%   \notice 
-% \end{verbatim}
-% 关于标题和考试须知中变量的设置，请参考\ref{sec:Variable}。
-%
-% \subsection{题目}
-% \label{sec:Example}
-% \DescribeMacro{\question}
-% \DescribeMacro{\stk}
-% \DescribeMacro{\onech}
-% \DescribeMacro{\part}
-% 在\verb"questions"环境中用\verb"\question"输入题目。
-% 在\verb"parts"环境中用\verb"\part"输入大题的小问。
-% 用\verb"\stk"、\verb"\mtk"
-% 和\verb"\ltk"输入填空题的答案。
-% 用\verb"\onech"、\verb"\twoch"
-% 和\verb"\fourch"输入选择题的选项。
-% 在\verb"questions"环境中用\verb"\tiankong"、\verb"\xuanze"
-% 和\verb"\jianda"分别显示填空题、选择题、简答题的提示语。
-% \begin{verbatim}
-% \begin{questions}
-%   \tiankong
-%   \question 这是第1道填空题\stk{答案，不显示答案时显示段横线}
-%   \question 这是第2道填空题\mtk{答案，不显示答案时显示段横线}
-%   \question 这是第2道填空题\ltk{答案，不显示答案时显示段横线}
-%   \xuanze
-%   \question 问题3是一道选择题,四个选项显示在一行
-%   \onech{选项1}{选项2}{选项3}{选项4}
-%   \question 问题3是一道选择题,四个选项显示在两行
-%   \twoch{选项1}{选项2}{选项3}{选项4}
-%   \question 问题3是一道选择题,四个选项显示在四行
-%   \fourch{选项1}{选项2}{选项3}{选项4}
-%   \jianda
-%   \question 问题4是一道简答题
-%   \begin{parts}
-%   \part 第1小问
-%   \part 第2小问
-%   \end{parts}
-% \end{questions}
-% \end{verbatim}
-% 关于填空题、选择题、简答题的提示语中几个变量的设置，请参考\ref{sec:Variable}。
-% 关于题目的更多内容，请参考\ref{sec:Environment}。
-%
-% \section{使用帮助}
-%
-% \subsection{选项}
-% \label{sec:Options}
-%
-% \changes{v0.2}{2011/07/27}{增加UTF8选项以支持pdflatex}
-% \changes{v0.4}{2015/10/10}{取消UTF8选项放弃支持pdflatex}
-
-% 宏包的选项用于改变一些缺省的设置。虽然缺省的设置基本能过满足一般用户的
-% 使用需要，但用户也可以根据自己的情况，使用这些选项。
-%
-% \begin{description}
-% \item[cs4size]     使用小四字号为缺省字体大小。
-% \item[c5size]      使用五号字为缺省字体大小（缺省选项）。
-% \item[answers]     在每一个问题后附上答案。
-% \item[marginline]  放置装订线。
-% \item[16kpaper]    使用16开纸张（缺省使用A4纸张）。
-% \item[noindent]    没有缩进。
-% \item[printbox]    显示评分框。
-% \end{description}
-%
-% \subsection{变量}
-% \label{sec:Variable}
-% \changes{v0.2}{2011/07/27}{试卷中改用英文标点符号}
-% 本宏包在题量和分值等方面均以高考试卷为模板，
-% 默认的变量值可以在\texttt{BHCexam.cfg}中设置，当然
-% 你也可以在使用相关命令之前使用以下命令进行更改。\\\\
-% \DescribeMacro{\biaoti}
-% 设置标题信息。
-% \begin{quote}
-% |\biaoti|\marg*{TEXT}
-% \end{quote}
-% \DescribeMacro{\fubiaoti}
-% 设置副标题，他会显示在标题下方。
-% \begin{quote}     
-% |\kemu|\marg*{TEXT}
-% \end{quote}
-% \DescribeMacro{\xinxi}
-% 设置总分和考试时间信息，\parg{num1}为总分，\parg{num2}为考试时间。
-% \begin{quote}
-% |\xinxi|\marg*{num1}\marg*{num2}
-% \end{quote}
-% \DescribeMacro{\settk}
-% 设置填空题的总分、题量和小分信息，\parg{num1}为总分，\parg{num2}为题量，
-% \parg{num3}为小分。
-% \begin{quote}
-% |\settk|\marg*{num1}\marg*{num2}\marg*{num3}
-% \end{quote}
-% \DescribeMacro{\setxz}
-% 设置选择题的总分、题量和小分信息，\parg{num1}为总分，\parg{num2}为题量，
-% \parg{num3}为小分。
-% \begin{quote}
-% |\setxz|\marg*{num1}\marg*{num2}\marg*{num3}
-% \end{quote}
-% \DescribeMacro{\setjd}
-% 设置简答题的总分、题量和小分信息，\parg{num1}为总分，\parg{num2}为题量。
-% \begin{quote}
-% |\setjd|\marg*{num1}\marg*{num2}\marg*{num3}
-% \end{quote}
-
-% \subsection{环境}
-% \label{sec:Environment} 
-% 经常使用的环境有\verb"questions"环境、\verb"parts"环境，
-% 关于它们的简单介绍，请参考\ref{sec:Example}，这里做一点补充说明，
-% 更详细的介绍，请参考~\exam~文档。\\\\
-% 在排版简答题时需要用\verb"\part"命令输入各小问的分值，宏包会自动算出总分并显示在该简答题的第一行。
-% 当该道简答题没有小问时，则要用\verb"\question"命令输入该问题的分值。
-% 
-% \begin{verbatim}
-%  ...
-%  \jianda
-%  \question 这是一道简答题
-%  \begin{parts}[
-%  \part[3] 第1小问3分。
-%  \part[3] 第2小问3分。
-%  \part[3] 第3小问4分。
-%  \end{parts}
-%  \question[12] 这是一道没有小问的简答题，这道题有12分
-%  ... 
-% \end{verbatim}
-% 在\verb"\question"后新建\verb"solution"环境，在其中输入该问题的解答，
-% 在不显示答案的情况下，该问题后会预留答题空间。
-% \begin{verbatim}
-%  ...
-%  \jianda
-%  \question 这是一道简答题
-%  \begin{solution}
-%  这是解答,不显示答案的情况下则这个问题后预留答题空间。
-%  \end{solution}
-%  ... 
-% \end{verbatim}
-%
-% \subsection{常用命令}
-% 
-% \DescribeMacro{\newpage}
-% 每道问题的间距是弹性设置的，你只要在想换页的地方输入\verb"\newpage"命令，
-% 则上一页的各问题间距会自动调整到最美观的效果。\\
-% \DescribeMacro{\mininotice}
-% 在一行内输出精简的考试注意事项。\\
-% \DescribeMacro{\printmalol}
-% 在当前页为正面时，在左边插入装订线（仅在使用marginline选项时有效）。\\
-% \DescribeMacro{\printmalol}
-% 在当前页为反面时，在左边插入装订线（仅在使用marginline选项时有效）。\\
-%
-% \StopEventually{}
-%
-% \section{源代码说明}
-%
-% \subsection{选项}
-%
-%
-% 处理~\BHCexam~文档类的选项
-%
-% \begin{macro}{\@sixteenkpaper}
-% 16k纸张大小设置,缺省选项为a4paper
-%    \begin{macrocode}
-%<*cls>
-\newif\if@sixteenkpaper \@sixteenkpaperfalse
-\DeclareOption{16kpaper}{\@sixteenkpapertrue}
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-%
-% \begin{macro}{\@marginline}
-% 是否有装订线
-%    \begin{macrocode}
-%<*cls>
-\newif\if@marginline \@marginlinefalse
-\DeclareOption{marginline}{\@marginlinetrue}
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-%
-% 不缩进，缺省为缩进
-%    \begin{macrocode}
-%<*cls>
-\newif\if@noindent \@noindentfalse
-\DeclareOption{noindent}{\@noindenttrue}
-%</cls>
-%    \end{macrocode}
-%
-% 显示答案的方式，缺省不显示答案
-%    \begin{macrocode}
-%<cls>\DeclareOption{answers}{\PassOptionsToClass{\CurrentOption}{exam}}
-%    \end{macrocode}
-
-% \begin{macro}{\@printbox}
-% 显示计分框，缺省为不显示。
-%    \begin{macrocode}
-%<*cls>
-\newif\if@printbox \@printboxfalse
-\DeclareOption{printbox}{\@printboxtrue}
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-%
-% 把没有定义的选项传递给缺省的文档类
-%    \begin{macrocode}
-%<cls>\DeclareOption*{\PassOptionsToClass{\CurrentOption}{exam}}
-%    \end{macrocode}
-%
-% 处理选项
-%    \begin{macrocode}
-%<cls>\ProcessOptions
-%    \end{macrocode}
-%
-% 装入缺省的文档类
-%    \begin{macrocode}
-%<cls>\LoadClass[addpoints]{exam}
-%    \end{macrocode}
-% 
-% 导入ctex类的实现
-% \changes{v0.4}{2015/10/10}{修正ctex宏包实现}
-%    \begin{macrocode}
-%<*cls>
-\RequirePackage{ctex}
-%</cls>
-%    \end{macrocode}
-% \subsection{宏包}
-%
-% \begin{macro}{\RequirePackage}
-% 我们需要使用的一些宏包
-%    \begin{macrocode}
-%<*cls>
-\RequirePackage{ifpdf,ifxetex}
-\RequirePackage{amsmath,amssymb,amsthm,bm,bbding,pifont,dsfont}
-\RequirePackage{mathtools}
-\RequirePackage{paralist,cases,tabularx}
-\RequirePackage{pstricks,pst-plot,xcolor,graphicx}
-%</cls>
-%    \end{macrocode}
-% 
-% 用geometry宏包进行页面设置
-% \changes{v0.2}{2011/07/27}{改用geometry宏包实现纸张设置}
-% \changes{v0.3}{2014/02/18}{修正了纸张的尺寸}
-%    \begin{macrocode}
-%<*cls>
-\if@marginline
-\marginparwidth = 2cm
-\if@sixteenkpaper
-\RequirePackage[papersize={184mm,260mm},hmargin={3cm,2cm},
-vmargin={2cm,2cm},marginparsep=0.5cm,hoffset=0cm,voffset=0cm,
-footnotesep=0.5cm,headsep=0.5cm,twoside]{geometry}
-\else
-\RequirePackage[paper=a4paper,hmargin={3cm,2cm},vmargin={2cm,2cm},
-marginparsep=0.5cm,hoffset=0cm,voffset=0cm,footnotesep=0.5cm,
-headsep=0.5cm,twoside]{geometry}
-\fi
-\else
-\if@sixteenkpaper
-\RequirePackage[papersize={184mm,260mm},hmargin={2cm,2cm},
-vmargin={2cm,2cm},marginparsep=0.5cm,hoffset=0cm,voffset=0cm,
-footnotesep=0.5cm,headsep=0.5cm]{geometry}
-\else
-\RequirePackage[papersize={210mm,297mm},hmargin={2cm,2cm},vmargin={2cm,2cm},
-marginparsep=0.5cm,hoffset=0cm,voffset=0cm,footnotesep=0.5cm,
-headsep=0.5cm]{geometry}
-\fi
-\fi
-%</cls>
-%    \end{macrocode}
-%\end{macro}
-%
-% \subsection{自定义设置}
-%行距、页眉、页脚
-%    \begin{macrocode}
-%<*cls>
-\renewcommand{\baselinestretch}{1.5}
-\pagestyle{headandfoot}
-\header{}{}{}
-\footer{}{\small \quad 第~\thepage~页(共~\numpages~页)}{}
-%</cls>
-%    \end{macrocode}
-%
-% 分值显示
-%    \begin{macrocode}
-%<*cls>
-\pointname{分}
-\pointformat{\kaishu (\thepoints)}
-%</cls>
-%    \end{macrocode}
-%
-% 问题的显示
-%    \begin{macrocode}
-%<*cls>
-\renewcommand{\questionshook}{
-  \settowidth{\leftmargin}{22.\hskip\labelsep}
-  \if@noindent \setlength\leftmargin{0pt} \fi
-}
-\renewcommand{\thepartno}{\arabic{partno}}
-\renewcommand{\partlabel}{(\thepartno)}
-\renewcommand{\partshook}{
-  \settowidth{\leftmargin}{(3).\hskip\labelsep}
-  \if@noindent \setlength\leftmargin{0pt} \fi
-}
-%</cls>
-%    \end{macrocode}
-%
-% 解答的显示
-%    \begin{macrocode}
-%<*cls>
-\newif\if@cancelspace \@cancelspacetrue
-\renewcommand{\solutiontitle}{\noindent 解：\noindent}
-\renewenvironment{solution}%
-  {%
-    \ifprintanswers
-	%\unskip
-      \begingroup
-      \Solution@Emphasis
-      \begin{TheSolution}%
-    \else
-      \if@cancelspace
-        %\unskip
-      \else
-        \par
-        \penalty 0
-        \vfill%
-        \if@printbox \if@houpinfen \houpinfen \fi \fi
-      \fi
-      \setbox\z@\vbox\bgroup
-    \fi
-  }{%
-    \ifprintanswers
-      \end{TheSolution}%
-      \endgroup
-    \else
-      \egroup
-    \fi
-  }%
-%</cls>
-%    \end{macrocode}
-%
-% \subsection{新的命令和环境}
-%
-% \begin{macro}{\printmlor}
-% \begin{macro}{\printmlol}
-% \changes{v0.1}{2011/07/23}{支持在首页插入装订线}
-% \changes{v0.2}{2011/07/27}{手动在指定页插入左(右)装订线}
-% 装订线
-%    \begin{macrocode}
-%<*cls>
-\if@marginline
-\newsavebox{\zdxl}
-\sbox{\zdxl}{
-\begin{minipage}{0.7\paperheight}
-\begin{center}
-\heiti 班级\underline{\hspace{15ex}} \quad
-姓名 \underline{\hspace{15ex}} \quad
-学号 \underline{\hspace{15ex}} \quad \\
-\vspace{3ex}
-\dotfill 装 \dotfill 订 \dotfill 线 \dotfill
-\end{center}
-\end{minipage}
-}
-\newsavebox{\zdxr}
-\sbox{\zdxr}{
-\begin{minipage}{0.7\paperheight}
-\begin{center}
-\heiti \hfill 请 \hfill 不 \hfill 要 \hfill 在 \hfill
- 装 \hfill 订 \hfill 线 \hfill 内 \hfill 答 \hfill 题 \hfill \\
-\vspace{3ex}
-\dotfill 装 \dotfill 订 \dotfill 线 \dotfill
-\end{center}
-\end{minipage}
-}
-\newcommand{\printmlol}{
-\marginpar{\rotatebox{90}{\usebox{\zdxl}}}
-}
-\newcommand{\printmlor}{
-\marginpar{\rotatebox{-90}{\usebox{\zdxr}}}
-}
-\reversemarginpar
-\fi
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-% \end{macro}
-%
-% \begin{macro}{\biaoti}
-% \begin{macro}{\fubiaoti}
-% 标题
-%    \begin{macrocode}
-%<*cls>
-\newcommand\biaoti[1]{\def\@biaoti{#1}}
-\newcommand\fubiaoti[1]{\def\@fubiaoti{#1}}
-\renewcommand\maketitle{
-  \begin{center}{\heiti \Large{\@biaoti}}\end{center}
-  \begin{center}{\heiti \Large{\@fubiaoti}}\end{center}
-}
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-% \end{macro}
-%
-% \begin{macro}{\mininotice}
-% \begin{macro}{\xinxi}
-% 一行内显示考试时间和考试总分
-%    \begin{macrocode}
-%<*cls>
-\newcommand\xinxi[2]{
-  \def\@zongfen{#1}
-  \def\@shijian{#2}
-}
-\newcommand\mininotice{
-  \begin{center}{
-    \kaishu (本试卷满分~\@zongfen~分, 考试时间~\@shijian~分钟)}
-  \end{center}
-}
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-% \end{macro}
-%
-% \begin{macro}{\notice}
-% 注意事项
-%    \begin{macrocode}
-%<*cls>
-\newcommand{\notice}{
-  \heiti 注意事项： \songti
-  \begin{enumerate}
-  \item 答卷前, 考生务必将姓名、高考准考证号、校验码等填写清楚.
-  \item 本试卷共~\numquestions{}~道试题, 满分~\@zongfen~分,考试时间~\@shijian~分钟.
-  \end{enumerate}
-}
-%</cls>
-%    \end{macrocode}
-%\end{macro}
-%
-% \begin{macro}{\pingfen}
-% 前评分框
-%    \begin{macrocode}
-%<*cls>
-\newlength\@boxwidth
-\setlength\@boxwidth{0ex}
-\if@printbox \setlength\@boxwidth{18ex} \fi
-\newcommand\pinfen{
-  \heiti
-  \begin{minipage}{\@boxwidth}
-  \begin{tabular}{|c|c|}
-  \hline
-  得分 & 评卷人\\
-  \hline
-      &       \\
-  \hline
-  \end{tabular}
-  \end{minipage}
-}
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-%
-% \begin{macro}{\houpinfen}
-% 后评分框
-%    \begin{macrocode}
-%<*cls>
-\newcommand{\houpinfen}{
-  \hfill
-  \begin{tabular}{|l|l|}
-    \hline
-    得分 & \hspace*{1.5cm}\\
-    \hline
-  \end{tabular}
-  \bigskip
-}
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-% 
-%
-% 一些参数及变量
-%    \begin{macrocode}
-%<*cfg>
-\def\@biaoti{2011年全国普通高等学校招生统一考试(上海卷)}
-\def\@fubiaoti{数学(理科)}
-\def\@zongfen{150}
-\def\@shijian{120}
-\def\@tiankong@zongfen{56}
-\def\@tiankong@tishu{14}
-\def\@tiankong@fen{4}
-\def\@xuanze@zongfen{16}
-\def\@xuanze@tishu{4}
-\def\@xuanze@fen{4}
-\def\@jianda@zongfen{78}
-\def\@jianda@tishu{5}
-%</cfg>
-%<*cls>
-\newcounter{@dati}
-\newif\if@houpinfen \@houpinfenfalse
-%</cls>
-%    \end{macrocode}
-%
-% 填空题提示语
-% \begin{macro}{\tiankong}
-%    \begin{macrocode}
-%<*cls>
-\newcommand\settk[3]{
-  \def\@tiankong@zongfen{#1}
-  \def\@tiankong@tishu{#2}
-  \def\@tiankong@fen{#3}
-}
-\newcommand\tiankong{
-  \@houpinfenfalse
-  \stepcounter{@dati}
-  \fullwidth{
-    \if@printbox \pinfen \fi
-    \begin{minipage}{\textwidth-\@boxwidth}
-    \heiti \chinese{@dati}. 填空题：本大题共~\@tiankong@tishu~题，满分~\@tiankong@zongfen~分. 请在横线上方填写最终的、最准确的、最完整的结果. 每题填写正确得~\@tiankong@fen~分，否则一律得0分.
-    \end{minipage}
-  }
-}
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-%
-% \begin{macro}{\stk}
-% \begin{macro}{\mtk}
-% \begin{macro}{\ltk}
-% \changes{v0.3}{2014/02/18}{设置三种长度的横线，并支持在横线上输出答案}
-% 填空题横线
-%    \begin{macrocode}
-%<*cls>
-\newcommand{\stk}[1]{
-  \ifprintanswers
-    \underline{~#1~}
-  \else
-    \underline{~\hspace{1cm}~}
-  \fi}
-\newcommand{\mtk}[1]{
-  \ifprintanswers
-    \underline{~#1~}
-  \else
-    \underline{~\hspace{2cm}~}
-  \fi}
-\newcommand{\ltk}[1]{
-  \ifprintanswers
-    \underline{~#1~}
-  \else
-    \underline{~\hspace{3cm}~}
-  \fi}
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-% \end{macro}
-% \end{macro}
-%
-% 选择题提示语
-% \begin{macro}{\xuanze}
-%    \begin{macrocode}
-%<*cls>
-\newcommand\setxz[3]{
-  \def\@xuanze@zongfen{#1}
-  \def\@xuanze@tishu{#2}
-  \def\@xuanze@fen{#3}
-}
-\newcommand\xuanze{
-  \@houpinfenfalse
-  \stepcounter{@dati}
-  \fullwidth{
-    \if@printbox \pinfen \fi
-    \begin{minipage}{\textwidth-\@boxwidth}
-      \heiti \chinese{@dati}. 选择题：本大题共~\@xuanze@tishu~题，满分~\@xuanze@zongfen~分. 请选择你认为最正确的答案（每小题有且只有一个）写在括号内. 每题填写正确得~\@xuanze@fen~分，否则得0分.
-    \end{minipage}
-  }
-}
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-%
-% \begin{macro}{\onech}
-% \begin{macro}{\twoch}
-% \begin{macro}{\fourch}
-% \changes{v0.3}{2014/02/18}{分一行、两行、四行显示选项}
-% 选择题的选项
-%    \begin{macrocode}
-%<*cls>
-\newcommand{\onech}[4]{
-\renewcommand\arraystretch{1.4}
-\begin{tabularx}{\linewidth}{XXXX}
-\setlength\tabcolsep{0pt}
-(A) #1 & (B) #2 & (C) #3 & (D) #4 \\
-\end{tabularx}
-\unskip \unskip}
-\newcommand{\twoch}[4]{
-\renewcommand\arraystretch{1.4}
-\begin{tabularx}{\linewidth}{XX}
-\setlength\tabcolsep{0pt}
-(A) #1 & (B) #2 \\
-(C) #3 & (D) #4
-\end{tabularx}
-\unskip \unskip}
-\newcommand{\fourch}[4]{
-\renewcommand\arraystretch{1.4}
-\begin{tabularx}{\linewidth}{X}
-\setlength\tabcolsep{0pt}
-(A) #1 \\
-(B) #2 \\
-(C) #3 \\
-(D) #4 \\
-\end{tabularx}
-\unskip \unskip}
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-% \end{macro}
-% \end{macro}
-%
-% 简答题提示语
-% \begin{macro}{\jianda}
-%    \begin{macrocode}
-%<*cls>
-\newcommand\setjd[2]{
-  \def\@jianda@zongfen{#1}
-  \def\@jianda@tishu{#2}
-}
-\newcommand\jianda{
-  \@cancelspacefalse
-  \@houpinfentrue
-  \stepcounter{@dati}
-  \fullwidth{
-    \if@printbox \pinfen \fi
-    \begin{minipage}{\textwidth-\@boxwidth}
-      \heiti \chinese{@dati}. 简答题：本大题共~\@jianda@tishu~题，满分~\@jianda@zongfen~分. 请在题后空处写出必要的推理计算过程.
-    \end{minipage}
-  }
-}
-%</cls>
-%    \end{macrocode}
-% \end{macro}
-%
-% 数学运算符号、单位
-%    \begin{macrocode}
-%<*cls>
-\delimitershortfall-1sp
-\newcommand\abs[1]{\left|#1\right|}
-\renewcommand{\geq}{\geqslant}
-\renewcommand{\ge}{\geqslant}
-\renewcommand{\leq}{\leqslant}
-\renewcommand{\le}{\leqslant}
-%<\cls>
-%    \end{macrocode}
-%
-% 读取配置文件
-%    \begin{macrocode}
-%<*cls>
-\AtEndOfPackage{\makeatletter\input{BHCexam.cfg}\makeatother}
-%</cls>
-%    \end{macrocode}
-%
-%
-
-% \Finale
-%
-% \setcounter{IndexColumns}{2}
-% \IndexPrologue{\section*{索引}}
-%
-% \GlossaryPrologue{\section*{版本更新}}
-%
-% \PrintIndex \PrintChanges
-\endinput
-
-\EnableCrossrefs
diff --git a/macros/latex/contrib/bhcexam/BHCexam.ins b/macros/latex/contrib/bhcexam/BHCexam.ins
deleted file mode 100644
index 16f9828539..0000000000
--- a/macros/latex/contrib/bhcexam/BHCexam.ins
+++ /dev/null
@@ -1,107 +0,0 @@
-%%

-%% This file will generate fast loadable files and documentation

-%% driver files from the doc files in this package when run through

-%% LaTeX or TeX. It works with Docstrip versions after April 1998.

-%%

-%% Copyright (C) 2011 by Charles Bao <charley792@gmail.com>

-%%

-%% This file is part of the BHCexam package project.

-%% ---------------------------------------------------

-%%

-%% It may be distributed under the conditions of the LaTeX Project Public

-%% License, either version 1.2 of this license or (at your option) any

-%% later version. The latest version of this license is in

-%%    http://www.latex-project.org/lppl.txt

-%% and version 1.2 or later is part of all distributions of LaTeX

-%% version 1999/12/01 or later.

-%%

-%% In particular, NO PERMISSION is granted to modify the contents of this

-%% file since it contains the legal notices that are placed in the files

-%% it generates.

-%%

-%%

-%%

-%%

-%% --------------- start of docstrip commands ------------------

-%%

-

-\input docstrip

-

-\preamble

-

-This is a generated file.

-

-Copyright (C) 2011 by Charles Bao <charley792@gmail.com>

-

-This file was generated from file(s) of the BHCexam package project.

-----------------------------------------------------------------------

-

-It may be distributed and/or modified under the

-conditions of the LaTeX Project Public License, either version 1.2

-of this license or (at your option) any later version.

-The latest version of this license is in

-   http://www.latex-project.org/lppl.txt

-and version 1.2 or later is part of all distributions of LaTeX

-version 1999/12/01 or later.

-

-This file may only be distributed together with a copy of original

-source files. You may however distribute original source files

-without such generated files. Copying of this file is authorized

-only if either:

-(1) you make absolutely no changes to your copy, including name; OR

-(2) if you do make changes, you first rename it to some other name.

-

-To produce the documentation run the original source files ending

-with `.dtx' through LaTeX.

-

-\endpreamble

-

-\declarepreamble\cfgpreamble

-

-This is a generated file.

-

-Copyright (C) 2014 by Charles Bao <charley792@gmail.com>

-

-This file was generated from file(s) of the BHCexam package project.

-----------------------------------------------------------------------

-

-It may be distributed and/or modified under the

-conditions of the LaTeX Project Public License, either version 1.2

-of this license or (at your option) any later version.

-The latest version of this license is in

-   http://www.latex-project.org/lppl.txt

-and version 1.2 or later is part of all distributions of LaTeX

-version 1999/12/01 or later.

-

-This is the configuration file of the BHCexam package with LaTeX2e.

-

-\endpreamble

-

-\askonceonly

-\keepsilent

-\usedir{tex/latex/BHCexam}

-

-\generate{\file{BHCexam.cls}{\from{BHCexam.dtx}{cls}}

-          \usepreamble\cfgpreamble

-          \file{BHCexam.cfg}{\from{BHCexam.dtx}{cfg}}

-          }

-

-\ifToplevel{

-\Msg{***********************************************************}

-\Msg{*}

-\Msg{* To finish the installation you have to move the following}

-\Msg{* files into a directory searched by TeX:}

-\Msg{*}

-\Msg{* The recommended directory is TEXMF/tex/latex/BHCexam}

-\Msg{*}

-\Msg{* \space\space BHCexam.cls}

-\Msg{* \space\space BHCexam.cfg}

-\Msg{*}

-\Msg{* To produce the documentation run the files ending with}

-\Msg{* `.dtx' through LaTeX.}

-\Msg{*}

-\Msg{* Happy TeXing}

-\Msg{***********************************************************}

-}

-

-\endbatchfile

diff --git a/macros/latex/contrib/bhcexam/Makefile b/macros/latex/contrib/bhcexam/Makefile
deleted file mode 100644
index 667c242491..0000000000
--- a/macros/latex/contrib/bhcexam/Makefile
+++ /dev/null
@@ -1,88 +0,0 @@
-PACKAGE = BHCexam
-########################################################################
-## LaTeX2e Makefile
-##
-## Update the following defines for your local configuration, 
-##
-TEXMFDIR   = ~/texmf/tex/latex/BHCexam
-GZIP      = gzip
-XELATEX  = xelatex
-PDFLATEX  = pdflatex
-MAKEINDEX = makeindex
-########################################################################
-## make [all]		Generates the class(.cls) file, the configuration(.cfg)
-##                        file and the documentation (.pdf).  If you don't have 
-##                        the required MAKEINDEX (along with `gglo.ist' and
-##                        `gind.ist'), then change the first dependency
-##			  of "all" from "fullpdf" to "pdf" below.
-## make [un]install 	Install or uninstall the class(.cls) file and 
-##                        the configuration(.cfg) file.
-## make [very]clean	Clean out various auxillary files.  "veryclean"
-##			  cleans out more stuff.
-########################################################################
-## make [full]doc       Generate the documentation.  The "fulldoc" version
-##			  adds the change log and the cross-references.
-## make idx		Generate the change log and the cross-references
-##			  (for fulldoc -- requires MAKEINDEX).
-## make cls 		Generate the class(.cls) file and the
-##                        configuration(.cfg) file.
-
-########################################################################
-## make test		Run test file(s)
-## make src		Builds a src distribution (.tar.gz) file.
-## make distribtion	Builds a distribution (.tar.gz) file.
-########################################################################
-
-all:		veryclean cls fulldoc test clean
-
-install:	cls
-		cp $(PACKAGE).{cls,cfg} $(TEXMFDIR)
-uninstall:	; rm $(TEXMFDIR)/$(PACKAGE).{cls,cfg}
-clean:		; -rm -f *.dvi *.log *.aux *.lof *.lot *.toc 
-		-rm -f *.idx *.ind *.glo *.gls *.ilg *.out *~
-veryclean:	; -rm -f *.dvi *.log *.aux *.lof *.lot *.toc 
-		-rm -f *.idx *.ind *.glo *.gls *.ilg *.out
-		-rm -f *.idx *.ind *.glo *.gls *.ilg *.out
-		-rm -f *.sty *.cls *.pdf *pk *.cfg *.tar *~
-
-
-doc:		; $(XELATEX) $(PACKAGE).dtx
-
-cls:		; $(XELATEX) $(PACKAGE).ins
-
-fulldoc:	doc $(PACKAGE).gls $(PACKAGE).ind 
-		$(XELATEX) $(PACKAGE).dtx
-
-src:		; mkdir $(PACKAGE)
-		cp -p README Makefile $(PACKAGE)
-		cp -p $(PACKAGE).dtx $(PACKAGE).ins $(PACKAGE)
-		cp -p test*.tex $(PACKAGE)
-		tar -cvf $(PACKAGE)-src.tar ./$(PACKAGE) 
-		rm -rf $(PACKAGE)
-		$(GZIP) -9 $(PACKAGE)-src.tar
-
-distribution:	veryclean cls fulldoc test clean
-		mkdir $(PACKAGE)
-		cp -p $(PACKAGE).cls $(PACKAGE).cfg $(PACKAGE)
-		cp -p *.tex $(PACKAGE)
-		cp -p *.pdf $(PACKAGE)
-		tar -cvf $(PACKAGE).tar ./$(PACKAGE) 
-		rm -rf $(PACKAGE)
-		$(GZIP) -9 $(PACKAGE).tar
-
-$(PACKAGE).gls:	doc
-		$(MAKEINDEX) -s gglo.ist -o $(PACKAGE).gls $(PACKAGE).glo
-
-$(PACKAGE).ind:	doc
-		$(MAKEINDEX) -s gind.ist -o $(PACKAGE).ind $(PACKAGE).idx
-
-test:		cls
-		$(XELATEX) test1.tex 
-		$(XELATEX) test1.tex 
-		$(XELATEX) test2.tex 
-		$(XELATEX) test2.tex 
-		$(XELATEX) test3.tex 
-		$(XELATEX) test3.tex 
-		$(XELATEX) test4.tex 
-		$(XELATEX) test4.tex 
-
diff --git a/macros/latex/contrib/bhcexam/README b/macros/latex/contrib/bhcexam/README
deleted file mode 100644
index 73804a8a61..0000000000
--- a/macros/latex/contrib/bhcexam/README
+++ /dev/null
@@ -1,21 +0,0 @@
-This is version 0.4 of the BHCexam document class, dated Oct 10, 2015.
-
-The BHCexam document class attempts to make it easy for even a LaTeX novice to prepare exams.
-
-To generate the document class files from .dtx file:
-make cls;
-
-To install the document class files to ~/texmf
-make install;
-
-To generate the user's guide document (with index)
-make [full]doc;
-
-To generate test page
-make test
-
---------------------------------------------------------------------
-
-This work may be distributed and/or modified under the conditions of
-the LaTeX Project Public License, either version 1.3 of this license
-or (at your option) any later version.
diff --git a/macros/latex/contrib/bhcexam/test1.tex b/macros/latex/contrib/bhcexam/test1.tex
deleted file mode 100644
index 01fc0b2035..0000000000
--- a/macros/latex/contrib/bhcexam/test1.tex
+++ /dev/null
@@ -1,143 +0,0 @@
-\documentclass[printbox,marginline,noindent,adobefonts]{BHCexam}
-\begin{document}
-\printmlol
-\maketitle
-
-\begin{questions}
-\tiankong
-\question 已知~$\bm{a}=(k,-9)$、$\bm{b}=(-1,k)$, $\bm{a}$~与~$\bm{b}$~为平行向量,
-    则~$k=$\stk{$\pm3$}.
-
-\question 若函数~$f(x)=x^{6m^2-5m-4}\,(m\in\mathbb{Z})$~的图像关于~$y$~轴对称,
-    且~$f(2)<f(6)$, 则~$f(x)$~的解析式为\stk{$f(x)=x^{-4}$}.
-
-\question 若~$f(x+1)=x^2\,(x\leq0)$, 则~$f^{-1}(1)=$\stk{0}.
-
-\question 在~$b\text{g}$~糖水中含糖~$a\text{g}$\,($b>a>0$), 若再添加~$m\text{g}$~糖~($m>0$),
-
-
-\question 已知~$f(x)=1-\textbf{c}_8^1x+\textbf{c}_8^2x^2-\textbf{c}_8^3x^3+\cdots+\textbf{c}_8^8x^8$,
-    则~$f\big(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\textbf{i}\big)$~的值是\ltk{$-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\textbf{i}$}.
-
-\question 自然数~$1,2,3,\ldots,10$~的方差记为~$\sigma^2$,
-    其中的偶数~$2,4,6,8,10$~的方差记为~$\sigma_1^2$,
-    则~$\sigma^2$~与~$\sigma_1^2$~的大小关系为~$\sigma^2$\stk{$>$}$\sigma_1^2$.
-
-\question 若~$\theta$~为三角形的一个内角, 且~$\sin\theta+\cos\theta=\dfrac{2}{3}$,
-    则方程~$x^2\csc\theta-y^2\sec\theta=1$~表示的曲线的焦点坐标是\stk{$\big(\pm\dfrac{\sqrt{6}}{3},0\big)$}.
-
-
-\question 高为~$h$~的棱锥被平行于棱锥底面的截得棱台侧面积是
-    原棱锥的侧面积的~$\dfrac{5}{9}$,
-    则截得的棱台的体积与原棱锥的体积之比是\stk{$19:27$}.
-
-\question 以椭圆~$\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$~的右焦点为圆心,
-    且与双曲线~$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$~的渐近线相切的圆方程是\mtk{$(x-5)^2+y^2=16$}.
-
-
-\question 若~$\sqrt{\,\sin x}$~是有理数且~$x$~不是~$\dfrac{\pi}{6}$~的整数倍,
-    则~$x$~可能取的值是\mtk{$\arcsin\dfrac{1}{4}$ 等}.（只要求写出一个）
-
-\question 马路上有编号~1~到~10~的~10~盏路灯, 为节约用电又不影响照明,
-    可以关掉其中的~3~盏, 但又不能同时关掉相邻的两盏, 也不能关掉两端的路灯,
-    满足条件的关灯方法有\stk{$20$}种.
-
-
-\question 以椭圆~$\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$~的右焦点为圆心,
-    且与双曲线~$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$~的渐近线相切的圆方程是\mtk{$(x-5)^2+y^2=16$}.
-
-\question 若~$\sqrt{\,\sin x}$~是有理数且~$x$~不是~$\dfrac{\pi}{6}$~的整数倍,
-    则~$x$~可能取的值是\mtk{$\arcsin\dfrac{1}{4}$ 等}.（只要求写出一个）
-
-\question 马路上有编号~1~到~10~的~10~盏路灯, 为节约用电又不影响照明,
-    可以关掉其中的~3~盏, 但又不能同时关掉相邻的两盏, 也不能关掉两端的路灯,
-    满足条件的关灯方法有\stk{$20$}种.
-
-
-\question 以椭圆~$\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$~的右焦点为圆心,
-    且与双曲线~$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$~的渐近线相切的圆方程是\mtk{$(x-5)^2+y^2=16$}.
-
-\newpage
-
-\xuanze
-\question 已知集合~$A=\{x\mid \abs{x-1}<3 \}$,
-集合~$B=\{y| y=x^2+2x+1,x\in\mathbb{R}\}$, 则~$A\cap
-\complement_U B$~为\stk{C}.
-\twoch{$[\,0,4)$}{$(-\infty,-2\,]\cup[4,+\infty)$}{$(-2,0)$}{$(0,4)$}
-
-\question 若~$a$、$b$~是直线, $\alpha$、$\beta$~是平面,
-则以下命题中真命题是\stk{D}.\\
-\fourch{若~$a$、$b$~异面, $a\subset\alpha$,$b\subset\beta$, 且~$a\perp b$, 则~$\alpha\perp\beta$}{若~$a\parallel b$, $a\subset\alpha$, $b\subset\beta$,则~$\alpha\parallel\beta$}{若~$a\parallel \alpha$,
-$b\subset\beta$, 则~$a$、$b$ 异面}{若~$a\perp b$, $a\perp\alpha$,$b\perp\beta$, 则~$\alpha\perp\beta$}
-
-\question 已知集合~$A=\{x\mid \abs{x-1}<3 \}$,
-集合~$B=\{y| y=x^2+2x+1,x\in\mathbb{R}\}$, 则~$A\cap
-\complement_U B$~为\stk{C}.
-\twoch{$[\,0,4)$}{$(-\infty,-2\,]\cup[4,+\infty)$}{$(-2,0)$}{$(0,4)$}
-
-\question 若~$a$、$b$~是直线, $\alpha$、$\beta$~是平面,
-则以下命题中真命题是\stk{D}.\\
-\fourch{若~$a$、$b$~异面, $a\subset\alpha$,$b\subset\beta$, 且~$a\perp b$, 则~$\alpha\perp\beta$}{若~$a\parallel b$, $a\subset\alpha$, $b\subset\beta$,则~$\alpha\parallel\beta$}{若~$a\parallel \alpha$,
-$b\subset\beta$, 则~$a$、$b$ 异面}{若~$a\perp b$, $a\perp\alpha$,$b\perp\beta$, 则~$\alpha\perp\beta$}
-
-\newpage
-\jianda
-\question 已知复数~$z$ 满足：$\abs{z}-z^*=\dfrac{10}{1-w\textbf{i}}$（其中~$z^*$
-是~$z$ 的共轭复数）.
-\begin{parts}
-\part[7] 求复数~$z$；
-\part[7] 若复数~$w=\cos\theta+\textbf{i}\sin\theta\,(\theta\in\mathbb{R})$, 求~$\abs{z-2}$ 的取值范围.
-\end{parts}
-
-\begin{solution}
-\begin{parts}
-\part $z=3+4\textbf{i}$
-\part $\abs{z-w}\in[4,6]$
-\end{parts}
-\end{solution}
-
-\question[14] 函数~$f(x)=4\sin\dfrac{\pi}{12}x\cdot\sin
-    \left(\dfrac{\pi}{2}+\dfrac{\pi}{12}x\right),x\in[a,a+1]$,
-    其中常数~$a\in[0,5]$, 求函数~$f(x)$ 的最大值~$g(a)$.
-
-\begin{solution}
-略
-\end{solution}
-
-\newpage
-
-\question[16] 函数~$f(x)=4\sin\dfrac{\pi}{12}x\cdot\sin
-    \left(\dfrac{\pi}{2}+\dfrac{\pi}{12}x\right),x\in[a,a+1]$,
-    其中常数~$a\in[0,5]$, 求函数~$f(x)$ 的最大值~$g(a)$.
-
-\begin{solution}
-略
-\end{solution}
-
-\newpage
-\question 已知复数~$z$ 满足：$\abs{z}-z^*=\dfrac{10}{1-w\textbf{i}}$（其中~$z^*$
-是~$z$ 的共轭复数）.
-\begin{parts}
-\part[8] 求复数~$z$；
-\part[8] 若复数~$w=\cos\theta+\textbf{i}\sin\theta\,(\theta\in\mathbb{R})$, 求~$\abs{z-2}$ 的取值范围.
-\end{parts}
-
-\begin{solution}
-\begin{parts}
-\part $z=3+4\textbf{i}$
-\part $\abs{z-w}\in[4,6]$
-\end{parts}
-\end{solution}
-
-\newpage
-
-\question[18] 函数~$f(x)=4\sin\dfrac{\pi}{12}x\cdot\sin
-    \left(\dfrac{\pi}{2}+\dfrac{\pi}{12}x\right),x\in[a,a+1]$,
-    其中常数~$a\in[0,5]$, 求函数~$f(x)$ 的最大值~$g(a)$.
-
-\begin{solution}
-略
-\end{solution}
-
-\end{questions}
-\end{document}
diff --git a/macros/latex/contrib/bhcexam/test2.tex b/macros/latex/contrib/bhcexam/test2.tex
deleted file mode 100644
index 8dac567787..0000000000
--- a/macros/latex/contrib/bhcexam/test2.tex
+++ /dev/null
@@ -1,144 +0,0 @@
-\documentclass[16kpaper]{BHCexam}
-\begin{document}
-
-\maketitle
-\mininotice
-
-\begin{questions}
-\tiankong
-\question 已知~$\bm{a}=(k,-9)$、$\bm{b}=(-1,k)$, $\bm{a}$~与~$\bm{b}$~为平行向量,
-    则~$k=$\stk{$\pm3$}.
-
-\question 若函数~$f(x)=x^{6m^2-5m-4}\,(m\in\mathbb{Z})$~的图像关于~$y$~轴对称,
-    且~$f(2)<f(6)$, 则~$f(x)$~的解析式为\stk{$f(x)=x^{-4}$}.
-
-\question 若~$f(x+1)=x^2\,(x\leq0)$, 则~$f^{-1}(1)=$\stk{0}.
-
-\question 在~$b\text{g}$~糖水中含糖~$a\text{g}$\,($b>a>0$), 若再添加~$m\text{g}$~糖~($m>0$),
-
-
-\question 已知~$f(x)=1-\textbf{c}_8^1x+\textbf{c}_8^2x^2-\textbf{c}_8^3x^3+\cdots+\textbf{c}_8^8x^8$,
-    则~$f\big(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\textbf{i}\big)$~的值是\ltk{$-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\textbf{i}$}.
-
-\question 自然数~$1,2,3,\ldots,10$~的方差记为~$\sigma^2$,
-    其中的偶数~$2,4,6,8,10$~的方差记为~$\sigma_1^2$,
-    则~$\sigma^2$~与~$\sigma_1^2$~的大小关系为~$\sigma^2$\stk{$>$}$\sigma_1^2$.
-
-\question 若~$\theta$~为三角形的一个内角, 且~$\sin\theta+\cos\theta=\dfrac{2}{3}$,
-    则方程~$x^2\csc\theta-y^2\sec\theta=1$~表示的曲线的焦点坐标是\stk{$\big(\pm\dfrac{\sqrt{6}}{3},0\big)$}.
-
-
-\question 高为~$h$~的棱锥被平行于棱锥底面的截得棱台侧面积是
-    原棱锥的侧面积的~$\dfrac{5}{9}$,
-    则截得的棱台的体积与原棱锥的体积之比是\stk{$19:27$}.
-
-\question 以椭圆~$\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$~的右焦点为圆心,
-    且与双曲线~$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$~的渐近线相切的圆方程是\mtk{$(x-5)^2+y^2=16$}.
-
-
-\question 若~$\sqrt{\,\sin x}$~是有理数且~$x$~不是~$\dfrac{\pi}{6}$~的整数倍,
-    则~$x$~可能取的值是\mtk{$\arcsin\dfrac{1}{4}$ 等}.（只要求写出一个）
-
-\question 马路上有编号~1~到~10~的~10~盏路灯, 为节约用电又不影响照明,
-    可以关掉其中的~3~盏, 但又不能同时关掉相邻的两盏, 也不能关掉两端的路灯,
-    满足条件的关灯方法有\stk{$20$}种.
-
-
-\question 以椭圆~$\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$~的右焦点为圆心,
-    且与双曲线~$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$~的渐近线相切的圆方程是\mtk{$(x-5)^2+y^2=16$}.
-
-\question 若~$\sqrt{\,\sin x}$~是有理数且~$x$~不是~$\dfrac{\pi}{6}$~的整数倍,
-    则~$x$~可能取的值是\mtk{$\arcsin\dfrac{1}{4}$ 等}.（只要求写出一个）
-
-\question 马路上有编号~1~到~10~的~10~盏路灯, 为节约用电又不影响照明,
-    可以关掉其中的~3~盏, 但又不能同时关掉相邻的两盏, 也不能关掉两端的路灯,
-    满足条件的关灯方法有\stk{$20$}种.
-
-
-\question 以椭圆~$\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$~的右焦点为圆心,
-    且与双曲线~$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$~的渐近线相切的圆方程是\mtk{$(x-5)^2+y^2=16$}.
-
-\newpage
-
-\xuanze
-\question 已知集合~$A=\{x\mid \abs{x-1}<3 \}$,
-集合~$B=\{y| y=x^2+2x+1,x\in\mathbb{R}\}$, 则~$A\cap
-\complement_U B$~为\stk{C}.
-\twoch{$[\,0,4)$}{$(-\infty,-2\,]\cup[4,+\infty)$}{$(-2,0)$}{$(0,4)$}
-
-\question 若~$a$、$b$~是直线, $\alpha$、$\beta$~是平面,
-则以下命题中真命题是\stk{D}.\\
-\fourch{若~$a$、$b$~异面, $a\subset\alpha$,$b\subset\beta$, 且~$a\perp b$, 则~$\alpha\perp\beta$}{若~$a\parallel b$, $a\subset\alpha$, $b\subset\beta$,则~$\alpha\parallel\beta$}{若~$a\parallel \alpha$,
-$b\subset\beta$, 则~$a$、$b$ 异面}{若~$a\perp b$, $a\perp\alpha$,$b\perp\beta$, 则~$\alpha\perp\beta$}
-
-\question 已知集合~$A=\{x\mid \abs{x-1}<3 \}$,
-集合~$B=\{y| y=x^2+2x+1,x\in\mathbb{R}\}$, 则~$A\cap
-\complement_U B$~为\stk{C}.
-\twoch{$[\,0,4)$}{$(-\infty,-2\,]\cup[4,+\infty)$}{$(-2,0)$}{$(0,4)$}
-
-\question 若~$a$、$b$~是直线, $\alpha$、$\beta$~是平面,
-则以下命题中真命题是\stk{D}.\\
-\fourch{若~$a$、$b$~异面, $a\subset\alpha$,$b\subset\beta$, 且~$a\perp b$, 则~$\alpha\perp\beta$}{若~$a\parallel b$, $a\subset\alpha$, $b\subset\beta$,则~$\alpha\parallel\beta$}{若~$a\parallel \alpha$,
-$b\subset\beta$, 则~$a$、$b$ 异面}{若~$a\perp b$, $a\perp\alpha$,$b\perp\beta$, 则~$\alpha\perp\beta$}
-
-\newpage
-\jianda
-\question 已知复数~$z$ 满足：$\abs{z}-z^*=\dfrac{10}{1-w\textbf{i}}$（其中~$z^*$
-是~$z$ 的共轭复数）.
-\begin{parts}
-\part[7] 求复数~$z$；
-\part[7] 若复数~$w=\cos\theta+\textbf{i}\sin\theta\,(\theta\in\mathbb{R})$, 求~$\abs{z-2}$ 的取值范围.
-\end{parts}
-
-\begin{solution}
-\begin{parts}
-\part $z=3+4\textbf{i}$
-\part $\abs{z-w}\in[4,6]$
-\end{parts}
-\end{solution}
-
-\question[14] 函数~$f(x)=4\sin\dfrac{\pi}{12}x\cdot\sin
-    \left(\dfrac{\pi}{2}+\dfrac{\pi}{12}x\right),x\in[a,a+1]$,
-    其中常数~$a\in[0,5]$, 求函数~$f(x)$ 的最大值~$g(a)$.
-
-\begin{solution}
-略
-\end{solution}
-
-\newpage
-
-\question[16] 函数~$f(x)=4\sin\dfrac{\pi}{12}x\cdot\sin
-    \left(\dfrac{\pi}{2}+\dfrac{\pi}{12}x\right),x\in[a,a+1]$,
-    其中常数~$a\in[0,5]$, 求函数~$f(x)$ 的最大值~$g(a)$.
-
-\begin{solution}
-略
-\end{solution}
-
-\newpage
-\question 已知复数~$z$ 满足：$\abs{z}-z^*=\dfrac{10}{1-w\textbf{i}}$（其中~$z^*$
-是~$z$ 的共轭复数）.
-\begin{parts}
-\part[8] 求复数~$z$；
-\part[8] 若复数~$w=\cos\theta+\textbf{i}\sin\theta\,(\theta\in\mathbb{R})$, 求~$\abs{z-2}$ 的取值范围.
-\end{parts}
-
-\begin{solution}
-\begin{parts}
-\part $z=3+4\textbf{i}$
-\part $\abs{z-w}\in[4,6]$
-\end{parts}
-\end{solution}
-
-\newpage
-
-\question[18] 函数~$f(x)=4\sin\dfrac{\pi}{12}x\cdot\sin
-    \left(\dfrac{\pi}{2}+\dfrac{\pi}{12}x\right),x\in[a,a+1]$,
-    其中常数~$a\in[0,5]$, 求函数~$f(x)$ 的最大值~$g(a)$.
-
-\begin{solution}
-略
-\end{solution}
-
-\end{questions}
-\end{document}
diff --git a/macros/latex/contrib/bhcexam/test3.tex b/macros/latex/contrib/bhcexam/test3.tex
deleted file mode 100644
index 0748babf7a..0000000000
--- a/macros/latex/contrib/bhcexam/test3.tex
+++ /dev/null
@@ -1,144 +0,0 @@
-\documentclass[answers]{BHCexam}
-\begin{document}
-
-\maketitle
-\mininotice
-
-\begin{questions}
-\tiankong
-\question 已知~$\bm{a}=(k,-9)$、$\bm{b}=(-1,k)$, $\bm{a}$~与~$\bm{b}$~为平行向量,
-    则~$k=$\stk{$\pm3$}.
-
-\question 若函数~$f(x)=x^{6m^2-5m-4}\,(m\in\mathbb{Z})$~的图像关于~$y$~轴对称,
-    且~$f(2)<f(6)$, 则~$f(x)$~的解析式为\stk{$f(x)=x^{-4}$}.
-
-\question 若~$f(x+1)=x^2\,(x\leq0)$, 则~$f^{-1}(1)=$\stk{0}.
-
-\question 在~$b\text{g}$~糖水中含糖~$a\text{g}$\,($b>a>0$), 若再添加~$m\text{g}$~糖~($m>0$),
-
-
-\question 已知~$f(x)=1-\textbf{c}_8^1x+\textbf{c}_8^2x^2-\textbf{c}_8^3x^3+\cdots+\textbf{c}_8^8x^8$,
-    则~$f\big(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\textbf{i}\big)$~的值是\ltk{$-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\textbf{i}$}.
-
-\question 自然数~$1,2,3,\ldots,10$~的方差记为~$\sigma^2$,
-    其中的偶数~$2,4,6,8,10$~的方差记为~$\sigma_1^2$,
-    则~$\sigma^2$~与~$\sigma_1^2$~的大小关系为~$\sigma^2$\stk{$>$}$\sigma_1^2$.
-
-\question 若~$\theta$~为三角形的一个内角, 且~$\sin\theta+\cos\theta=\dfrac{2}{3}$,
-    则方程~$x^2\csc\theta-y^2\sec\theta=1$~表示的曲线的焦点坐标是\stk{$\big(\pm\dfrac{\sqrt{6}}{3},0\big)$}.
-
-
-\question 高为~$h$~的棱锥被平行于棱锥底面的截得棱台侧面积是
-    原棱锥的侧面积的~$\dfrac{5}{9}$,
-    则截得的棱台的体积与原棱锥的体积之比是\stk{$19:27$}.
-
-\question 以椭圆~$\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$~的右焦点为圆心,
-    且与双曲线~$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$~的渐近线相切的圆方程是\mtk{$(x-5)^2+y^2=16$}.
-
-
-\question 若~$\sqrt{\,\sin x}$~是有理数且~$x$~不是~$\dfrac{\pi}{6}$~的整数倍,
-    则~$x$~可能取的值是\mtk{$\arcsin\dfrac{1}{4}$ 等}.（只要求写出一个）
-
-\question 马路上有编号~1~到~10~的~10~盏路灯, 为节约用电又不影响照明,
-    可以关掉其中的~3~盏, 但又不能同时关掉相邻的两盏, 也不能关掉两端的路灯,
-    满足条件的关灯方法有\stk{$20$}种.
-
-
-\question 以椭圆~$\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$~的右焦点为圆心,
-    且与双曲线~$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$~的渐近线相切的圆方程是\mtk{$(x-5)^2+y^2=16$}.
-
-\question 若~$\sqrt{\,\sin x}$~是有理数且~$x$~不是~$\dfrac{\pi}{6}$~的整数倍,
-    则~$x$~可能取的值是\mtk{$\arcsin\dfrac{1}{4}$ 等}.（只要求写出一个）
-
-\question 马路上有编号~1~到~10~的~10~盏路灯, 为节约用电又不影响照明,
-    可以关掉其中的~3~盏, 但又不能同时关掉相邻的两盏, 也不能关掉两端的路灯,
-    满足条件的关灯方法有\stk{$20$}种.
-
-
-\question 以椭圆~$\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$~的右焦点为圆心,
-    且与双曲线~$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$~的渐近线相切的圆方程是\mtk{$(x-5)^2+y^2=16$}.
-
-\newpage
-
-\xuanze
-\question 已知集合~$A=\{x\mid {x-1}<3 \}$,
-集合~$B=\{y| y=x^2+2x+1,x\in\mathbb{R}\}$, 则~$A\cap
-\complement_U B$~为\stk{C}.
-\twoch{$[\,0,4)$}{$(-\infty,-2\,]\cup[4,+\infty)$}{$(-2,0)$}{$(0,4)$}
-
-\question 若~$a$、$b$~是直线, $\alpha$、$\beta$~是平面,
-则以下命题中真命题是\stk{D}.\\
-\fourch{若~$a$、$b$~异面, $a\subset\alpha$,$b\subset\beta$, 且~$a\perp b$, 则~$\alpha\perp\beta$}{若~$a\parallel b$, $a\subset\alpha$, $b\subset\beta$,则~$\alpha\parallel\beta$}{若~$a\parallel \alpha$,
-$b\subset\beta$, 则~$a$、$b$ 异面}{若~$a\perp b$, $a\perp\alpha$,$b\perp\beta$, 则~$\alpha\perp\beta$}
-
-\question 已知集合~$A=\{x\mid {x-1}<3 \}$,
-集合~$B=\{y| y=x^2+2x+1,x\in\mathbb{R}\}$, 则~$A\cap
-\complement_U B$~为\stk{C}.
-\twoch{$[\,0,4)$}{$(-\infty,-2\,]\cup[4,+\infty)$}{$(-2,0)$}{$(0,4)$}
-
-\question 若~$a$、$b$~是直线, $\alpha$、$\beta$~是平面,
-则以下命题中真命题是\stk{D}.\\
-\fourch{若~$a$、$b$~异面, $a\subset\alpha$,$b\subset\beta$, 且~$a\perp b$, 则~$\alpha\perp\beta$}{若~$a\parallel b$, $a\subset\alpha$, $b\subset\beta$,则~$\alpha\parallel\beta$}{若~$a\parallel \alpha$,
-$b\subset\beta$, 则~$a$、$b$ 异面}{若~$a\perp b$, $a\perp\alpha$,$b\perp\beta$, 则~$\alpha\perp\beta$}
-
-\newpage
-\jianda
-\question 已知复数~$z$ 满足：${z}-z^*=\dfrac{10}{1-w\textbf{i}}$（其中~$z^*$
-是~$z$ 的共轭复数）.
-\begin{parts}
-\part[7] 求复数~$z$；
-\part[7] 若复数~$w=\cos\theta+\textbf{i}\sin\theta\,(\theta\in\mathbb{R})$, 求~${z-2}$ 的取值范围.
-\end{parts}
-
-\begin{solution}
-\begin{parts}
-\part $z=3+4\textbf{i}$
-\part ${z-w}\in[4,6]$
-\end{parts}
-\end{solution}
-
-\question[14] 函数~$f(x)=4\sin\dfrac{\pi}{12}x\cdot\sin
-    \left(\dfrac{\pi}{2}+\dfrac{\pi}{12}x\right),x\in[a,a+1]$,
-    其中常数~$a\in[0,5]$, 求函数~$f(x)$ 的最大值~$g(a)$.
-
-\begin{solution}
-略
-\end{solution}
-
-\newpage
-
-\question[16] 函数~$f(x)=4\sin\dfrac{\pi}{12}x\cdot\sin
-    \left(\dfrac{\pi}{2}+\dfrac{\pi}{12}x\right),x\in[a,a+1]$,
-    其中常数~$a\in[0,5]$, 求函数~$f(x)$ 的最大值~$g(a)$.
-
-\begin{solution}
-略
-\end{solution}
-
-\newpage
-\question 已知复数~$z$ 满足：${z}-z^*=\dfrac{10}{1-w\textbf{i}}$（其中~$z^*$
-是~$z$ 的共轭复数）.
-\begin{parts}
-\part[8] 求复数~$z$；
-\part[8] 若复数~$w=\cos\theta+\textbf{i}\sin\theta\,(\theta\in\mathbb{R})$, 求~${z-2}$ 的取值范围.
-\end{parts}
-
-\begin{solution}
-\begin{parts}
-\part $z=3+4\textbf{i}$
-\part ${z-w}\in[4,6]$
-\end{parts}
-\end{solution}
-
-\newpage
-
-\question[18] 函数~$f(x)=4\sin\dfrac{\pi}{12}x\cdot\sin
-    \left(\dfrac{\pi}{2}+\dfrac{\pi}{12}x\right),x\in[a,a+1]$,
-    其中常数~$a\in[0,5]$, 求函数~$f(x)$ 的最大值~$g(a)$.
-
-\begin{solution}
-略
-\end{solution}
-
-\end{questions}
-\end{document}
diff --git a/macros/latex/contrib/bhcexam/test4.tex b/macros/latex/contrib/bhcexam/test4.tex
deleted file mode 100644
index 17a5ef8f9b..0000000000
--- a/macros/latex/contrib/bhcexam/test4.tex
+++ /dev/null
@@ -1,144 +0,0 @@
-\documentclass[UTF8,printbox]{BHCexam}
-\begin{document}
-
-\maketitle
-\notice
-
-\begin{questions}
-\tiankong
-\question 已知~$\bm{a}=(k,-9)$、$\bm{b}=(-1,k)$, $\bm{a}$~与~$\bm{b}$~为平行向量,
-    则~$k=$\stk{$\pm3$}.
-
-\question 若函数~$f(x)=x^{6m^2-5m-4}\,(m\in\mathbb{Z})$~的图像关于~$y$~轴对称,
-    且~$f(2)<f(6)$, 则~$f(x)$~的解析式为\stk{$f(x)=x^{-4}$}.
-
-\question 若~$f(x+1)=x^2\,(x\leq0)$, 则~$f^{-1}(1)=$\stk{0}.
-
-\question 在~$b\text{g}$~糖水中含糖~$a\text{g}$\,($b>a>0$), 若再添加~$m\text{g}$~糖~($m>0$),
-
-
-\question 已知~$f(x)=1-\textbf{c}_8^1x+\textbf{c}_8^2x^2-\textbf{c}_8^3x^3+\cdots+\textbf{c}_8^8x^8$,
-    则~$f\big(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\textbf{i}\big)$~的值是\ltk{$-\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\textbf{i}$}.
-
-\question 自然数~$1,2,3,\ldots,10$~的方差记为~$\sigma^2$,
-    其中的偶数~$2,4,6,8,10$~的方差记为~$\sigma_1^2$,
-    则~$\sigma^2$~与~$\sigma_1^2$~的大小关系为~$\sigma^2$\stk{$>$}$\sigma_1^2$.
-
-\question 若~$\theta$~为三角形的一个内角, 且~$\sin\theta+\cos\theta=\dfrac{2}{3}$,
-    则方程~$x^2\csc\theta-y^2\sec\theta=1$~表示的曲线的焦点坐标是\stk{$\big(\pm\dfrac{\sqrt{6}}{3},0\big)$}.
-
-
-\question 高为~$h$~的棱锥被平行于棱锥底面的截得棱台侧面积是
-    原棱锥的侧面积的~$\dfrac{5}{9}$,
-    则截得的棱台的体积与原棱锥的体积之比是\stk{$19:27$}.
-
-\question 以椭圆~$\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$~的右焦点为圆心,
-    且与双曲线~$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$~的渐近线相切的圆方程是\mtk{$(x-5)^2+y^2=16$}.
-
-
-\question 若~$\sqrt{\,\sin x}$~是有理数且~$x$~不是~$\dfrac{\pi}{6}$~的整数倍,
-    则~$x$~可能取的值是\mtk{$\arcsin\dfrac{1}{4}$ 等}.（只要求写出一个）
-
-\question 马路上有编号~1~到~10~的~10~盏路灯, 为节约用电又不影响照明,
-    可以关掉其中的~3~盏, 但又不能同时关掉相邻的两盏, 也不能关掉两端的路灯,
-    满足条件的关灯方法有\stk{$20$}种.
-
-
-\question 以椭圆~$\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$~的右焦点为圆心,
-    且与双曲线~$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$~的渐近线相切的圆方程是\mtk{$(x-5)^2+y^2=16$}.
-
-\question 若~$\sqrt{\,\sin x}$~是有理数且~$x$~不是~$\dfrac{\pi}{6}$~的整数倍,
-    则~$x$~可能取的值是\mtk{$\arcsin\dfrac{1}{4}$ 等}.（只要求写出一个）
-
-\question 马路上有编号~1~到~10~的~10~盏路灯, 为节约用电又不影响照明,
-    可以关掉其中的~3~盏, 但又不能同时关掉相邻的两盏, 也不能关掉两端的路灯,
-    满足条件的关灯方法有\stk{$20$}种.
-
-
-\question 以椭圆~$\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$~的右焦点为圆心,
-    且与双曲线~$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$~的渐近线相切的圆方程是\mtk{$(x-5)^2+y^2=16$}.
-
-\newpage
-
-\xuanze
-\question 已知集合~$A=\{x\mid \abs{x-1}<3 \}$,
-集合~$B=\{y| y=x^2+2x+1,x\in\mathbb{R}\}$, 则~$A\cap
-\complement_U B$~为\stk{C}.
-\twoch{$[\,0,4)$}{$(-\infty,-2\,]\cup[4,+\infty)$}{$(-2,0)$}{$(0,4)$}
-
-\question 若~$a$、$b$~是直线, $\alpha$、$\beta$~是平面,
-则以下命题中真命题是\stk{D}.\\
-\fourch{若~$a$、$b$~异面, $a\subset\alpha$,$b\subset\beta$, 且~$a\perp b$, 则~$\alpha\perp\beta$}{若~$a\parallel b$, $a\subset\alpha$, $b\subset\beta$,则~$\alpha\parallel\beta$}{若~$a\parallel \alpha$,
-$b\subset\beta$, 则~$a$、$b$ 异面}{若~$a\perp b$, $a\perp\alpha$,$b\perp\beta$, 则~$\alpha\perp\beta$}
-
-\question 已知集合~$A=\{x\mid \abs{x-1}<3 \}$,
-集合~$B=\{y| y=x^2+2x+1,x\in\mathbb{R}\}$, 则~$A\cap
-\complement_U B$~为\stk{C}.
-\twoch{$[\,0,4)$}{$(-\infty,-2\,]\cup[4,+\infty)$}{$(-2,0)$}{$(0,4)$}
-
-\question 若~$a$、$b$~是直线, $\alpha$、$\beta$~是平面,
-则以下命题中真命题是\stk{D}.\\
-\fourch{若~$a$、$b$~异面, $a\subset\alpha$,$b\subset\beta$, 且~$a\perp b$, 则~$\alpha\perp\beta$}{若~$a\parallel b$, $a\subset\alpha$, $b\subset\beta$,则~$\alpha\parallel\beta$}{若~$a\parallel \alpha$,
-$b\subset\beta$, 则~$a$、$b$ 异面}{若~$a\perp b$, $a\perp\alpha$,$b\perp\beta$, 则~$\alpha\perp\beta$}
-
-\newpage
-\jianda
-\question 已知复数~$z$ 满足：$\abs{z}-z^*=\dfrac{10}{1-w\textbf{i}}$（其中~$z^*$
-是~$z$ 的共轭复数）.
-\begin{parts}
-\part[7] 求复数~$z$；
-\part[7] 若复数~$w=\cos\theta+\textbf{i}\sin\theta\,(\theta\in\mathbb{R})$, 求~$\abs{z-2}$ 的取值范围.
-\end{parts}
-
-\begin{solution}
-\begin{parts}
-\part $z=3+4\textbf{i}$
-\part $\abs{z-w}\in[4,6]$
-\end{parts}
-\end{solution}
-
-\question[14] 函数~$f(x)=4\sin\dfrac{\pi}{12}x\cdot\sin
-    \left(\dfrac{\pi}{2}+\dfrac{\pi}{12}x\right),x\in[a,a+1]$,
-    其中常数~$a\in[0,5]$, 求函数~$f(x)$ 的最大值~$g(a)$.
-
-\begin{solution}
-略
-\end{solution}
-
-\newpage
-
-\question[16] 函数~$f(x)=4\sin\dfrac{\pi}{12}x\cdot\sin
-    \left(\dfrac{\pi}{2}+\dfrac{\pi}{12}x\right),x\in[a,a+1]$,
-    其中常数~$a\in[0,5]$, 求函数~$f(x)$ 的最大值~$g(a)$.
-
-\begin{solution}
-略
-\end{solution}
-
-\newpage
-\question 已知复数~$z$ 满足：$\abs{z}-z^*=\dfrac{10}{1-w\textbf{i}}$（其中~$z^*$
-是~$z$ 的共轭复数）.
-\begin{parts}
-\part[8] 求复数~$z$；
-\part[8] 若复数~$w=\cos\theta+\textbf{i}\sin\theta\,(\theta\in\mathbb{R})$, 求~$\abs{z-2}$ 的取值范围.
-\end{parts}
-
-\begin{solution}
-\begin{parts}
-\part $z=3+4\textbf{i}$
-\part $\abs{z-w}\in[4,6]$
-\end{parts}
-\end{solution}
-
-\newpage
-
-\question[18] 函数~$f(x)=4\sin\dfrac{\pi}{12}x\cdot\sin
-    \left(\dfrac{\pi}{2}+\dfrac{\pi}{12}x\right),x\in[a,a+1]$,
-    其中常数~$a\in[0,5]$, 求函数~$f(x)$ 的最大值~$g(a)$.
-
-\begin{solution}
-略
-\end{solution}
-
-\end{questions}
-\end{document}
diff --git a/macros/latex/contrib/proflycee/doc/ProfLycee-doc.pdf b/macros/latex/contrib/proflycee/doc/ProfLycee-doc.pdf
index a2dff6588f..950f44259d 100644
--- a/macros/latex/contrib/proflycee/doc/ProfLycee-doc.pdf
+++ b/macros/latex/contrib/proflycee/doc/ProfLycee-doc.pdf
diff --git a/macros/latex/contrib/proflycee/doc/ProfLycee-doc.tex b/macros/latex/contrib/proflycee/doc/ProfLycee-doc.tex
index 5f7796fa90..5036e35a4f 100644
--- a/macros/latex/contrib/proflycee/doc/ProfLycee-doc.tex
+++ b/macros/latex/contrib/proflycee/doc/ProfLycee-doc.tex
@@ -12,6 +12,7 @@
 \usepackage{tkz-euclide}
 \usetikzlibrary{hobby}
 \usepackage{siunitx}
+\usepackage{fancyvrb}
 \usepackage{fancyhdr}
 %fancy
 \fancyhf{}
@@ -54,8 +55,8 @@
 \setlength{\parindent}{0pt}
 \definecolor{LightGray}{gray}{0.9}
 
-\def\PLversion{1.1.8}
-\def\PLdate{01 Août 2022}
+\def\PLversion{1.2}
+\def\PLdate{10 Août 2022}
 
 \tcbset{vignettes/.style={%
 		nobeforeafter,box align=base,boxsep=0pt,enhanced,sharp corners=all,rounded corners=southeast,%
@@ -225,6 +226,8 @@ Une commande pour déterminer des paramètres ($a$, $b$, $r$ et $r^2$) d'une ré
 
 Quelques commandes pour convertir bin/dec/hex avec certains détails.
 
+Une commande pour, en \TikZ, créer un pixelart avec correction éventuelle.
+
 \vspace{1.5cm}
 
 \hfill{}\textit{Merci à Anne pour ses retours et sa relecture !}
@@ -2494,15 +2497,15 @@ La commande étant \og autonome \fg{}, elle va pouvoir être intégrée dans des
 
 \begin{codetex}[listing only]
 \begin{tikzpicture}
-	\begin{axis}[
-		/pgf/number format/.cd,use comma,xmin = 1992, xmax = 2012,ymin = 1640, ymax = 1730,
-		width = 0.7\textwidth,height = 0.35\textwidth,xtick distance = 2,ytick distance = 10,
-		grid = both,minor tick num = 1,major grid style = {lightgray},minor grid style = {lightgray!25},
+	\begin{axis}[/pgf/number format/.cd,
+		use comma,xmin = 1992, xmax = 2012,ymin = 1640, ymax = 1730,width = 0.7\textwidth,
+		height = 0.35\textwidth,xtick distance = 2,ytick distance = 10,grid = both,
+		minor tick num = 1,major grid style = {lightgray},minor grid style = {lightgray!25},
 		xlabel = {\small Année ($x$)},ylabel = {\small Altitude du glacier (en m) ($y$)},
 		x tick label style={/pgf/number format/.cd, set thousands separator={}},
 		y tick label style={/pgf/number format/.cd, set thousands separator={}},
 		legend cell align = {left},legend pos = north east]
-		\addplot[teal, only marks,forget plot] table{
+		\addplot[teal, only marks] table{
 			X Y
 			1994 1718
 			1995 1710
@@ -2927,8 +2930,334 @@ Les valeurs \Cle{mainlevee=5mm et 0.6pt} donnent des résultats -- à mon sens -
 
 \newpage
 
+\section{PixelART via un fichier csv, en \TikZ}
+
+\subsection{Introduction}
+
+\begin{codeidee}
+L'idée est de \textit{proposer}, dans un environnement \TikZ, une commande permettant de générer des grilles PixelART.
+
+Les données sont \textit{lues} à partir d'un fichier \textsf{csv}, externe au fichier \textsf{tex} ou déclaré en interne grâce à l'environnement \ctex{filecontents}.
+\end{codeidee}
+
+\begin{codeinfo}
+Avant toute chose, quelques petites infos sur les données au format \textsf{csv}, surtout dans l'optique de sa lecture et de son traitement par \ctex{ProfLycee} :
+
+\begin{itemize}
+	\item le fichier de données \textsf{csv} doit être formaté avec le séparateur décimal \og , \fg ;
+	\item chaque ligne doit se terminer par \og \texttt{,ZZ} \fg{} (choix arbitraire de l'auteur, sinon il y a un \textsf{bug} non réglé pour le moment\ldots) ;
+	\item des cases vides seront codées par \og \texttt{-} \fg.
+\end{itemize}
+
+Le fichier \textsf{csv} peut être déclaré directement dans le fichier \textsf{tex}, grâce à l'environnement \ctex{filecontents} (intégré en natif sur les dernières versions de \LaTeX) :
+
+\begin{Verbatim}[tabsize=2,fontsize=\scriptsize]
+\begin{filecontents*}{<nomfichier>.csv}
+A,B,C,D,ZZ
+A,B,D,C,ZZ
+B,A,C,D,ZZ
+B,A,D,C,ZZ
+\end{filecontents*}
+\end{Verbatim}
+
+À la compilation, le fichier \textsf{<nomfichier>.csv} sera créé automatiquement, cependant toute modification dans le fichier \textsf{tex} ne sera pas répercutée dans le fichier\ldots
+\end{codeinfo}
+
+\subsection{Package csvsimple et option}
+
+\begin{codeinfo}
+Le package \textit{central} est ici \ctex{csvsimple}, qui permet de lire et traiter le fichier \textsf{csv}.
+
+Il est \og disponible \fg{} en version \LaTeXe{} ou en version \LaTeXIII. Par défaut, \ctex{ProfLycee} le charge en version \LaTeXIII{}, mais une \Cle{option} est disponible pour une \textit{rétro-compatibilité} avec la version \LaTeXe.
+
+\smallskip
+
+L'option \Cle{csvii} permet de passer l'appel au package en version \LaTeXe.
+\end{codeinfo}
+
+\begin{codetex}[listing only]
+\usepackage{ProfLycee}                     %chargement du package version 3
+%qui charge :
+%\RequirePackage{expl3}
+%\RequirePackage[l3]{csvsimple}
+
+\usepackage[csvii]{ProfLycee}              %chargement du package version 2
+%qui charge :
+%\RequirePackage[legacy]{csvsimple}
+\end{codetex}
+
+\subsection{Exemple simple, clés et options}
+
+\begin{codetex}[listing only]
+%déclaration du fichier csv
+\begin{filecontents*}{basique.csv}
+A,B,C,D,ZZ
+A,B,D,C,ZZ
+B,A,C,D,ZZ
+B,A,D,C,ZZ
+C,A,B,D,ZZ
+\end{filecontents*}
+
+\begin{tikzpicture}%avec lettres
+	\PLpixelart[codes=ABCD,style=\large\sffamily]{basique.csv}
+\end{tikzpicture}
+\begin{tikzpicture}%avec chiffres
+	\PLpixelart[codes=ABCD,symboles={45,22,1,7},symb,style=\large\sffamily]{basique.csv}
+\end{tikzpicture}
+\begin{tikzpicture}%avec correction
+	\PLpixelart[codes=ABCD,couleurs={Black,Green,Yellow,Red},correction]{basique.csv}
+\end{tikzpicture}
+\end{codetex}
+
+\begin{codesortie}
+\begin{filecontents*}{basique.csv}
+A,B,C,D,ZZ
+A,B,D,C,ZZ
+B,A,D,C,ZZ
+C,A,B,D,ZZ
+\end{filecontents*}
+
+\begin{center}
+	\renewcommand\arraystretch{1.25}
+	\begin{tabular}{|c|c|c|c|}
+		\hline
+		\multicolumn{4}{|c|}{Notice} \\ \hline
+		A & B & C & D \\ \hline
+		45 & 22 & 1 & 7 \\ \hline
+		Noir & Vert & Jaune & Rouge \\ \hline
+	\end{tabular}
+\end{center}
+
+\hfill\begin{tikzpicture}
+	%avec lettres
+	\PLpixelart[codes=ABCD,style=\large\sffamily]{basique.csv}
+\end{tikzpicture}
+\begin{tikzpicture}
+	%avec chiffres
+	\PLpixelart[codes=ABCD,symboles={45,22,1,7},symb,style=\large\sffamily]{basique.csv}
+\end{tikzpicture}
+\begin{tikzpicture}
+	%avec correction
+	\PLpixelart[codes=ABCD,couleurs={Black,Green,Yellow,Red},correction]{basique.csv}
+\end{tikzpicture}\hfill~
+\end{codesortie}
+
+\begin{codeinfo}
+La commande \ctex{PLpixelart} nécessite de connaître :
+
+\begin{itemize}
+	\item le fichier \textsf{csv} à traiter ;
+	\item la liste (en fait sous forme de chaîne) des codes utilisés dans le fichier \textsf{csv} (comme \ctex{234679} ou \ctex{ABCDJK}\ldots);
+	\item la liste des symboles (éventuellement !) à afficher dans les cases s'il y a ambiguïté, comme \ctex{25,44,12} ou \ctex{AA,AB,AC};
+	\item la liste des couleurs (si la correction est demandée), dans le même ordre que la liste des caractères.
+\end{itemize}
+\end{codeinfo}
+
+\begin{codetex}[listing only]
+%environnement tikz
+\PLpixelart[<clés>]{<fichier>.csv}
+\end{codetex}
+
+\begin{codecles}
+Quelques \Cle{Clés} sont nécessaires au bon fonctionnement de la commande :
+
+\begin{itemize}
+	\item la clé \Cle{codes} contient la \textsf{chaîne} des codes \textit{simples} du  fichier \textsf{csv} ;
+	\item la clé \Cle{couleurs} qui contient la \textsf{liste} des couleurs associées ;
+	\item la clé \Cle{symboles} qui contient la \textsf{liste éventuelles} des caractères alternatifs à afficher dans les cases ;
+	\item la clé booléenne \Cle{correction} qui permet de colorier le PixelART ;\hfill{}défaut \Cle{false}
+	\item la clé booléenne \Cle{symb} qui permet d'afficher les caractères \textit{alternatifs} ;\hfill{}défaut \Cle{false}
+	\item la clé \Cle{style} qui permet de spécifier le style des caractères. \hfill{}défaut \Cle{scriptsize}
+\end{itemize}
+\end{codecles}
+
+\begin{codetex}[listing only]
+%codes simples et sans ambiguïté
+%une case vide sera codée par -
+
+\begin{filecontents*}{perroquet.csv}
+-,-,-,-,-,-,4,4,4,4,-,-,-,-,-,-,ZZ
+-,-,-,-,4,4,1,1,1,1,4,4,-,-,-,-,ZZ
+-,-,-,4,1,1,1,1,1,1,1,1,4,-,-,-,ZZ
+-,-,4,1,1,1,1,1,1,1,1,1,1,4,-,-,ZZ
+-,-,4,1,1,1,1,1,1,1,1,1,1,4,-,-,ZZ
+-,4,1,9,9,1,1,1,1,1,1,9,9,1,4,-,ZZ
+-,4,9,9,9,9,4,4,4,4,9,9,9,9,4,-,ZZ
+-,4,9,4,9,9,4,4,4,4,9,4,9,9,4,-,ZZ
+-,4,1,9,9,9,4,4,4,4,9,9,9,1,4,-,ZZ
+-,-,4,1,1,9,4,4,4,4,9,1,1,4,-,-,ZZ
+-,-,4,1,1,1,4,4,4,4,1,1,1,4,-,-,ZZ
+-,-,-,4,1,1,1,4,4,1,1,1,4,-,-,-,ZZ
+-,-,4,3,1,1,1,1,1,1,1,1,3,4,-,-,ZZ
+-,4,6,3,1,1,1,1,1,1,1,1,3,6,4,-,ZZ
+-,4,6,6,1,1,1,1,1,1,1,1,6,6,4,-,ZZ
+-,4,6,6,1,1,1,1,1,1,1,1,6,6,4,-,ZZ
+-,4,6,4,1,1,1,4,4,1,1,1,4,6,4,-,ZZ
+2,2,4,2,4,4,4,2,2,4,4,4,2,4,2,2,ZZ
+2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,ZZ
+2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,ZZ
+-,-,-,-,-,4,1,1,1,1,4,-,-,-,-,-,ZZ
+-,-,-,-,-,-,4,1,1,4,-,-,-,-,-,-,ZZ
+-,-,-,-,-,-,-,4,4,-,-,-,-,-,-,-,ZZ
+\end{filecontents*}
+
+\begin{tikzpicture}[x=0.35cm,y=0.35cm]
+	\PLpixelart[codes=123469,style=\ttfamily]{perroquet.csv}
+\end{tikzpicture}
+
+\begin{tikzpicture}[x=0.35cm,y=0.35cm]
+	\PLpixelart[codes=123469,couleurs={Red,Brown,Yellow,Black,Blue,White},correction]{perroquet.csv}
+\end{tikzpicture}
+\end{codetex}
+
+\subsection{Exemples complémentaires}
+
+\begin{codeinfo}
+Les symboles affichés dans les cases sont situés aux nœuds de coordonnées $(c\,;\,-l)$ où $l$ et $c$ sont les numéros de ligne et de colonne correspondants à la position de la donnée dans le fichier \textsf{csv}.
+\end{codeinfo}
+
+\begin{codesortie}
+\begin{filecontents*}{perroquet.csv}
+-,-,-,-,-,-,4,4,4,4,-,-,-,-,-,-,ZZ
+-,-,-,-,4,4,1,1,1,1,4,4,-,-,-,-,ZZ
+-,-,-,4,1,1,1,1,1,1,1,1,4,-,-,-,ZZ
+-,-,4,1,1,1,1,1,1,1,1,1,1,4,-,-,ZZ
+-,-,4,1,1,1,1,1,1,1,1,1,1,4,-,-,ZZ
+-,4,1,9,9,1,1,1,1,1,1,9,9,1,4,-,ZZ
+-,4,9,9,9,9,4,4,4,4,9,9,9,9,4,-,ZZ
+-,4,9,4,9,9,4,4,4,4,9,4,9,9,4,-,ZZ
+-,4,1,9,9,9,4,4,4,4,9,9,9,1,4,-,ZZ
+-,-,4,1,1,9,4,4,4,4,9,1,1,4,-,-,ZZ
+-,-,4,1,1,1,4,4,4,4,1,1,1,4,-,-,ZZ
+-,-,-,4,1,1,1,4,4,1,1,1,4,-,-,-,ZZ
+-,-,4,3,1,1,1,1,1,1,1,1,3,4,-,-,ZZ
+-,4,6,3,1,1,1,1,1,1,1,1,3,6,4,-,ZZ
+-,4,6,6,1,1,1,1,1,1,1,1,6,6,4,-,ZZ
+-,4,6,6,1,1,1,1,1,1,1,1,6,6,4,-,ZZ
+-,4,6,4,1,1,1,4,4,1,1,1,4,6,4,-,ZZ
+2,2,4,2,4,4,4,2,2,4,4,4,2,4,2,2,ZZ
+2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,ZZ
+2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,ZZ
+-,-,-,-,-,4,1,1,1,1,4,-,-,-,-,-,ZZ
+-,-,-,-,-,-,4,1,1,4,-,-,-,-,-,-,ZZ
+-,-,-,-,-,-,-,4,4,-,-,-,-,-,-,-,ZZ
+\end{filecontents*}
+
+\begin{center}
+	\renewcommand\arraystretch{1.25}
+	\begin{tabular}{|c|c|c|c|c|c|}
+		\hline
+		\multicolumn{6}{|c|}{Notice} \\ \hline
+		1 & 2 & 3 & 5 & 6 & 9 \\ \hline
+		Rouge & Marron & Jaune & Noir & Bleu & Blanc \\ \hline
+	\end{tabular}
+\end{center}
+
+\begin{center}
+	\begin{tikzpicture}[x=0.35cm,y=0.35cm]
+		\PLpixelart[codes=123469,style=\ttfamily]{perroquet.csv}
+	\end{tikzpicture}
+	\begin{tikzpicture}[x=0.35cm,y=0.35cm]
+		\PLpixelart[codes=123469,couleurs={Red,Brown,Yellow,Black,Blue,White},correction]{perroquet.csv}
+	\end{tikzpicture}
+\end{center}
+
+\begin{center}
+	\begin{tikzpicture}[x=0.35cm,y=0.35cm]
+		\draw[very thin,gray,xstep=1,ystep=1] (0,0) grid (17,-24) ;
+		\foreach \x in {0,1,...,17} \draw[very thin,gray] (\x,-3pt)--(\x,3pt) node[above,font=\scriptsize\sffamily] {\x} ;
+		\foreach \y in {0,-1,...,-24} \draw[very thin,gray] (3pt,\y)--(-3pt,\y) node[left,font=\scriptsize\sffamily] {\y} ;
+		\PLpixelart[codes=123469,couleurs={Red,Brown,Yellow,Black,Blue,White},correction]{perroquet.csv}
+		\filldraw[Blue] (14,-1) circle[radius=1] ;
+		\filldraw[Yellow] (14,-1) circle[radius=0.8] ;
+	\end{tikzpicture}
+\end{center}
+\end{codesortie}
+
+\pagebreak
+
+\begin{codetex}[listing only]
+%codes avec ambiguïté
+
+\begin{filecontents*}{cap.csv}
+-,-,-,-,-,-,-,-,D,-,D,-,D,-,-,-,-,-,-,-,-,-,ZZ
+-,D,D,-,-,-,-,D,D,D,D,D,D,-,-,D,D,D,D,-,-,-,ZZ
+D,-,-,D,-,D,D,F,F,F,F,F,F,D,D,-,-,-,-,D,-,-,ZZ
+-,D,-,-,D,F,F,F,-,-,F,F,F,F,F,D,-,D,D,-,-,-,ZZ
+-,-,D,D,F,F,F,-,F,F,-,F,F,F,F,F,D,D,-,-,-,-,ZZ
+-,-,-,D,F,F,F,F,F,F,F,F,F,F,F,F,D,-,-,-,-,-,ZZ
+-,-,-,D,F,J,J,J,J,J,J,J,F,F,F,F,D,-,-,-,-,-,ZZ
+-,-,-,D,J,-,-,-,J,-,-,-,J,J,F,F,D,-,-,-,-,-,ZZ
+-,-,-,D,J,-,D,-,J,-,D,-,J,J,B,B,D,-,-,-,-,-,ZZ
+-,-,-,D,J,-,-,-,J,-,-,-,J,J,B,B,D,-,-,-,-,-,ZZ
+-,-,-,D,C,J,J,J,J,J,J,J,J,C,C,C,D,-,-,-,-,-,ZZ
+-,-,-,D,C,C,C,C,C,C,C,C,C,C,C,D,D,D,-,-,-,-,ZZ
+-,-,-,D,C,C,C,D,D,D,D,D,D,C,D,A,A,A,D,-,-,-,ZZ
+-,-,-,D,F,C,C,C,C,C,C,C,C,D,A,-,-,-,A,D,-,-,ZZ
+-,-,-,D,F,C,F,C,C,C,C,F,D,A,-,A,A,A,-,A,D,-,ZZ
+-,-,D,C,F,F,F,F,C,C,F,D,A,-,A,F,F,F,A,-,A,D,ZZ
+-,-,D,C,F,F,F,F,F,F,F,D,A,-,A,F,-,F,A,-,A,D,ZZ
+-,-,D,A,D,-,A,-,A,-,A,D,A,-,A,F,F,F,A,-,A,D,ZZ
+-,-,-,D,D,-,A,-,A,-,A,-,D,A,-,A,A,A,-,A,D,-,ZZ
+-,-,-,-,-,D,D,F,D,D,D,D,F,D,A,-,-,-,A,D,-,-,ZZ
+-,-,-,-,-,-,D,A,D,-,-,D,-,-,D,A,A,A,D,-,-,-,ZZ
+-,-,-,-,-,-,D,D,D,-,-,D,D,D,D,D,D,D,-,-,-,-,ZZ
+\end{filecontents*}
+
+\begin{tikzpicture}[x=0.35cm,y=0.35cm]
+	\PLpixelart[codes=ABCDFJ,symboles={1,2,3,4,6,10},symb,style=\tiny\sffamily]{cap.csv}
+\end{tikzpicture}
+
+\begin{tikzpicture}[x=0.35cm,y=0.35cm]
+	\PLpixelart[codes=ABCDFJ,couleurs={Red,Brown,Yellow,Black,Blue,Gray},correction]{cap.csv}
+\end{tikzpicture}
+\end{codetex}
+
+\begin{codesortie}
+\begin{filecontents*}{cap.csv}
+-,-,-,-,-,-,-,-,D,-,D,-,D,-,-,-,-,-,-,-,-,-,ZZ
+-,D,D,-,-,-,-,D,D,D,D,D,D,-,-,D,D,D,D,-,-,-,ZZ
+D,-,-,D,-,D,D,F,F,F,F,F,F,D,D,-,-,-,-,D,-,-,ZZ
+-,D,-,-,D,F,F,F,-,-,F,F,F,F,F,D,-,D,D,-,-,-,ZZ
+-,-,D,D,F,F,F,-,F,F,-,F,F,F,F,F,D,D,-,-,-,-,ZZ
+-,-,-,D,F,F,F,F,F,F,F,F,F,F,F,F,D,-,-,-,-,-,ZZ
+-,-,-,D,F,J,J,J,J,J,J,J,F,F,F,F,D,-,-,-,-,-,ZZ
+-,-,-,D,J,-,-,-,J,-,-,-,J,J,F,F,D,-,-,-,-,-,ZZ
+-,-,-,D,J,-,D,-,J,-,D,-,J,J,B,B,D,-,-,-,-,-,ZZ
+-,-,-,D,J,-,-,-,J,-,-,-,J,J,B,B,D,-,-,-,-,-,ZZ
+-,-,-,D,C,J,J,J,J,J,J,J,J,C,C,C,D,-,-,-,-,-,ZZ
+-,-,-,D,C,C,C,C,C,C,C,C,C,C,C,D,D,D,-,-,-,-,ZZ
+-,-,-,D,C,C,C,D,D,D,D,D,D,C,D,A,A,A,D,-,-,-,ZZ
+-,-,-,D,F,C,C,C,C,C,C,C,C,D,A,-,-,-,A,D,-,-,ZZ
+-,-,-,D,F,C,F,C,C,C,C,F,D,A,-,A,A,A,-,A,D,-,ZZ
+-,-,D,C,F,F,F,F,C,C,F,D,A,-,A,F,F,F,A,-,A,D,ZZ
+-,-,D,C,F,F,F,F,F,F,F,D,A,-,A,F,-,F,A,-,A,D,ZZ
+-,-,D,A,D,-,A,-,A,-,A,D,A,-,A,F,F,F,A,-,A,D,ZZ
+-,-,-,D,D,-,A,-,A,-,A,-,D,A,-,A,A,A,-,A,D,-,ZZ
+-,-,-,-,-,D,D,F,D,D,D,D,F,D,A,-,-,-,A,D,-,-,ZZ
+-,-,-,-,-,-,D,A,D,-,-,D,-,-,D,A,A,A,D,-,-,-,ZZ
+-,-,-,-,-,-,D,D,D,-,-,D,D,D,D,D,D,D,-,-,-,-,ZZ
+\end{filecontents*}
+
+\begin{center}
+	\begin{tikzpicture}[x=0.35cm,y=0.35cm]
+		\PLpixelart[codes=ABCDFJ,symboles={1,2,3,4,6,10},symb,style=\tiny\sffamily]{cap.csv}
+	\end{tikzpicture}
+	~~
+	\begin{tikzpicture}[x=0.35cm,y=0.35cm]
+		\PLpixelart[codes=ABCDFJ,couleurs={Red,Brown,Yellow,Black,Blue,Gray},correction]{cap.csv}
+	\end{tikzpicture}
+\end{center}
+\end{codesortie}
+
+\newpage
+
 \section{Historique}
 
+{\small \bverb|v1.2  |~:~~~~Correction d'un méchant bug sur Pixelart
+
+{\small \bverb|v1.1.9|~:~~~~Pixelart en \TikZ
+
 {\small \bverb|v1.1.8|~:~~~~Style "Mainlevée" basique pour \TikZ
 
 {\small \bverb|v1.1.7|~:~~~~Conversions bin/hex/dec (basées sur \ctex{xintbinhex}) avec quelques détails
diff --git a/macros/latex/contrib/proflycee/doc/basique.csv b/macros/latex/contrib/proflycee/doc/basique.csv
new file mode 100644
index 0000000000..b15c24713c
--- /dev/null
+++ b/macros/latex/contrib/proflycee/doc/basique.csv
@@ -0,0 +1,4 @@
+A,B,C,D,ZZ
+A,B,D,C,ZZ
+B,A,D,C,ZZ
+C,A,B,D,ZZ
diff --git a/macros/latex/contrib/proflycee/doc/cap.csv b/macros/latex/contrib/proflycee/doc/cap.csv
new file mode 100644
index 0000000000..850d0c2152
--- /dev/null
+++ b/macros/latex/contrib/proflycee/doc/cap.csv
@@ -0,0 +1,22 @@
+-,-,-,-,-,-,-,-,D,-,D,-,D,-,-,-,-,-,-,-,-,-,ZZ
+-,D,D,-,-,-,-,D,D,D,D,D,D,-,-,D,D,D,D,-,-,-,ZZ
+D,-,-,D,-,D,D,F,F,F,F,F,F,D,D,-,-,-,-,D,-,-,ZZ
+-,D,-,-,D,F,F,F,-,-,F,F,F,F,F,D,-,D,D,-,-,-,ZZ
+-,-,D,D,F,F,F,-,F,F,-,F,F,F,F,F,D,D,-,-,-,-,ZZ
+-,-,-,D,F,F,F,F,F,F,F,F,F,F,F,F,D,-,-,-,-,-,ZZ
+-,-,-,D,F,J,J,J,J,J,J,J,F,F,F,F,D,-,-,-,-,-,ZZ
+-,-,-,D,J,-,-,-,J,-,-,-,J,J,F,F,D,-,-,-,-,-,ZZ
+-,-,-,D,J,-,D,-,J,-,D,-,J,J,B,B,D,-,-,-,-,-,ZZ
+-,-,-,D,J,-,-,-,J,-,-,-,J,J,B,B,D,-,-,-,-,-,ZZ
+-,-,-,D,C,J,J,J,J,J,J,J,J,C,C,C,D,-,-,-,-,-,ZZ
+-,-,-,D,C,C,C,C,C,C,C,C,C,C,C,D,D,D,-,-,-,-,ZZ
+-,-,-,D,C,C,C,D,D,D,D,D,D,C,D,A,A,A,D,-,-,-,ZZ
+-,-,-,D,F,C,C,C,C,C,C,C,C,D,A,-,-,-,A,D,-,-,ZZ
+-,-,-,D,F,C,F,C,C,C,C,F,D,A,-,A,A,A,-,A,D,-,ZZ
+-,-,D,C,F,F,F,F,C,C,F,D,A,-,A,F,F,F,A,-,A,D,ZZ
+-,-,D,C,F,F,F,F,F,F,F,D,A,-,A,F,-,F,A,-,A,D,ZZ
+-,-,D,A,D,-,A,-,A,-,A,D,A,-,A,F,F,F,A,-,A,D,ZZ
+-,-,-,D,D,-,A,-,A,-,A,-,D,A,-,A,A,A,-,A,D,-,ZZ
+-,-,-,-,-,D,D,F,D,D,D,D,F,D,A,-,-,-,A,D,-,-,ZZ
+-,-,-,-,-,-,D,A,D,-,-,D,-,-,D,A,A,A,D,-,-,-,ZZ
+-,-,-,-,-,-,D,D,D,-,-,D,D,D,D,D,D,D,-,-,-,-,ZZ
diff --git a/macros/latex/contrib/proflycee/doc/perroquet.csv b/macros/latex/contrib/proflycee/doc/perroquet.csv
new file mode 100644
index 0000000000..4bcbe1d191
--- /dev/null
+++ b/macros/latex/contrib/proflycee/doc/perroquet.csv
@@ -0,0 +1,23 @@
+-,-,-,-,-,-,4,4,4,4,-,-,-,-,-,-,ZZ
+-,-,-,-,4,4,1,1,1,1,4,4,-,-,-,-,ZZ
+-,-,-,4,1,1,1,1,1,1,1,1,4,-,-,-,ZZ
+-,-,4,1,1,1,1,1,1,1,1,1,1,4,-,-,ZZ
+-,-,4,1,1,1,1,1,1,1,1,1,1,4,-,-,ZZ
+-,4,1,9,9,1,1,1,1,1,1,9,9,1,4,-,ZZ
+-,4,9,9,9,9,4,4,4,4,9,9,9,9,4,-,ZZ
+-,4,9,4,9,9,4,4,4,4,9,4,9,9,4,-,ZZ
+-,4,1,9,9,9,4,4,4,4,9,9,9,1,4,-,ZZ
+-,-,4,1,1,9,4,4,4,4,9,1,1,4,-,-,ZZ
+-,-,4,1,1,1,4,4,4,4,1,1,1,4,-,-,ZZ
+-,-,-,4,1,1,1,4,4,1,1,1,4,-,-,-,ZZ
+-,-,4,3,1,1,1,1,1,1,1,1,3,4,-,-,ZZ
+-,4,6,3,1,1,1,1,1,1,1,1,3,6,4,-,ZZ
+-,4,6,6,1,1,1,1,1,1,1,1,6,6,4,-,ZZ
+-,4,6,6,1,1,1,1,1,1,1,1,6,6,4,-,ZZ
+-,4,6,4,1,1,1,4,4,1,1,1,4,6,4,-,ZZ
+2,2,4,2,4,4,4,2,2,4,4,4,2,4,2,2,ZZ
+2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,ZZ
+2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,ZZ
+-,-,-,-,-,4,1,1,1,1,4,-,-,-,-,-,ZZ
+-,-,-,-,-,-,4,1,1,4,-,-,-,-,-,-,ZZ
+-,-,-,-,-,-,-,4,4,-,-,-,-,-,-,-,ZZ
diff --git a/macros/latex/contrib/proflycee/tex/ProfLycee.sty b/macros/latex/contrib/proflycee/tex/ProfLycee.sty
index 7e32293bc5..1f3e210420 100644
--- a/macros/latex/contrib/proflycee/tex/ProfLycee.sty
+++ b/macros/latex/contrib/proflycee/tex/ProfLycee.sty
@@ -3,7 +3,9 @@
 % or later, see http://www.latex-project.org/lppl.txtf
 
 \NeedsTeXFormat{LaTeX2e}
-\ProvidesPackage{ProfLycee}[2022/08/01 - v1.1.8 - Aide pour l'utilisation de LaTeX en lycée]
+\ProvidesPackage{ProfLycee}[2022/08/10 - v1.2 - Aide pour l'utilisation de LaTeX en lycée]
+% 1.2	Correction d'un méchant bug avec pixelart, erf...
+% 1.1.9	Commande pour du pixelart en tikz (avec option csvii pour compatibilité)
 % 1.1.8	Ajout d'un style mainlevee pour tikz
 % 1.1.7	Ajout de commandes pour des conversions bin/dec/hex
 % 1.1.6	Ajout d'une commande pour déterminer les paramètres d'un régression linéaire par moindres carrés
@@ -29,6 +31,8 @@
 \DeclareOption{nominted}{\@nomintedtrue}
 \newif\if@build \@buildfalse
 \DeclareOption{build}{\@buildtrue}
+\newif\if@csvii \@csviifalse
+\DeclareOption{csvii}{\@csviitrue}
 \DeclareOption*{}
 \ProcessOptions\relax
 
@@ -49,6 +53,12 @@
 \RequirePackage{listofitems}
 \RequirePackage{tabularray}
 \RequirePackage{fontawesome5}
+\if@csvii
+	\RequirePackage[legacy]{csvsimple}
+\else
+	\RequirePackage{expl3}
+	\RequirePackage[l3]{csvsimple}
+\fi
 %librairies tikz
 \usetikzlibrary{calc}
 \usetikzlibrary{decorations}
@@ -1444,4 +1454,71 @@
 	mainlevee/.default={5mm et 0.6pt}
 }
 
+%==============PixelArt TikZ=========== OK !!
+\defKV[pixelarttkz]{%
+	codes=\def\PATlettres{#1},%
+	couleurs=\def\PATcouleurs{#1},%
+	symboles=\def\PATchiffres{#1},%
+	style=\def\PATtaille{#1}
+}
+
+\setKVdefault[pixelarttkz]{%
+	style=\scriptsize,%
+	correction=false,%
+	symb=false
+}
+
+\newcommand\PLpixelart[2][]{%
+	\useKVdefault[pixelarttkz]
+	\setKV[pixelarttkz]{#1}
+	%affichage du corrigé
+	\ifboolKV[pixelarttkz]{correction}
+	{%
+		\csvloop{file=#2,no head,command={%
+				\foreach \l [count=\n] in \csvline {%
+					\setsepchar{,}
+					\readlist\LCPA\PATcouleurs
+					\StrPosition{\PATlettres}{\l}[\pixcnt]%
+					\xintifboolexpr{\pixcnt>0}%
+					{%
+						\itemtomacro\LCPA[\pixcnt]\pixcol%
+						\draw[fill=\pixcol] ({\n-0.5},{-\thecsvrow+0.5}) rectangle ({\n+0.5},{-\thecsvrow-0.5}) ;%
+					}%
+					{}
+				}
+			}
+		}
+	}
+	{%
+		\ifboolKV[pixelarttkz]{symb}
+		{%affichage du pixelart versions chiffres si besoin
+			\csvloop{file=#2,no head,command={%
+					\foreach \l [count=\n] in \csvline {%
+						\setsepchar{,}
+						\readlist\LCNA\PATchiffres
+						\StrPosition{\PATlettres}{\l}[\pixpos]%
+						\xintifboolexpr{\pixpos>0}%
+						{%
+							\itemtomacro\LCNA[\pixpos]\pixchf%
+							\draw ({\n-0.5},{-\thecsvrow+0.5}) rectangle ({\n+0.5},{-\thecsvrow-0.5}) node[inner sep=0,midway,font=\PATtaille] {\pixchf} ;%
+						}
+						{}%
+					}
+				}
+			}
+		}
+		{%affichage du pixelart versions lettres ou chiffres simples (<=9)
+			\csvloop{file=#2,no head,command={%
+					\foreach \l [count=\n] in \csvline {%
+						\IfSubStr{\PATlettres}{\l}%
+						{%
+							\draw ({\n-0.5},{-\thecsvrow+0.5}) rectangle ({\n+0.5},{-\thecsvrow-0.5}) node[inner sep=0,midway,font=\PATtaille] {\l};}
+						{}
+					}
+				}
+			}
+		}
+	}
+}
+
 \endinput
\ No newline at end of file
diff --git a/macros/latex/contrib/simplebnf/LICENSE b/macros/latex/contrib/simplebnf/LICENSE
index 3a0de6d06f..720f8374ba 100644
--- a/macros/latex/contrib/simplebnf/LICENSE
+++ b/macros/latex/contrib/simplebnf/LICENSE
@@ -1,6 +1,6 @@
 The MIT License (MIT)
 
-Copyright © 2019-2020 Jay Lee <jaeho.lee@snu.ac.kr>
+Copyright © 2019-2022 Jay Lee <jaeho.lee@snu.ac.kr>
 
 Permission is hereby granted, free of charge, to any person obtaining
 a copy of this software and associated documentation files (the "Software"),
diff --git a/macros/latex/contrib/simplebnf/simplebnf-doc.tex b/macros/latex/contrib/simplebnf/simplebnf-doc.tex
index 30ea889f18..99ac96780b 100644
--- a/macros/latex/contrib/simplebnf/simplebnf-doc.tex
+++ b/macros/latex/contrib/simplebnf/simplebnf-doc.tex
@@ -32,10 +32,10 @@
 
 \title{%
   \textsf{simplebnf} --- A simple package to format Backus-Naur form%
-  \footnote{This file describes v0.3.0.}}
+  \footnote{This file describes v0.3.1.}}
 \author{Jay Lee\footnote{E-mail: %
   \href{mailto:jaeho.lee@snu.ac.kr}{\texttt{jaeho.lee@snu.ac.kr}}}}
-\date{2022/05/07}
+\date{2022/05/08}
 
 \begin{document}
 \maketitle
@@ -43,9 +43,24 @@
 This package provides a simple way to typeset grammars written in Backus-Naur form (BNF).
 
 \begin{presentcommand}
-  \cmd{bnfexpr} \cmd{bnfannot}
+  \cmd{SimpleBNFDefEq}
 \end{presentcommand}
-These commands are wrappers around \cmd{texttt} and \cmd{textit} respectively.
+This command is used to typeset the definition symbol separate a nonterminal from its productions. It defaults to \SimpleBNFDefEq. It can be redefined to customized output using \verb|RenewDocumentCommand|.
+
+\begin{presentcommand}
+  \cmd{SimpleBNFDefOr}
+\end{presentcommand}
+This command is used to typeset the separator symbol between productions. It defaults to \SimpleBNFDefOr. It can be redefined to customized output using \verb|RenewDocumentCommand|.
+
+\begin{presentcommand}
+  \cmd{bnfexpr}
+\end{presentcommand}
+This command is used when typesetting the BNF nonterminal and productions. It defaults to a wrappers around \cmd{texttt}. It can be redefined to customized output using \verb|RenewDocumentCommand|.
+
+\begin{presentcommand}
+ \cmd{bnfannot}
+\end{presentcommand}
+This command is used when typesetting the annotations on nonterminals and productions. It defaults to a wrappers around \cmd{textit}. It can be redefined to customized output using \verb|RenewDocumentCommand|.
 
 \begin{presentcommand}
   \env{bnfgrammar}{text}
diff --git a/macros/latex/contrib/simplebnf/simplebnf.sty b/macros/latex/contrib/simplebnf/simplebnf.sty
index fd7b707ff7..ae4482c29b 100644
--- a/macros/latex/contrib/simplebnf/simplebnf.sty
+++ b/macros/latex/contrib/simplebnf/simplebnf.sty
@@ -13,18 +13,16 @@
 \RequirePackage{mathtools}
 \ProvidesExplPackage
   {simplebnf}
-  {2022/05/07}
-  {0.3.0}
+  {2022/05/08}
+  {0.3.1}
   {A simple package to format Backus–Naur form}
 
 \cs_generate_variant:Nn \regex_split:nnNTF {nVNTF}
 \cs_generate_variant:Nn \regex_split:NnN {NVN}
 
-\tl_new:N \g__simplebnf_defeq_tl
-\tl_gset:Nn \g__simplebnf_defeq_tl { \ensuremath{\Coloneqq} }
+\NewDocumentCommand\SimpleBNFDefEq{}{\ensuremath{\Coloneqq}}
 
-\tl_new:N \g__simplebnf_defor_tl
-\tl_gset:Nn \g__simplebnf_defor_tl { \ensuremath{|} }
+\NewDocumentCommand\SimpleBNFDefOr{}{\ensuremath{|}}
 
 \seq_new:N \l__input_seq
 \seq_new:N \l__term_seq
@@ -48,7 +46,7 @@
       \bool_set_false:N \l__first_rhs
     }
     {
-      \tl_put_right:Nn \l__table_tl { \\ && \g__simplebnf_defor_tl & }
+      \tl_put_right:Nn \l__table_tl { \\ && \SimpleBNFDefOr & }
     }
 
   \tl_set:Nn \l_tmpa_tl { #1 }
@@ -58,23 +56,30 @@
     {
       \seq_pop_left:NNT \l_tmpa_seq \l_tmpa_tl
         {
-          \regex_replace_all:NnN \g_simplebnf_rhs_nb_r { \c{g__simplebnf_defor_tl} } \l_tmpa_tl
-          \tl_put_right:Nx \l__table_tl { \bnfexpr { \l_tmpa_tl } }
+          \regex_replace_all:NnN \g_simplebnf_rhs_nb_r { \c{SimpleBNFDefOr} } \l_tmpa_tl
+          % Expand only the local temporary variable.
+          \tl_put_right:No \l__table_tl
+             {
+               \exp_after:wN\bnfexpr\exp_after:wN{\l_tmpa_tl} &
+             }
         }
 
-      \tl_put_right:Nn \l__table_tl { & }
-
       \seq_pop_left:NNT \l_tmpa_seq \l_tmpb_tl
         {
-          \tl_set:Nx \l_tmpb_tl { \bnfannot { \l_tmpb_tl } }
-          \tl_put_right:NV \l__table_tl \l_tmpb_tl
+          \regex_replace_once:nnN { ^\s+ } {} \l_tmpb_tl
+          \tl_put_right:No \l__table_tl
+            {
+              \exp_after:wN\bnfannot\exp_after:wN{\l_tmpb_tl}
+            }
         }
     }
     {
-      \regex_replace_all:NnN \g_simplebnf_rhs_nb_r { \c{g__simplebnf_defor_tl} } \l_tmpa_tl
+      \regex_replace_all:NnN \g_simplebnf_rhs_nb_r { \c{SimpleBNFDefOr} } \l_tmpa_tl
 
-      \tl_put_right:Nx \l__table_tl { \bnfexpr { \l_tmpa_tl } }
-      \tl_put_right:Nn \l__table_tl { & }
+      \tl_put_right:No \l__table_tl
+         {
+           \exp_after:wN\bnfexpr\exp_after:wN{\l_tmpa_tl}
+         }
     }
 }
 
@@ -89,20 +94,21 @@
   \regex_split:nVNTF { : } \l_tmpa_tl \l_tmpa_seq
     {
       \seq_pop_right:NN \l_tmpa_seq \l_tmpa_tl
-      \tl_put_right:Nx \l__table_tl
+      \regex_replace_once:nnN { ^\s+ } {} \l_tmpa_tl
+      \tl_put_right:No \l__table_tl
         {
-          \bnfannot{\l_tmpa_tl}
+          \exp_after:wN\bnfannot\exp_after:wN{\l_tmpa_tl} &
         }
       \seq_pop_left:NN \l_tmpa_seq \l_tmpa_tl
-      \tl_put_right:Nx \l__table_tl
+      \tl_put_right:No \l__table_tl
         {
-          & \bnfexpr { \l_tmpa_tl }
+          \exp_after:wN\bnfexpr\exp_after:wN{\l_tmpa_tl}
         }
     }
     {
-      \tl_put_right:Nx \l__table_tl
+      \tl_put_right:No \l__table_tl
         {
-          & \bnfexpr { \l_tmpa_tl }
+          \exp_after:wN&\exp_after:wN\bnfexpr\exp_after:wN{\l_tmpa_tl}
         }
     }
 }
@@ -150,9 +156,9 @@
             \seq_pop_left:NN \l__term_seq \l__keypairs_tl
 
             \simplebnf_typeset_lhs:n{\l__term_tl}
-            \tl_put_right:Nx \l__table_tl
+            \tl_put_right:Nn \l__table_tl
               {
-                & \g__simplebnf_defeq_tl &
+                & \SimpleBNFDefEq &
               }
             %% \l__keypairs_seq - (rhs:annot | rhs)...
             \regex_split:NVN \g_simplebnf_rhs_newline_r \l__keypairs_tl \l__keypairs_seq
@@ -191,7 +197,7 @@
 
 %% The MIT License (MIT)
 %%
-%% Copyright © 2019-2020 Jay Lee <jaeho.lee@snu.ac.kr>
+%% Copyright © 2019-2022 Jay Lee <jaeho.lee@snu.ac.kr>
 %%
 %% Permission is hereby granted, free of charge, to any person obtaining
 %% a copy of this software and associated documentation files (the "Software"),
diff --git a/macros/luatex/generic/luatexja/README b/macros/luatex/generic/luatexja/README
index d37dc3db67..6cca0c1099 100644
--- a/macros/luatex/generic/luatexja/README
+++ b/macros/luatex/generic/luatexja/README
@@ -1,4 +1,4 @@
-The LuaTeX-ja Package 20220807.0
+The LuaTeX-ja Package 20220810.0
 --------------------------------
 Copyright (c) 2011--2022 The LuaTeX-ja project
 License: modified BSD (see COPYING)
@@ -42,9 +42,9 @@ Installation
 2. If you must/want to install manually:
 
    a. Download the source archive from CTAN,
-      or tagged as 20220807.0 in the Git repository by
+      or tagged as 20220810.0 in the Git repository by
       <http://git.osdn.jp/view?p=luatex-ja/luatexja.git
-                              ;a=snapshot;h=refs/tags/20220807.0;sf=tgz>
+                              ;a=snapshot;h=refs/tags/20220810.0;sf=tgz>
 
    b. Extract the archive and process following three files by LuaLaTeX
       to generate classes for Japanese typesetting:
@@ -93,4 +93,4 @@ Notes
   This file and ltj-kinsoku.lua are not used anymore.
   (Do not remove ltj-kinsoku.tex.)
 
-Last commit date: Sun Aug 7 12:20:08 2022 +0900
+Last commit date: Wed Aug 10 11:32:01 2022 +0900
diff --git a/macros/luatex/generic/luatexja/doc/luatexja-en.pdf b/macros/luatex/generic/luatexja/doc/luatexja-en.pdf
index 5ee7e4047d..4ec7df2004 100644
--- a/macros/luatex/generic/luatexja/doc/luatexja-en.pdf
+++ b/macros/luatex/generic/luatexja/doc/luatexja-en.pdf
diff --git a/macros/luatex/generic/luatexja/doc/luatexja-ja.pdf b/macros/luatex/generic/luatexja/doc/luatexja-ja.pdf
index 2ed0625eb1..8742f56b0a 100644
--- a/macros/luatex/generic/luatexja/doc/luatexja-ja.pdf
+++ b/macros/luatex/generic/luatexja/doc/luatexja-ja.pdf
diff --git a/macros/luatex/generic/luatexja/src/ltj-direction.lua b/macros/luatex/generic/luatexja/src/ltj-direction.lua
index 565190be2d..4a0ad3ab06 100644
--- a/macros/luatex/generic/luatexja/src/ltj-direction.lua
+++ b/macros/luatex/generic/luatexja/src/ltj-direction.lua
@@ -334,7 +334,9 @@ do
          while x and getid(x)==id_glue and getsubtype(x)==3 do
             node_remove(hd,x); node_free(x); x = node_next(hd)
          end
-         if #wh==1 then wh[1]=nil else wh[#wh], wh[1]=nil, wh[#wh] end
+         --if gc~='vtop' then
+         --    if #wh==1 then wh[1]=nil else wh[#wh], wh[1]=nil, wh[#wh] end
+         --end
       end
       for i=1,#wh do
          hd = node_remove(hd, wh[i]); node_free(wh[i]); wh[i] = nil
diff --git a/macros/luatex/generic/luatexja/src/luatexja.sty b/macros/luatex/generic/luatexja/src/luatexja.sty
index 5eab1c4e35..7353d8aa08 100644
--- a/macros/luatex/generic/luatexja/src/luatexja.sty
+++ b/macros/luatex/generic/luatexja/src/luatexja.sty
@@ -27,7 +27,7 @@
 \newif\ifltj@disablejfam
 
 %%%% VERSION
-\def\LuaTeXjaversion{20220807.0}
+\def\LuaTeXjaversion{20220810.0}
 
 %% Check if LaTeX is used.
 \begingroup\expandafter\expandafter\expandafter\endgroup
diff --git a/macros/xetex/latex/bhcexam/BHCexam.cls b/macros/xetex/latex/bhcexam/BHCexam.cls
new file mode 100644
index 0000000000..1648653a50
--- /dev/null
+++ b/macros/xetex/latex/bhcexam/BHCexam.cls
@@ -0,0 +1,550 @@
+% BHCexam.cls
+%
+% An exam class designed for mathematics teacher in China.
+% Since 2016, it was used by mathcrowd.cn ( an opensource math exam database) as the default class to export exam papers.
+%
+
+%% BHCexam.cls
+%% Copyright (c) 2011-2022 BAO HONG CHANG
+%
+% This work may be distributed and/or modified under the
+% conditions of the LaTeX Project Public License, either version 1.3
+% of this license or (at your option) any later version.
+% The latest version of this license is in
+%   http://www.latex-project.org/lppl.txt
+% and version 1.3 or later is part of all distributions of LaTeX
+% version 2003/12/01 or later.
+%
+% This work has the LPPL maintenance status "author-maintained".
+% 
+% This work consists of the files BHCexam.cls.
+% Documation on https://docs.mathcrowd.cn/advances/bhcexam.html
+% Support on https://github.com/mathedu4all/bhcexam/issues
+%
+
+%%% BAO HONG CHANG
+%%% Mathcrowd Inc.
+%%% Shanghai
+%%% charles@mathcrowd.cn
+
+% The newest version of this documentclass should always be available
+% from my web page: https://github.com/mathedu4all/bhcexam
+
+\def\fileversion{1.7}
+\def\filedate{2022/07/29}
+
+\NeedsTeXFormat{LaTeX2e}
+\ProvidesClass{BHCexam}[\filedate\space Version \fileversion\space by
+ BAO HONG CHANG]
+ 
+ %                         *****************
+ %                         **   宏包选项   **
+ %                         *****************
+ 
+ 
+ % 文档类支持以下宏包选项:
+ 
+ % answers 显示解析
+ % adobe 使用adobe字体
+ % ubuntu 使用ubuntu字体
+ % windows 使用windows字体
+ % fandol 使用fandol字体，随texlive默认安装
+ % mac 使用mac字体
+ % list 以列表项目格式显示试题
+ % twocolumn 使用A3纸张并分栏
+  
+
+\newif\if@printanswers \@printanswersfalse
+\DeclareOption{answers}{\@printanswerstrue}
+
+\newcommand\@fontset{windows}
+\DeclareOption{adobe}{\renewcommand\@fontset{adobe}}
+\DeclareOption{ubuntu}{\renewcommand\@fontset{ubuntu}}
+\DeclareOption{mac}{\renewcommand\@fontset{mac}}
+\DeclareOption{windows}{\renewcommand\@fontset{windows}}
+\DeclareOption{fandol}{\renewcommand\@fontset{fandol}}
+
+\newif\if@twocolumn  \@twocolumnfalse
+\DeclareOption{twocolumn}{\@twocolumntrue}
+
+\newif\if@list  \@listfalse
+\DeclareOption{list}{\@listtrue}
+
+
+\DeclareOption*{\PassOptionsToClass{\CurrentOption}{article}}
+\ProcessOptions\relax
+\LoadClass{article}
+
+\RequirePackage[fontset = \@fontset, punct=kaiming]{ctex}
+\ctexset{linestretch = 4, autoindent = 0pt}
+
+
+%                    *****************************
+%                    **        加载其他宏包       **
+%                    *****************************
+
+\RequirePackage{tabularx, ifthen} % 排选项用
+\RequirePackage{xcolor,graphicx,caption} % 彩色、图片、图释
+\RequirePackage{geometry,fancyhdr} % 纸张、边距、页眉、页脚
+\RequirePackage{etoolbox} 
+\RequirePackage{amsmath,amsmath,amssymb} % ams数学相关
+\RequirePackage{unicode-math}
+\RequirePackage{pifont} % 带圈数字\ding
+\RequirePackage{bbding} % 图案
+\RequirePackage{romannum} % 罗马数字
+\RequirePackage{enumitem}
+
+
+
+%                         *****************
+%                         **   页面设置   **
+%                         *****************
+
+
+\AtBeginDocument{
+	\pagenumbering{arabic}  % 使用阿拉伯数字页码
+	\renewcommand{\parallel}{\mathrel{\mathpalette\new@parallel\relax}} % 重定义平行符号
+	\newcommand{\new@parallel}[2]{%
+		\begingroup
+		\sbox\z@{高度}% get the height of an uppercase letter
+		\resizebox{!}{\ht\z@}{\raisebox{\depth}{$\m@th#1/\mkern-10mu/$}}%
+		\endgroup
+	}
+}
+
+
+%
+% 双栏显示
+%
+\if@twocolumn
+\geometry{a3paper,landscape, twocolumn,columnsep=40mm,left=50mm,right=30mm,top=35mm,bottom=25mm,headheight=20mm}
+\else
+\geometry{a4paper,left=30mm,right=30mm,top=35mm,bottom=25mm, headheight=20mm}
+\fi
+
+%
+% 图释
+%
+\captionsetup[figure]{font=small,belowskip=0pt}
+
+
+
+%                         *****************
+%                         **    试卷头    **
+%                         *****************
+
+
+
+\gdef\@subtitle{}
+\gdef\@notice{}
+\gdef\@author{}
+\gdef\@date{}
+
+
+
+\renewcommand{\headrulewidth}{0pt}
+\renewcommand{\title}[1]{\gdef\@title{#1}}
+\newcommand{\subtitle}[1]{\gdef\@subtitle{#1}}
+\newcommand{\notice}[1]{\def\@notice{#1}}
+\renewcommand{\author}[1]{\gdef\@author{#1}}
+\renewcommand{\date}[1]{\gdef\@date{#1}}
+
+\renewcommand\maketitle{\begingroup
+	\renewcommand{\baselinestretch}{2}
+	\newpage
+	\begin{center}
+		\heiti \Large
+		\@title \par
+		\ifdefempty{\@subtitle}{}{
+			\@subtitle \par
+		}
+	
+		\ifdefempty{\@author}{}{
+			\songti \normalsize
+			\@author \par
+		}
+	
+		\ifdefempty{\@notice}{}{
+			\setlength{\fboxsep}{1em}
+			\vspace{0.5\baselineskip}
+			\fbox{
+				\parbox{0.5\linewidth}{
+					\kaishu \normalsize 
+					\@notice \par
+				}
+			}
+			\vspace{0.5\baselineskip}
+		}
+	
+
+	\end{center}
+	\renewcommand{\baselinestretch}{1}
+	\ifdefempty{\@date}{}{
+		\begin{flushright}
+			\songti \small
+			\@date \par
+		\end{flushright}
+	}
+	\songti \normalsize
+\endgroup}
+
+\renewcommand\arraystretch{1.5}
+\renewcommand{\baselinestretch}{1.5}
+
+%                    ***************************
+%                    **    试题、解答环境定义   **
+%                    ***************************
+
+\newcounter{Group}
+\newcounter{Question}
+\newcounter{Example}
+\newcounter{Exercise}
+\newcounter{Method}[Question]
+
+\newif\if@showscore
+\@showscorefalse
+\newif\if@showskip
+\@showskipfalse
+\newif\if@showparen
+\@showparenfalse
+\newif\if@resetcounter
+\@resetcounterfalse
+\newlength{\myvertspace}
+
+
+% 定义题组环境
+\newenvironment{groups}{
+	\if@list
+		\list{\heiti\chinese{Group}.}{\usecounter{Group}}
+	\else
+		\par \begingroup \par
+	\fi
+}{
+	\if@list
+		\endlist
+	\else
+		\par \endgroup \par
+	\fi
+}
+
+% 定义新增题组命令
+\newcommand{\group}[2]{
+	\if@list
+		\item \heiti{#1} \quad \kaishu \small #2 \songti \normalsize \par
+	\else
+		\stepcounter{Group}
+		\par \heiti{\par \chinese{Group} 、#1} \quad \small{\kaishu #2} \songti \normalsize \par
+	\fi
+
+}
+
+% 定义试题环境
+\newenvironment{questions}[1][]{
+	\@tfor \@opt :=#1\do
+		{\if\@opt s\global\@showscoretrue\fi
+		 \if\@opt t\global\@showskiptrue\fi
+	 	 \if\@opt p\global\@showparentrue\fi
+ 	 	 \if\@opt r\global\@resetcountertrue\fi}
+  	\if@resetcounter
+  		\setcounter{Question}{0}
+	 \fi
+	\if@list
+		\list{\arabic{Question}.}{\setlength{\leftmargin}{0pt}}
+	\else
+		\par \begingroup \par
+	\fi
+}{
+	\if@list
+		\endlist
+	\else
+		\par \endgroup \par
+	\fi
+	
+	\global\@showscorefalse
+	\global\@showskipfalse
+	\global\@showparenfalse
+}
+
+% 定义新增试题命令
+\newcommand{\question}[1][0]{
+	\stepcounter{Question}
+	\if@list
+		\item \if@showscore \kaishu ( #1 分) \songti \fi
+	\else
+		\vspace{2mm}
+
+		\arabic{Question}. 
+		\if@showscore \kaishu ( #1 分) \songti \fi
+	\fi
+}
+
+% 定义新增例题命令
+\newcommand{\example}[1][0]{
+	\stepcounter{Example}
+	\if@list
+		\item[例题\arabic{Example}. ]
+		\if@showscore \kaishu ( #1 分) \songti \fi
+	\else
+		\vspace{2mm}
+		\par 例题\arabic{Example}. 
+		\if@showscore \kaishu ( #1 分) \songti \fi
+	\fi
+}
+
+% 定义新增习题命令
+\newcommand{\exercise}[1][0]{
+	\stepcounter{Exercise}
+	\if@list
+		\item[习题\arabic{Exercise}. ]
+		\if@showscore \kaishu ( #1 分) \songti \fi
+	\else
+		\vspace{2mm}
+		\par 习题\arabic{Exercise}. 
+		\if@showscore \kaishu ( #1 分) \songti \fi
+	\fi
+}
+
+% 定义小问环境
+\newlist{subquestions}{enumerate}{2}
+\setlist[subquestions,1]{label=(\arabic*)}
+\setlist[subquestions,2]{label=(\roman*)}
+
+% 定义新增小问命令
+\newcommand{\subquestion}{\item} 
+
+\newenvironment{solution}[1]{
+	\setlength{\myvertspace}{#1}
+	\par \if@printanswers \par \color{red} \begingroup \else \if@showskip \vspace*{\myvertspace} \fi \setbox\z@\vbox\bgroup\fi \songti
+}{
+	\par \if@printanswers \endgroup \color{black} \else \egroup \fi \par
+}
+
+\newcommand{\hint}{
+	\par \fbox{\heiti{提示}} \par \songti
+}
+\newcommand{\method}{
+	\stepcounter{Method}
+	\vspace{2mm}
+	\par \fbox{\heiti{解法\chinese{Method}}} \par \songti
+}
+	
+\newcommand{\methodonly}{
+	\par \fbox{\heiti{解答}} \par \songti
+}
+
+\newcommand{\score}[2]{
+	\par
+	\dotfill 本步骤 #1 分， 累计 #2 分
+	\par}
+
+\makeatletter
+\newcommand{\build}[2]{\leavevmode
+	\count@=\z@ \toks@={}%
+	\loop\ifnum\count@<\numexpr#1\relax
+	\toks@=\expandafter{\the\toks@#2}%
+	\advance\count@\@ne
+	\repeat
+	\the\toks@}
+\makeatletter
+
+\newlength{\keylength}
+\newcommand{\key}[1]{
+	\if@printanswers
+	\underline{~~#1~~}
+	\else
+	\settowidth{\keylength}{~~#1~~}
+	\build{13}{\hskip1sp\kern-1sp\hbox to 0.1\keylength{\hrulefill}}
+	\fi}
+
+\newlength{\choicelengtha}
+\newlength{\choicelengthb}
+\newlength{\choicelengthc}
+\newlength{\choicelengthd}
+\newlength{\choicelengthe}
+\newlength{\maxlength}
+
+% 三个选项
+\newcommand{\threechoices}[3]{
+	\if@showparen \dotfill (\qquad) \fi
+	\par
+	\settowidth{\choicelengtha}{A.~#1~~~}
+	\settowidth{\choicelengthb}{B.~#2~~~}
+	\settowidth{\choicelengthc}{C.~#3~~~}
+
+	\ifthenelse{\lengthtest{\choicelengtha>\choicelengthb}}{\setlength{\maxlength}{\choicelengtha}}{\setlength{\maxlength}{\choicelengthb}}
+	\ifthenelse{\lengthtest{\choicelengthc>\maxlength}}{\setlength{\maxlength}{\choicelengthc}}{}
+	\ifthenelse{\lengthtest{\maxlength>0.4\linewidth}}
+	{
+		\begin{tabularx}{\linewidth}{X}
+			\setlength\tabcolsep{0pt}
+			A.~#1~~~\\
+			B.~#2~~~\\
+			C.~#3~~~\\
+		\end{tabularx}
+	}%
+	{
+		\ifthenelse{\lengthtest{\maxlength>0.2\linewidth}}
+		{
+			\begin{tabularx}{\linewidth}{XX}
+				\setlength\tabcolsep{0pt}
+				A.~#1~~~ & B.~#2~~~\\
+				C.~#3~~~ & \\
+			\end{tabularx}
+		}%
+		{
+			\begin{tabularx}{\linewidth}{XXXX}
+				\setlength\tabcolsep{0pt}
+				A.~#1~~~  & B.~#2~~~ & C.~#3~~~ &\\ 
+			\end{tabularx}
+		}
+	}
+	\unskip \unskip 
+}
+
+% 四个选项
+\newcommand{\fourchoices}[4]{
+	\if@showparen \dotfill (\qquad) \fi
+	\par
+	\settowidth{\choicelengtha}{A.~#1~~~}
+	\settowidth{\choicelengthb}{B.~#2~~~}
+	\settowidth{\choicelengthc}{C.~#3~~~}
+	\settowidth{\choicelengthd}{D.~#4~~~}
+	\ifthenelse{\lengthtest{\choicelengtha>\choicelengthb}}{\setlength{\maxlength}{\choicelengtha}}{\setlength{\maxlength}{\choicelengthb}}
+	\ifthenelse{\lengthtest{\choicelengthc>\maxlength}}{\setlength{\maxlength}{\choicelengthc}}{}
+	\ifthenelse{\lengthtest{\choicelengthd>\maxlength}}{\setlength{\maxlength}{\choicelengthd}}{}
+	\ifthenelse{\lengthtest{\maxlength>0.4\linewidth}}
+	{
+		\begin{tabularx}{\linewidth}{X}
+			\setlength\tabcolsep{0pt}
+			A.~#1~~~\\
+			B.~#2~~~\\
+			C.~#3~~~\\
+			D.~#4~~~\\
+		\end{tabularx}
+	}%
+	{
+		\ifthenelse{\lengthtest{\maxlength>0.2\linewidth}}
+		{
+			\begin{tabularx}{\linewidth}{XX}
+				\setlength\tabcolsep{0pt}
+				A.~#1~~~ & B.~#2~~~\\
+				C.~#3~~~ & D.~#4~~~\\
+			\end{tabularx}
+		}%
+		{
+			\begin{tabularx}{\linewidth}{XXXX}
+				\setlength\tabcolsep{0pt}
+				A.~#1~~~  & B.~#2~~~ & C.~#3~~~ & D.~#4~~~\\ 
+			\end{tabularx}
+		}%
+	}
+	\unskip \unskip 
+}
+
+% 五个选项
+\newcommand{\fivechoices}[5]{
+	\if@showparen \hfill \dotfill (\quad) \fi
+	\par
+	\settowidth{\choicelengtha}{A.~#1~~~}
+	\settowidth{\choicelengthb}{B.~#2~~~}
+	\settowidth{\choicelengthc}{C.~#3~~~}
+	\settowidth{\choicelengthd}{D.~#4~~~}
+	\settowidth{\choicelengthe}{E.~#5~~~}
+	\ifthenelse{\lengthtest{\choicelengtha>\choicelengthb}}{\setlength{\maxlength}{\choicelengtha}}{\setlength{\maxlength}{\choicelengthb}}
+	\ifthenelse{\lengthtest{\choicelengthc>\maxlength}}{\setlength{\maxlength}{\choicelengthc}}{}
+	\ifthenelse{\lengthtest{\choicelengthd>\maxlength}}{\setlength{\maxlength}{\choicelengthd}}{}
+	\ifthenelse{\lengthtest{\choicelengthe>\maxlength}}{\setlength{\maxlength}{\choicelengthe}}{}
+	\ifthenelse{\lengthtest{\maxlength>0.4\linewidth}}{
+		\begin{tabularx}{\linewidth}{X}
+			\setlength\tabcolsep{0pt}
+			A.~#1~~~\\
+			B.~#2~~~\\
+			C.~#3~~~\\
+			D.~#4~~~\\
+			E.~#5~~~\\
+		\end{tabularx}
+	}%
+	{
+		\ifthenelse{\lengthtest{\maxlength>0.2\linewidth}}
+		{
+			\begin{tabularx}{\linewidth}{XX}
+				\setlength\tabcolsep{0pt}
+				A.~#1~~~ & B.~#2~~~\\
+				C.~#3~~~ & D.~#4~~~\\
+				E.~#5~~~ &         \\
+			\end{tabularx}
+		}%
+		{
+			\begin{tabularx}{\linewidth}{XXXX}
+				\setlength\tabcolsep{0pt}
+				A.~#1~~~  & B.~#2~~~ & C.~#3~~~ & D.~#4~~~ \\
+				E.~#5~~~  &          &          &          \\
+			\end{tabularx}
+		}
+	}%
+	\unskip \unskip 
+}
+
+% 六个选项
+\newcommand{\sixchoices}[6]{
+	\if@showparen \hfill \dotfill (\quad) \fi
+	\par
+	\settowidth{\choicelengtha}{A.~#1~~~}
+	\settowidth{\choicelengthb}{B.~#2~~~}
+	\settowidth{\choicelengthc}{C.~#3~~~}
+	\settowidth{\choicelengthd}{D.~#4~~~}
+	\settowidth{\choicelengthe}{D.~#5~~~}
+	\ifthenelse{\lengthtest{\choicelengtha>\choicelengthb}}{\setlength{\maxlength}{\choicelengtha}}{\setlength{\maxlength}{\choicelengthb}}
+	\ifthenelse{\lengthtest{\choicelengthc>\maxlength}}{\setlength{\maxlength}{\choicelengthc}}{}
+	\ifthenelse{\lengthtest{\choicelengthd>\maxlength}}{\setlength{\maxlength}{\choicelengthd}}{}	\ifthenelse{\lengthtest{\choicelengthe>\maxlength}}{\setlength{\maxlength}{\choicelengthe}}{}
+	
+	\ifthenelse{\lengthtest{\maxlength>0.4\linewidth}}
+	{
+		\begin{tabularx}{\linewidth}{X}
+			\setlength\tabcolsep{0pt}
+			A.~#1~~~\\
+			B.~#2~~~\\
+			C.~#3~~~\\
+			D.~#4~~~\\
+			E.~#5~~~\\
+			F.~#6~~~\\
+		\end{tabularx}
+	}%
+	{
+		\ifthenelse{\lengthtest{\maxlength>0.2\linewidth}}
+		{
+			\begin{tabularx}{\linewidth}{XX}
+				\setlength\tabcolsep{0pt}
+				A.~#1~~~ & B.~#2~~~\\
+				C.~#3~~~ & D.~#4~~~\\
+				E.~#5~~~ & F.~#6~~~\\
+			\end{tabularx}
+		}%
+		{
+			\begin{tabularx}{\linewidth}{XXXX}
+				\setlength\tabcolsep{0pt}
+				A.~#1~~~  & B.~#2~~~ & C.~#3~~~ & D.~#4~~~ \\
+				E.~#2~~~  & F.~#6~~~ &          &          \\
+			\end{tabularx}
+		}%
+	}
+	\unskip \unskip 
+}
+
+
+%                    ***************************
+%                    **   CUSTOMIZED COMMAND  **
+%                    ***************************
+
+\newcommand\abs[1]{\left|#1\right|}
+\newcommand{\gt}{>}
+\newcommand{\lt}{<}
+\renewcommand{\geq}{\geqslant}
+\renewcommand{\ge}{\geqslant}
+\renewcommand{\leq}{\leqslant}
+\renewcommand{\le}{\leqslant}
+\renewenvironment{split}{\begin{aligned}}{\end{aligned}}
+
+
+
+\endinput
diff --git a/macros/xetex/latex/bhcexam/README-zh.md b/macros/xetex/latex/bhcexam/README-zh.md
new file mode 100644
index 0000000000..1d626ec24b
--- /dev/null
+++ b/macros/xetex/latex/bhcexam/README-zh.md
@@ -0,0 +1,52 @@
+# BHCexam
+
+[English](./README.md)
+
+BHCexam 是一款为中国数学老师设计的试卷排版文档类，目前已被 [橘子数学开源题库社区](https://www.mathcrowd.cn) 选用为默认文档类导出试卷pdf文件.
+
+你可以使用该文档类实现：
+
+* 将内容与样式分离
+* 使用一个参数生成教师版和学生版试卷;
+* 排版3-6个选项的选择题，可根据选项的长度自适应对齐;
+* 排版填空题，可根据答案的长度自适应设置横线长度;
+* 排版简答题，并以列表形式展示小问，并控制是否展示小问、缩进;
+* 对试题分组，对组内试题以列表呈现，并控制是否展示分值、留空、是否重新开始编号;
+* 更多 (见 [BHCexam 文档](http://docs.mathcrowd.cn/advances/bhcexam.html) )
+
+## 版本历史
+
+* **version 1.7** (2022.8)
+  * 针对TeXLive 2022, 修正`ctex`的字号、字体设置. 
+* **version 1.6** (2021.8)
+  * 不再使用`stix`数学字体
+  * `ctex` 设置 `punct = kaiming`
+* **version 1.5** (2020.6)
+  * `questions` 环境新增 `r` 选项，重置题组的编号;
+  * 在 `master` 分支中清理历史版本;
+  * 新增 `fandol` 宏包选项以支持 `fandol` 字体;
+* **version 1.4** (2020.5)
+  * 支持`subquestion`环境的嵌套;
+  * `\parallel` 命令重定义;
+* **version 1.3** (2020.3)
+  * 新增 `\sixchoices` , `\threechoices` 命令，以支持对3个和6个选项，并保持智能断行同及选项对齐.
+* **version 1.2** (2020.3)
+  *  支持苹果字体
+  *  使用 `stix` 数学字体
+  *  支持在选择题最后显示右对齐括号
+* **version 1.1** (2020.1)
+  * 新增对A3双栏版式的支持
+  * 新增列表样式的试题
+* **version 1.0** (2019.5)
+  * 弃用 `exam` 而改用 `article` 为基宏包类
+
+## 贡献者
+
+* Bao Hongchang - @mathedu4all,  charles@mathcrowd.cn
+* CamuseCao - @ CamuseCao, camusecao@gmail.com
+
+------
+
+This work may be distributed and/or modified under the conditions of
+the LaTeX Project Public License, either version 1.3 of this license
+or (at your option) any later version.
diff --git a/macros/xetex/latex/bhcexam/README.md b/macros/xetex/latex/bhcexam/README.md
new file mode 100644
index 0000000000..e555332373
--- /dev/null
+++ b/macros/xetex/latex/bhcexam/README.md
@@ -0,0 +1,28 @@
+# BHCexam
+
+[中文版](./README-zh.md)
+
+BHCexam is an exam class designed for mathematics teacher in China.
+
+Now it is used by [mathcrowd.cn](www.mathcrowd.cn) ( an opensource math exam database) as the default class to export exam papers in pdf.
+
+Using bhcexam you can
+
+* separate the format and the content very well;
+* export both teacher paper and student paper;
+* typeset multiple choice questions with 3-6 options keeping adaptively neat alignment;
+* typeset cloze questions with an adaptively underline;
+* typeset questions with subquestions in list;
+* group questions  in list to control whether to show score, leave spacing, initialize question number;
+* and more (see [BHCexam Documentation](http://docs.mathcrowd.cn/advances/bhcexam.html)).
+
+## Authors and Contributors
+
+* Bao Hongchang <charles@mathcrowd.cn>
+* CamuseCao <camusecao@gmail.com>
+
+------
+
+This work may be distributed and/or modified under the conditions of
+the LaTeX Project Public License, either version 1.3 of this license
+or (at your option) any later version.
diff --git a/macros/xetex/latex/bhcexam/examples/To7VxKXpZaOWO5Qq9qzeZGtlwYTwJ5KF.jpg b/macros/xetex/latex/bhcexam/examples/To7VxKXpZaOWO5Qq9qzeZGtlwYTwJ5KF.jpg
new file mode 100644
index 0000000000..6d3adabeb6
--- /dev/null
+++ b/macros/xetex/latex/bhcexam/examples/To7VxKXpZaOWO5Qq9qzeZGtlwYTwJ5KF.jpg
diff --git a/macros/xetex/latex/bhcexam/examples/example_student_paper.pdf b/macros/xetex/latex/bhcexam/examples/example_student_paper.pdf
new file mode 100644
index 0000000000..d6080b26ef
--- /dev/null
+++ b/macros/xetex/latex/bhcexam/examples/example_student_paper.pdf
diff --git a/macros/xetex/latex/bhcexam/examples/example_student_paper.tex b/macros/xetex/latex/bhcexam/examples/example_student_paper.tex
new file mode 100644
index 0000000000..2d398fcd6d
--- /dev/null
+++ b/macros/xetex/latex/bhcexam/examples/example_student_paper.tex
@@ -0,0 +1,551 @@
+\documentclass[list, windows]{BHCexam}
+\pagestyle{fancy}
+\fancyfoot[C]{\kaishu \small 第 \thepage 页 共 \pageref{lastpage} 页}
+\fancyhead[L]{\includegraphics[width=2cm]{qrcode.png}}
+\fancyhead[R]{\raisebox{0.5\height}{\includegraphics[height=1cm]{logo.png}}}
+\begin{document}
+\title{上海某高中 2017-2018 学年度第一学期}
+\subtitle{高一数学期中试卷}
+\notice{满分150分，120分钟完成，允许使用计算器，答案一律写在答题纸上}
+\author{微信关注公众号：橘子数学}
+\date{2017.11}
+\maketitle
+\begin{groups}
+\group{填空}{第1-6题每题4分，第7-12题每题5分.}
+
+\begin{questions}[p]
+\begin{minipage}{\linewidth}
+\question [4] 已知集合$U=\{1,2,3,4\}$,集合$A=\{1,2\}$,$B=\{2,3\}$,则$(A\cap\complement_UB) \cup (\complement_UA\cap B)=$\key{$\{1,3\}$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$A\cap\complement_UB=\{1\}$, $B\cap\complement_UA=\{3\}$.
+
+$(A\cap\complement_UB) \cup (\complement_UA\cap B)=\{1,3\}$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 设集合$M=\{x|0\lt x\leqslant{}3\}$,$N=\{x|0\lt x\leqslant{}2\}$,那么 “ $a\in{M}$ ” 是 “ $a\in{N}$ ” 的\key{必要非充分}条件.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+由$N \subsetneqq M$，可知 “ $a\in{M}$ ” 是 “ $a\in{N}$ ” 的必要非充分条件.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 函数$f(x)=\sqrt{x+1}+\dfrac{1}{2-x}$的定义域为\key{$[-1,2)\cup{}(2,+\infty)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+根据题意:$\left\{\begin{array}{l}{x+1\geqslant{}0}\\{2-x\neq{}0}\end{array}\right.$
+
+解得:$x\geqslant{}-1$且$x\neq{}2$
+
+定义域是:$[-1,2)\cup{}(2,+\infty)$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 已知集合$A=\{x \big| |x-a| \lt 1,x\in\symbf{R}\}$,
+$B=\{x\Big|\dfrac{2x-a}{x+1} \lt 1,x\in\symbf{R}\}$,
+且$A\cap B=\varnothing$,
+ 则实数$a$的取值范围是\key{$a \in (-\infty,-2]$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$A=(a-1,a+1)$，
+
+由$\dfrac{2x-a}{x+1} \lt 1$化简得$\dfrac{x-a-1}{x+1} \lt 0$.
+
+若$a+1 \gt -1$，不满足$A \cap B=\varnothing$.
+
+若$a+1 \le -1$，满足$A \cap B=\varnothing$.
+
+故$a\in(-\infty,-2]$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 已知$y=f(x), y=g(x)$是两个定义在$\symbf{R}$上的二次函数，其$x, y$的取值如下表所示: 
+\[\begin{array}{|c|c|c|c|c|}
+\hline
+x & 1 & 2 & 3 & 4 \\
+\hline
+f(x) & -3 & -4 & -3 & 0\\
+\hline
+g(x) & 0 & 1 & 0 & -3\\
+\hline
+\end{array}\]
+则不等式$f(g(x))\ge0$的解集为\key{$(-\infty,1]\cup[3,+\infty)$.}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+由表格可得$y=f(x)$图像开口向上且关于$x=2$对称，其零点为$x=4$和$x=0$.
+
+故不等式$f(g(x))\ge0$, 即$g(x)\in(-\infty,0]\cup[4,+\infty)$.
+
+又$y=g(x)$的图像开口向下且关于$x=2$对称，其最大值为$1$.
+
+故$g(x)\in(-\infty,0]$，解得$x\in(-\infty,1]\cup[3,+\infty)$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 关于$x$的不等式$2kx^2+kx+\frac{3}{8} \lt 0$的解集不为空集，则$k$的取值范围为\key{$k \in (-\infty,0)\cup(3,+\infty)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+若$k \lt 0 $, 二次函数$y=2kx^2+kx+\frac{3}{8}$图像开口向下，解集恒不为空集. 满足.
+
+若$k=0$, 不等式解集为$\varnothing$，不满足.
+
+若$k \gt 0$，二次函数$y=2kx^2+kx+\frac{3}{8}$图像开口向上，要求$\Delta=k^2-4 \cdot (2k) \cdot \frac{3}{8}=k^2-3k \gt 0$，解得$k\in(3,+\infty)$.
+
+综上所属，$k \in (-\infty,0)\cup(3,+\infty)$
+
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [5] 已知本张试卷的出卷人在公元$x^2$年时年龄为$x-8$岁，则出卷人的出生年份是\key{$1989$}.(假设出生当年的年龄为$1$岁)
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+设出卷人的出生年份是$y$.
+
+则有$x^2-y+1=x-8$.
+
+化简得$y=x^2-x+9$.
+
+由生活常识，
+
+$1952 \lt x^2-x+9 \lt 2000$, $x\in\symbf{N}$.
+
+解得$x=45$, 故$y=1989$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 若对任意$x\in\symbf{R}$,不等式$|x|\geqslant{}ax$恒成立,则实数$a$的取值范围是\key{$a \in [-1,1]$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+若$x \gt 0$，得$x \ge ax$， 解得$a \le 1$.
+
+若$x = 0$，得$0 \ge 0$成立.
+
+若$x \lt 0$，得$-x \ge ax$，解得$a \ge -1$.
+
+综上所述，$a \in [-1,1]$.
+\method
+画出$y=|x|$与$y=ax$的图像, 
+
+要使$y=ax$的图像恒在$y=|x|$下方，
+
+则$a\in[-1,1]$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 设常数$a\gt \ 0$,若$9x+\dfrac{a^2}{x}\geqslant{}a+1$对一切正实数$x$成立,则$a$的取值范围为\key{$\left[\dfrac{1}{5}, + \infty \right)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$\forall x\in{R}^{+},9x+\dfrac{a^2}{x}\geqslant{}2\sqrt{9x\cdot{}\dfrac{a^2}{x}}=6a$,
+
+$\therefore{}6a\geqslant{}a+1$,
+
+$\therefore{}a\geqslant{}\dfrac{1}{5}$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 设函数$f(x)=\begin{cases}x^2+2x+2,x\leqslant{}0\\-x^2,x\gt \ 0\end{cases}$,若$f(f(a))=2$,则$a=$\key{}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+若$a\leqslant{}0$, 
+则$f(a)=a^2+2a+2=(a+1)^2+1 \gt 0$, 
+而$f(f(a))=-(a^2+2a+2)^2=2$无解，舍去;
+
+若$a\gt \ 0$,
+则$f(a)=-a^2\lt 0$,
+故$f(f(a))=(-a^2)^2+2(-a^2)+2=2$,
+解得$a=\sqrt{2}$.
+
+综上所述, $a=\sqrt{2}$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 若二次函数$y=f(x)$对一切$x\in \text{R}$恒有${{x}^{2}}-2x+4\le f(x)\le 2{{x}^{2}}-4x+5$成立，且$f(5)=27$，则$f(11)=$\key{$153$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+画出不等式两边的两个二次函数图像，如图，可得未知的二次函数$y=f(x)$开口向上，以$(1,3)$为顶点，故可设函数解析式为$f(x)=a(x-1)^2+3$，将$f(5)=27$代入可得$a=\frac{3}{2}$，则$f(x)=\frac{3}{2}(x-1)^2+3$，故$f(11)=153$.
+\begin{center}
+\includegraphics[height=8cm]{./To7VxKXpZaOWO5Qq9qzeZGtlwYTwJ5KF.jpg}\\
+第11题
+\vspace{0.5cm}
+\end{center}
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 已知$f(x)=(a^2-5)x^2+2x+2$. 若不等式$f(x) \gt x$的解集为$A$. 已知$(0,1)\subseteq A$，则$a$的取值范围为\key{$a \in (-\infty,-\sqrt{2}]\cup[\sqrt{2},+\infty)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+由题意得，$(a^2-5)x^2+x+2 \gt 0$在$(0,1)$上恒成立.
+
+即$a^2-5 \gt -\frac{2}{x^2}-\frac{1}{x}$在$(0,1)$上恒成立.
+
+设$f(x)=-\frac{2}{x^2}-\frac{1}{x}=-2(\frac{1}{x}+\frac{1}{4})^2+\frac{1}{8}$, 
+
+又$\frac{1}{x}\in(1,+\infty)$, 故$f(x) \lt -3$.
+
+即$a^2-5 \ge -3$, 解得$a \in (-\infty,-\sqrt{2}]\cup[\sqrt{2},+\infty)$.
+
+\end{solution}
+\end{questions}
+
+\group{选择}{第13-16题每题5分.}
+
+\begin{questions}[p]
+\begin{minipage}{\linewidth}
+\question [5] 设$P,Q$为两个非空实数集,定义集合$P+Q=\{a+b|a\in{P},b\in{Q}\}$.若$P=\{0,2,5\}$,$Q=\{1,2,6\}$,则$P+Q$中元素的个数是\key{B}.
+\fourchoices{$9$}{$8$}{$7$}{$6$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$\because P=\{0,2,5\}$,$Q=\{1,2,6\}$,
+
+$\therefore P+Q=\{1,2,3,4,6,7,8,11\}$.
+
+故选B
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 不等式$(1+x)(1-|x|)\gt \ 0$的解集是\key{D}.
+\fourchoices{$\{x|0\leqslant{}x\lt 1\}$}{$\{x|x\lt 0$且$x\neq{}-1\}$}{$\{x|-1\lt x\lt 1\}$}{$\{x|x\lt 1$且$x\neq{}-1\}$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+求不等式$(1+x)(1-|x|)\gt \ 0$的解集,则分两种情况讨论:
+
+情况$1:\left\{\begin{array}{l}{1+x\gt \ 0\;\;}\\{1-|x|\gt \ 0}\end{array}\right.$
+
+即$\left\{\begin{array}{l}{x\gt \ -1}\\{-1\lt x\lt 1}\end{array}\right.$
+则$-1\lt x\lt 1$.
+
+情况$2:\left\{\begin{array}{l}{1+x\lt 0}\\{1-|x|\lt 0}\end{array}\right.$
+
+即$\left\{\begin{array}{l}{x\lt -1}\\{x\gt 1\text{或}x\lt -1}\end{array}\right.$
+则$x\lt -1$.
+
+两种情况取并集得$\{x|x\lt 1$且$x\neq{}-1\}$.
+
+故选D.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 已知三个不等式:$ab\gt \ 0$,$bc-ad\gt \ 0$,$\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0$(其中$a,b,c,d$均为实数),用其中两个不等式作为条件,余下的一个不等式作为结论组成一个命题,可组成的正确命题的个数是\key{D}.
+\fourchoices{$0$}{$1$}{$2$}{$3$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$\begin{cases}ab\gt \ 0, &\cdots(1)\\
+bc-ad\gt \ 0,&\cdots(2)
+\end{cases}$
+
+$(2)\div(1)$得$\dfrac{bc-ad}{ab}=\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0$.
+
+$\begin{cases}
+ab\gt \ 0,&\cdots(1)\\
+\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0,&\cdots(3)
+\end{cases}$
+
+$(1)\times{}(3)$得$ab(\dfrac{c}{a}-\dfrac{d}{b})=bc-ad\gt \ 0$.
+
+$\begin{cases}bc-ad\gt \ 0,&\cdots(2)\\
+\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0,&\cdots(3)
+\end{cases}$
+
+用反证法,显然$ab\neq{}0$,
+
+设$ab\lt 0$,$(2)$式同除$ab$得$\dfrac{c}{a}-\dfrac{d}{b}\lt 0$,矛盾,
+
+故$ab\gt \ 0$.
+故选D.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 设$a\gt \ 0$,$b\gt \ 0$,则以下不等式中不恒成立的是\key{B}.
+\fourchoices{$(a+b)(\dfrac{1}{a}+\dfrac{1}{b})\geqslant{}4$}{$a^{3}+b^{3}\geqslant{}2ab^{2}$}{$a^{2}+b^{2}+2\geqslant{}2a+2b$}{$\sqrt{|{a-b}|}\geqslant{}\sqrt{a}-\sqrt{b}$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$a\gt \ 0$,$b\gt \ 0$,
+
+$(a+b)(\dfrac{1}{a}+\dfrac{1}{b})\geqslant{}2\sqrt{ab}\cdot{}2\sqrt{\dfrac{1}{ab}}\geqslant{}4$,
+
+故A恒成立;
+
+$a^{3}+b^{3}\geqslant{}2ab^{2}$,
+
+取$a=\dfrac{1}{2}$,$b=\dfrac{2}{3}$,则B不成立;
+
+$a^{2}+b^{2}+2-(2a+2b)=(a-1)^{2}+(b-1)^{2}\geqslant{}0$,
+
+故C恒成立;
+
+若$a\lt b$,则$\sqrt{|{a-b}|}\geqslant{}\sqrt{a}-\sqrt{b}$恒成立,
+
+若$a\geqslant{}b$,则${(\sqrt{|{a-b}|})^2}-{(\sqrt{a}-\sqrt{b})^2}=2\sqrt{ab}\geqslant{}0$,
+
+故$\sqrt{|{a-b}|}\geqslant{}\sqrt{a}-\sqrt{b}$,
+即D恒成立.
+
+\end{solution}
+\end{questions}
+\group{简答}{第17-19题每题14分，第20题16分，第21题18分.}
+
+\begin{questions}[stp]
+\begin{minipage}{\linewidth}
+\question [14] 已知$\triangle ABC$为直角三角形, 记其两条直角边长分别为$a,b\in\symbf{R}^+$, 记面积为$S$, 周长为$C$. 若三角形面积为定值，其周长是否有最值，最大值还是最小值，何时取到，为多少（结果用$S$表示）？
+
+\end{minipage}
+\begin{solution}{6cm}
+\method
+$C=a+b+\sqrt{a^2+b^2}$,
+\score{2}{2}由$\sqrt{a^2+b^2} \ge \sqrt{2ab}$，$a+b \ge 2\sqrt{ab}$,
+
+得$C \ge  2\sqrt{ab}+\sqrt{2ab}$
+\score{6}{8}由$S=\frac{1}{2}ab$，
+
+得$C \ge (2+2\sqrt{2})\sqrt{S}$，
+\score{4}{12}当且仅当$a=b=\sqrt{2S}$时取到最小值.
+\score{2}{14}
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [14] 已知$a\in\textbf{R}$,若关于$x$的方程$x^2+x+|a-\dfrac{1}{4}|+|a|=0$有实根，求$a$的取值范围.
+
+\end{minipage}
+\begin{solution}{6cm}
+\method
+方程即$|a-\frac{1}{4}|+|a|=-x^2-x\in[0,\frac{1}{4}]$,
+
+即解不等式$|a-\frac{1}{4}|+|a| \le \frac{1}{4}$.
+
+当$a\le 0$，化简得$\frac{1}{4}-2a \le \frac{1}{4}$，解得$a=0$;
+
+当$0 \lt a\le \frac{1}{4}$, 化简得$\frac{1}{4} \le \frac{1}{4}$，解得$a\in(0,\frac{1}{4}]$;
+
+当$a \gt \frac{1}{4}$, 化简得$2a-\frac{1}{4}\le\frac{1}{4}$，无解.
+
+综上所述，$a\in [0,\frac{1}{4}]$
+\score{14}{14}
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [14] 先阅读下列不等式的证法，再解决后面的问题:
+\[
+\text{证明:} (a_1b_1+a_2b_2)^2 \le  (a_1^2+a_2^2)(b_1^2+b_2^2)\]
+
+证: 令$A=\sqrt{a_1^2+a_2^2}$, $B=\sqrt{b_1^2+b_2^2}$
+
+\[
+\begin{array}{rl}
+\frac{a_1b_1+a_2b_2}{\sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}}& =\frac{a_1b_1}{AB}+\frac{a_2b_2}{AB}\\
+& = \frac{a_1}{A}\cdot\frac{b_1}{B}+\frac{a_2}{A}\cdot\frac{b_2}{B} \\
+& \le \frac{1}{2}(\frac{a_1^2}{A^2}+\frac{b_1^2}{B^2})+ \frac{1}{2}(\frac{a_2^2}{A^2}+\frac{b_2^2}{B^2})\\
+& =\frac{1}{2}(\frac{a_1^2+a_2^2}{A^2}+\frac{b_1^2+b_2^2}{B^2})\\
+& =1
+\end{array}
+\]
+故$ (a_1b_1+a_2b_2)^2 \le  (a_1^2+a_2^2)(b_1^2+b_2^2)$.
+\begin{subquestions}
+    \subquestion 若$x_1,y_1,x_2,y_2\in\symbf{R^+}$，利用上述结论，证明:\[
+(x_1+x_2)(y_1+y_2) \ge (\sqrt{x_1y_1}+\sqrt{x_2y_2})^2
+\]
+    \subquestion 若$x_1,y_1,x_2,y_2,z_1,z_2\in\symbf{R^+}$，模仿上述证法并结合(1)的证法，证明:\[
+(x_1+x_2)(y_1+y_2)(z_1+z_2) \ge (\sqrt[3]{x_1y_1z_1}+\sqrt[3]{x_2y_2z_2})^3
+\]
+(提示：若$a,b,c\in\symbf{R^+}$，有$\frac{a^3+b^3+c^3}{3}\ge abc$,)
+\end{subquestions}
+\end{minipage}
+\begin{solution}{3cm}
+\method
+(1)设$a_1=\sqrt{x_1}$，$a_2=\sqrt{x_2}$，$b_1=\sqrt{y_1}$，$b_2=\sqrt{y_2}$.
+
+由\[ (a_1b_1+a_2b_2)^2 \le  (a_1^2+a_2^2)(b_1^2+b_2^2)\]
+
+得\[ (\sqrt{x_1}\sqrt{y_1}+\sqrt{x_2}\sqrt{y_2})^2 \le  [((\sqrt{x_1})^2+(\sqrt{x_2})^2)((\sqrt{y_1})^2+(\sqrt{y_2})^2)]\]
+
+即\[
+(x_1+x_2)(y_1+y_2) \ge (\sqrt{x_1y_1}+\sqrt{x_2y_2})^2
+\]
+\score{6}{6}(2) 设$a_1=\sqrt[3]{x_1}, b_1=\sqrt[3]{y_1}, c_1=\sqrt[3]{z_1}$, $a_2=\sqrt[3]{x_2}, b_2=\sqrt[3]{y_2}, c_2=\sqrt[3]{z_2}$.
+
+要证
+\[
+(x_1+x_2)(y_1+y_2)(z_1+z_2) \ge (\sqrt[3]{x_1y_1z_1}+\sqrt[3]{x_2y_2z_2})^3
+\]
+
+即证
+\[
+(a_1b_1c_1+a_2b_2c_2)^3 \le  (a_1^3+a_2^3)(b_1^3+b_2^3)(c_1^3+c_2^3)
+\]
+
+(不换元，直接证不扣分)
+\score{2}{8}令$A=\sqrt[3]{a_1^3+a_2^3}$，$B=\sqrt[3]{b_1^3+b_2^3}$, $C=\sqrt[3]{c_1^3+c_2^3}$.
+
+\[
+\begin{array}{rl}
+& \frac{a_1b_1c_1+a_2b_2c_2}{\sqrt[3]{a_1^3+a_2^3}\sqrt[3]{b_1^3+b_2^3}\sqrt[3]{c_1^3+c_2^3}}\\
+= & \frac{a_1b_1c_1}{ABC}+\frac{a_2b_2c_2}{ABC}\\
+= & \frac{a_1}{A}\cdot\frac{b_1}{B}\cdot\frac{c_1}{C}+\frac{a_2}{A}\cdot\frac{b_2}{B}\cdot\frac{c_2}{C}\\
+\le & \frac{1}{3}(\frac{a_1^3}{A^3}+\frac{b_1^3}{B^3}+\frac{c_1^3}{C^3})+ \frac{1}{3}(\frac{a_2^3}{A^3}+\frac{b_2^3}{B^3}+\frac{c_2^3}{C^3})\\
+= & \frac{1}{3}(\frac{a_1^3+a_2^3}{A^3}+\frac{b_1^3+b_2^3}{B^3}++\frac{c_1^3+c_2^3}{C^3})\\
+= & 1
+\end{array}
+\]
+
+故\[(a_1b_1c_1+a_2b_2c_2)^3 \le  (a_1^3+a_2^3)(b_1^3+b_2^3)(c_1^3+c_2^3)\]
+
+得证.
+\score{6}{14}
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [16] 公元2222年，有一种高危传染疾病在全球范围内蔓延，被感染者的潜伏期可以长达10年，期间会有约$0.05\%$的概率传染给他人，一旦发病三天内即死亡。某城市总人口约200万人，专家分析其中约有$1000$名感染者，为了防止疾病继续扩散，疾病预防控制中心现决定对全市人口进行血液检测以筛选出被感染者。由于检测试剂十分昂贵且数量有限，需要将血样混合后一起检测以节约试剂。已知感染者的检测结果为阳性, 未被感染者则为阴性. 阳性血样与阴性血样混合后的检测结果为阳性, 同为阴性或阳性的血样混合后结果不发生改变.
+\begin{subquestions}
+    \subquestion 若对全市人口进行平均分组，同一分组的血样将被混合到一起检测，若发现结果为阳性，则再在该分组内逐个检测排查. 设每个组$x$个人，那么最坏情况下，需要进行约多少次检测可以找到所有的被感染者? 在当前方案下，若要使检测的次数尽可能少，每个分组的最优人数是？
+    \subquestion 在(1)的检测方案中，对于检测结果为阳性的组采取逐一检测排查的方法并不是很好，或可将这些组的血样再进行一次分组混合血样检测，然后再进行逐一排查。仍然考虑最坏的情况，请问两次要如何分组，使检测总次数尽可能少？
+    \subquestion 在(2)的检测方案中，进行了两次分组混合血样检测。仍然考虑最坏情况，若再进行若干次分组混合血样检测，是否会使检测次数更少？请给出最优的检测方案.
+\end{subquestions}
+\end{minipage}
+\begin{solution}{6cm}
+\method
+设需要进行$y$次检测. 由每个组$x$个人，则有$\frac{2000000}{x}$个小组, 最坏情况是1000人被分配到了不同的组. 
+
+\[
+y=\begin{cases}
+\frac{2\cdot 10^6}{x}+2\cdot 10^6, & x \ge 2000 \\
+\frac{2\cdot 10^6}{x}+1000\cdot x, & x \lt 2000 \\
+\end{cases}
+\]
+
+(题目要求近似计算，可以不考虑整除，考虑整除不扣分)
+\score{2}{2}分组数小于1000的情况比逐个检测还要多，可以不用考虑，
+
+$y=\frac{2\times 10^6}{x}+1000\cdot x \ge 2\sqrt{2\cdot 10^9}$.
+
+当$x=\sqrt{2\times10^3}\approx45$人，检测次数最少.
+
+(结果为44-45都给分)
+\score{2}{4}(2) 设第二次分组每个组$w$人.
+
+\[
+\begin{array}{rl}
+y & = \frac{2\times10^6}{x}+\frac{1000x}{w}+1000w\\
+& \ge 3\sqrt[3]{2\times10^{12}}\\
+\end{array}
+\]
+
+当$\frac{2\times10^6}{x}=\frac{1000x}{w}=1000w$时，
+即$x=2000^\frac{2}{3}\approx159$,$w=2000^\frac{1}{3}\approx13$，
+检测次数最少.
+
+所以第一次分组159人一组，第二次分组13人一组.
+
+(也可以用两次基本不等式求得，允许$10\%$的误差.)
+\score{6}{10}(3) 若进行$k$次分组，设第$i$次分组时每组人数为$x_i$. 则总的检测次数为:
+\[
+\begin{array}{rl}
+y=&\dfrac{2\times10^6}{x_1}+\dfrac{1000x_1}{x_2}+\cdots\\
+&+\dfrac{1000x_{k-1}}{x_k}+1000x_k\\
+\ge & (k+1)\sqrt[k+1]{2\times10^{3k+2}}
+\end{array}
+\]
+
+利用计算器可得，
+
+1次分组， $y \ge 2\sqrt{2\cdot 10^9}\approx8.94\times10^4$;
+
+2次分组， $y \ge 3\sqrt[3]{2\cdot 10^{12}}\approx3.78\times10^4$;
+
+3次分组， $y \ge 4\sqrt[4]{2\cdot 10^{15}}\approx2.68\times10^4$;
+
+4次分组， $y \ge 5\sqrt[5]{2\cdot 10^{18}}\approx2.29\times10^4$;
+
+5次分组， $y \ge 6\sqrt[6]{2\cdot 10^{21}}\approx2.13\times10^4$;
+
+6次分组， $y \ge 7\sqrt[7]{2\cdot 10^{24}}\approx2.073\times10^4$;
+
+7次分组， $y \ge 8\sqrt[8]{2\cdot 10^{27}}\approx2.068\times10^4$;
+
+8次分组， $y \ge 9\sqrt[9]{2\cdot 10^{30}}\approx2.094\times10^4$;
+
+9次分组， $y \ge 10\sqrt[10]{2\cdot 10^{33}}\approx2.138\times10^4$;
+
+可见进行7次分组混合血样检测最优.
+
+其中第k次分组时$(2000)^{\frac{8-k}{8}}$人一组.
+\score{6}{16}
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [18] 已知函数$f(x)=a{{x}^{2}}-\frac{1}{2}x+c$（$a$、$c\in R$），满足$f(1)=0$，且$f(x)\ge 0$在$x\in R$时恒成立．
+\begin{subquestions}
+    \subquestion 求$a$、$c$的值；
+    \subquestion 若$h(x)=\frac{3}{4}{{x}^{2}}-bx+\frac{b}{2}-\frac{1}{4}$，解不等式$f(x)+h(x) \lt 0$；
+    \subquestion 是否存在实数$m$，使函数$g(x)=f(x)-mx$在区间$[m,m+2]$上有最小值$-5$？若存在，请求出$m$的值；若不存在，请说明理由．
+\end{subquestions}
+\end{minipage}
+\begin{solution}{6cm}
+\method
+由$f(1)=0$，得$a+c=\frac{1}{2}$,
+\score{1}{1}因为$f(x)\ge 0$在$x\in R$时恒成立，所以$a \gt 0$且△$=\frac{1}{4}-4ac\le 0$，$ac\ge \frac{1}{16}$，
+\score{1}{2}即$a\left( \frac{1}{2}-a \right)\ge \frac{1}{16}$，${{a}^{2}}-\frac{1}{2}a+\frac{1}{16}\le 0$，${{\left( a-\frac{1}{4} \right)}^{2}}\le 0$，所以$a=c=\frac{1}{4}$．
+\score{2}{4}(2) 由（1）得$f(x)=\frac{1}{4}{{x}^{2}}-\frac{1}{2}x+\frac{1}{4}$，由$f(x)+h(x) \lt 0$，得
+${{x}^{2}}-\left( b+\frac{1}{2} \right)x+\frac{b}{2} \lt 0$，即$(x-b)\left( x-\frac{1}{2} \right) \lt 0$，
+\score{3}{7}所以，当$b \lt \frac{1}{2}$时，原不等式解集为$(b,\frac{1}{2})$；
+当$b \gt \frac{1}{2}$时，原不等式解集为$(\frac{1}{2},b)$；
+当$b=\frac{1}{2}$时，原不等式解集为空集 ．
+\score{3}{10}(3) $g(x)=\frac{1}{4}{{x}^{2}}-\left( \frac{1}{2}+m \right)x+\frac{1}{4}$，
+\score{1}{11}$g(x)$的图像是开口向上的抛物线，对称轴为直线$x=2m+1$．
+假设存在实数$m$，使函数$g(x)$在区间$[m,m+2]$上有最小值$-5$．
+\ding{192}	当$2m+1 \lt m$，即$m \lt -1$时，函数$g(x)$在区间$[m,m+2]$上是增函数，所以$g(m)=-5$，即$\frac{1}{4}{{m}^{2}}-\left( \frac{1}{2}+m \right)m+\frac{1}{4}=-5$，解得$m=-3$或$m=\frac{7}{3}$，
+因为$m \lt -1$，所以$m=-3$；
+\score{2}{13}\ding{193}当$m\le 2m+1\le m+2$，即$-1\le m\le 1$时，函数$g(x)$的最小值为$g(2m+1)=-5$，即
+$\frac{1}{4}{{(2m+1)}^{2}}-\left( \frac{1}{2}+m \right)(2m+1)+\frac{1}{4}=-5$，解得$m=-\frac{1}{2}-\frac{\sqrt{21}}{2}$或$m=-\frac{1}{2}+\frac{\sqrt{21}}{2}$，均舍去；
+\score{2}{15}\ding{194}当$2m+1 \gt m+2$，即$m \gt 1$时，$g(x)$在区间$[m,m+2]$上是减函数，所以$g(m+2)=-5$，即$\frac{1}{4}{{(m+2)}^{2}}-\left( \frac{1}{2}+m \right)(m+2)+\frac{1}{4}=-5$，解得$m=-1-2\sqrt{2}$或$m=-1+2\sqrt{2}$，因$m \gt 1$，所以$m=-1+2\sqrt{2}$．
+\score{2}{17}综上，存在实数$m$，$m=-3$或$m=-1+2\sqrt{2}$时，函数$g(x)$在区间$[m,m+2]$上有最小值$-5$．
+\score{1}{18}
+\end{solution}
+\end{questions}
+\end{groups}
+\label{lastpage}
+\end{document}
diff --git a/macros/xetex/latex/bhcexam/examples/example_teacher_paper.pdf b/macros/xetex/latex/bhcexam/examples/example_teacher_paper.pdf
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+++ b/macros/xetex/latex/bhcexam/examples/example_teacher_paper.pdf
diff --git a/macros/xetex/latex/bhcexam/examples/example_teacher_paper.tex b/macros/xetex/latex/bhcexam/examples/example_teacher_paper.tex
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--- /dev/null
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@@ -0,0 +1,551 @@
+\documentclass[list, windows, answers]{BHCexam}
+\pagestyle{fancy}
+\fancyfoot[C]{\kaishu \small 第 \thepage 页 共 \pageref{lastpage} 页}
+\fancyhead[L]{\includegraphics[width=2cm]{qrcode.png}}
+\fancyhead[R]{\raisebox{0.5\height}{\includegraphics[height=1cm]{logo.png}}}
+\begin{document}
+\title{上海某高中 2017-2018 学年度第一学期}
+\subtitle{高一数学期中试卷}
+\notice{满分150分，120分钟完成，允许使用计算器，答案一律写在答题纸上}
+\author{微信关注公众号：橘子数学}
+\date{2017.11}
+\maketitle
+\begin{groups}
+\group{填空}{第1-6题每题4分，第7-12题每题5分.}
+
+\begin{questions}[p]
+\begin{minipage}{\linewidth}
+\question [4] 已知集合$U=\{1,2,3,4\}$,集合$A=\{1,2\}$,$B=\{2,3\}$,则$(A\cap\complement_UB) \cup (\complement_UA\cap B)=$\key{$\{1,3\}$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$A\cap\complement_UB=\{1\}$, $B\cap\complement_UA=\{3\}$.
+
+$(A\cap\complement_UB) \cup (\complement_UA\cap B)=\{1,3\}$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 设集合$M=\{x|0\lt x\leqslant{}3\}$,$N=\{x|0\lt x\leqslant{}2\}$,那么 “ $a\in{M}$ ” 是 “ $a\in{N}$ ” 的\key{必要非充分}条件.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+由$N \subsetneqq M$，可知 “ $a\in{M}$ ” 是 “ $a\in{N}$ ” 的必要非充分条件.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 函数$f(x)=\sqrt{x+1}+\dfrac{1}{2-x}$的定义域为\key{$[-1,2)\cup{}(2,+\infty)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+根据题意:$\left\{\begin{array}{l}{x+1\geqslant{}0}\\{2-x\neq{}0}\end{array}\right.$
+
+解得:$x\geqslant{}-1$且$x\neq{}2$
+
+定义域是:$[-1,2)\cup{}(2,+\infty)$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 已知集合$A=\{x \big| |x-a| \lt 1,x\in\symbf{R}\}$,
+$B=\{x\Big|\dfrac{2x-a}{x+1} \lt 1,x\in\symbf{R}\}$,
+且$A\cap B=\varnothing$,
+ 则实数$a$的取值范围是\key{$a \in (-\infty,-2]$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$A=(a-1,a+1)$，
+
+由$\dfrac{2x-a}{x+1} \lt 1$化简得$\dfrac{x-a-1}{x+1} \lt 0$.
+
+若$a+1 \gt -1$，不满足$A \cap B=\varnothing$.
+
+若$a+1 \le -1$，满足$A \cap B=\varnothing$.
+
+故$a\in(-\infty,-2]$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 已知$y=f(x), y=g(x)$是两个定义在$\symbf{R}$上的二次函数，其$x, y$的取值如下表所示: 
+\[\begin{array}{|c|c|c|c|c|}
+\hline
+x & 1 & 2 & 3 & 4 \\
+\hline
+f(x) & -3 & -4 & -3 & 0\\
+\hline
+g(x) & 0 & 1 & 0 & -3\\
+\hline
+\end{array}\]
+则不等式$f(g(x))\ge0$的解集为\key{$(-\infty,1]\cup[3,+\infty)$.}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+由表格可得$y=f(x)$图像开口向上且关于$x=2$对称，其零点为$x=4$和$x=0$.
+
+故不等式$f(g(x))\ge0$, 即$g(x)\in(-\infty,0]\cup[4,+\infty)$.
+
+又$y=g(x)$的图像开口向下且关于$x=2$对称，其最大值为$1$.
+
+故$g(x)\in(-\infty,0]$，解得$x\in(-\infty,1]\cup[3,+\infty)$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 关于$x$的不等式$2kx^2+kx+\frac{3}{8} \lt 0$的解集不为空集，则$k$的取值范围为\key{$k \in (-\infty,0)\cup(3,+\infty)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+若$k \lt 0 $, 二次函数$y=2kx^2+kx+\frac{3}{8}$图像开口向下，解集恒不为空集. 满足.
+
+若$k=0$, 不等式解集为$\varnothing$，不满足.
+
+若$k \gt 0$，二次函数$y=2kx^2+kx+\frac{3}{8}$图像开口向上，要求$\Delta=k^2-4 \cdot (2k) \cdot \frac{3}{8}=k^2-3k \gt 0$，解得$k\in(3,+\infty)$.
+
+综上所属，$k \in (-\infty,0)\cup(3,+\infty)$
+
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [5] 已知本张试卷的出卷人在公元$x^2$年时年龄为$x-8$岁，则出卷人的出生年份是\key{$1989$}.(假设出生当年的年龄为$1$岁)
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+设出卷人的出生年份是$y$.
+
+则有$x^2-y+1=x-8$.
+
+化简得$y=x^2-x+9$.
+
+由生活常识，
+
+$1952 \lt x^2-x+9 \lt 2000$, $x\in\symbf{N}$.
+
+解得$x=45$, 故$y=1989$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 若对任意$x\in\symbf{R}$,不等式$|x|\geqslant{}ax$恒成立,则实数$a$的取值范围是\key{$a \in [-1,1]$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+若$x \gt 0$，得$x \ge ax$， 解得$a \le 1$.
+
+若$x = 0$，得$0 \ge 0$成立.
+
+若$x \lt 0$，得$-x \ge ax$，解得$a \ge -1$.
+
+综上所述，$a \in [-1,1]$.
+\method
+画出$y=|x|$与$y=ax$的图像, 
+
+要使$y=ax$的图像恒在$y=|x|$下方，
+
+则$a\in[-1,1]$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 设常数$a\gt \ 0$,若$9x+\dfrac{a^2}{x}\geqslant{}a+1$对一切正实数$x$成立,则$a$的取值范围为\key{$\left[\dfrac{1}{5}, + \infty \right)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$\forall x\in{R}^{+},9x+\dfrac{a^2}{x}\geqslant{}2\sqrt{9x\cdot{}\dfrac{a^2}{x}}=6a$,
+
+$\therefore{}6a\geqslant{}a+1$,
+
+$\therefore{}a\geqslant{}\dfrac{1}{5}$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 设函数$f(x)=\begin{cases}x^2+2x+2,x\leqslant{}0\\-x^2,x\gt \ 0\end{cases}$,若$f(f(a))=2$,则$a=$\key{}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+若$a\leqslant{}0$, 
+则$f(a)=a^2+2a+2=(a+1)^2+1 \gt 0$, 
+而$f(f(a))=-(a^2+2a+2)^2=2$无解，舍去;
+
+若$a\gt \ 0$,
+则$f(a)=-a^2\lt 0$,
+故$f(f(a))=(-a^2)^2+2(-a^2)+2=2$,
+解得$a=\sqrt{2}$.
+
+综上所述, $a=\sqrt{2}$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 若二次函数$y=f(x)$对一切$x\in \text{R}$恒有${{x}^{2}}-2x+4\le f(x)\le 2{{x}^{2}}-4x+5$成立，且$f(5)=27$，则$f(11)=$\key{$153$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+画出不等式两边的两个二次函数图像，如图，可得未知的二次函数$y=f(x)$开口向上，以$(1,3)$为顶点，故可设函数解析式为$f(x)=a(x-1)^2+3$，将$f(5)=27$代入可得$a=\frac{3}{2}$，则$f(x)=\frac{3}{2}(x-1)^2+3$，故$f(11)=153$.
+\begin{center}
+\includegraphics[height=8cm]{./To7VxKXpZaOWO5Qq9qzeZGtlwYTwJ5KF.jpg}\\
+第11题
+\vspace{0.5cm}
+\end{center}
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 已知$f(x)=(a^2-5)x^2+2x+2$. 若不等式$f(x) \gt x$的解集为$A$. 已知$(0,1)\subseteq A$，则$a$的取值范围为\key{$a \in (-\infty,-\sqrt{2}]\cup[\sqrt{2},+\infty)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+由题意得，$(a^2-5)x^2+x+2 \gt 0$在$(0,1)$上恒成立.
+
+即$a^2-5 \gt -\frac{2}{x^2}-\frac{1}{x}$在$(0,1)$上恒成立.
+
+设$f(x)=-\frac{2}{x^2}-\frac{1}{x}=-2(\frac{1}{x}+\frac{1}{4})^2+\frac{1}{8}$, 
+
+又$\frac{1}{x}\in(1,+\infty)$, 故$f(x) \lt -3$.
+
+即$a^2-5 \ge -3$, 解得$a \in (-\infty,-\sqrt{2}]\cup[\sqrt{2},+\infty)$.
+
+\end{solution}
+\end{questions}
+
+\group{选择}{第13-16题每题5分.}
+
+\begin{questions}[p]
+\begin{minipage}{\linewidth}
+\question [5] 设$P,Q$为两个非空实数集,定义集合$P+Q=\{a+b|a\in{P},b\in{Q}\}$.若$P=\{0,2,5\}$,$Q=\{1,2,6\}$,则$P+Q$中元素的个数是\key{B}.
+\fourchoices{$9$}{$8$}{$7$}{$6$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$\because P=\{0,2,5\}$,$Q=\{1,2,6\}$,
+
+$\therefore P+Q=\{1,2,3,4,6,7,8,11\}$.
+
+故选B
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 不等式$(1+x)(1-|x|)\gt \ 0$的解集是\key{D}.
+\fourchoices{$\{x|0\leqslant{}x\lt 1\}$}{$\{x|x\lt 0$且$x\neq{}-1\}$}{$\{x|-1\lt x\lt 1\}$}{$\{x|x\lt 1$且$x\neq{}-1\}$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+求不等式$(1+x)(1-|x|)\gt \ 0$的解集,则分两种情况讨论:
+
+情况$1:\left\{\begin{array}{l}{1+x\gt \ 0\;\;}\\{1-|x|\gt \ 0}\end{array}\right.$
+
+即$\left\{\begin{array}{l}{x\gt \ -1}\\{-1\lt x\lt 1}\end{array}\right.$
+则$-1\lt x\lt 1$.
+
+情况$2:\left\{\begin{array}{l}{1+x\lt 0}\\{1-|x|\lt 0}\end{array}\right.$
+
+即$\left\{\begin{array}{l}{x\lt -1}\\{x\gt 1\text{或}x\lt -1}\end{array}\right.$
+则$x\lt -1$.
+
+两种情况取并集得$\{x|x\lt 1$且$x\neq{}-1\}$.
+
+故选D.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 已知三个不等式:$ab\gt \ 0$,$bc-ad\gt \ 0$,$\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0$(其中$a,b,c,d$均为实数),用其中两个不等式作为条件,余下的一个不等式作为结论组成一个命题,可组成的正确命题的个数是\key{D}.
+\fourchoices{$0$}{$1$}{$2$}{$3$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$\begin{cases}ab\gt \ 0, &\cdots(1)\\
+bc-ad\gt \ 0,&\cdots(2)
+\end{cases}$
+
+$(2)\div(1)$得$\dfrac{bc-ad}{ab}=\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0$.
+
+$\begin{cases}
+ab\gt \ 0,&\cdots(1)\\
+\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0,&\cdots(3)
+\end{cases}$
+
+$(1)\times{}(3)$得$ab(\dfrac{c}{a}-\dfrac{d}{b})=bc-ad\gt \ 0$.
+
+$\begin{cases}bc-ad\gt \ 0,&\cdots(2)\\
+\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0,&\cdots(3)
+\end{cases}$
+
+用反证法,显然$ab\neq{}0$,
+
+设$ab\lt 0$,$(2)$式同除$ab$得$\dfrac{c}{a}-\dfrac{d}{b}\lt 0$,矛盾,
+
+故$ab\gt \ 0$.
+故选D.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 设$a\gt \ 0$,$b\gt \ 0$,则以下不等式中不恒成立的是\key{B}.
+\fourchoices{$(a+b)(\dfrac{1}{a}+\dfrac{1}{b})\geqslant{}4$}{$a^{3}+b^{3}\geqslant{}2ab^{2}$}{$a^{2}+b^{2}+2\geqslant{}2a+2b$}{$\sqrt{|{a-b}|}\geqslant{}\sqrt{a}-\sqrt{b}$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$a\gt \ 0$,$b\gt \ 0$,
+
+$(a+b)(\dfrac{1}{a}+\dfrac{1}{b})\geqslant{}2\sqrt{ab}\cdot{}2\sqrt{\dfrac{1}{ab}}\geqslant{}4$,
+
+故A恒成立;
+
+$a^{3}+b^{3}\geqslant{}2ab^{2}$,
+
+取$a=\dfrac{1}{2}$,$b=\dfrac{2}{3}$,则B不成立;
+
+$a^{2}+b^{2}+2-(2a+2b)=(a-1)^{2}+(b-1)^{2}\geqslant{}0$,
+
+故C恒成立;
+
+若$a\lt b$,则$\sqrt{|{a-b}|}\geqslant{}\sqrt{a}-\sqrt{b}$恒成立,
+
+若$a\geqslant{}b$,则${(\sqrt{|{a-b}|})^2}-{(\sqrt{a}-\sqrt{b})^2}=2\sqrt{ab}\geqslant{}0$,
+
+故$\sqrt{|{a-b}|}\geqslant{}\sqrt{a}-\sqrt{b}$,
+即D恒成立.
+
+\end{solution}
+\end{questions}
+\group{简答}{第17-19题每题14分，第20题16分，第21题18分.}
+
+\begin{questions}[stp]
+\begin{minipage}{\linewidth}
+\question [14] 已知$\triangle ABC$为直角三角形, 记其两条直角边长分别为$a,b\in\symbf{R}^+$, 记面积为$S$, 周长为$C$. 若三角形面积为定值，其周长是否有最值，最大值还是最小值，何时取到，为多少（结果用$S$表示）？
+
+\end{minipage}
+\begin{solution}{6cm}
+\method
+$C=a+b+\sqrt{a^2+b^2}$,
+\score{2}{2}由$\sqrt{a^2+b^2} \ge \sqrt{2ab}$，$a+b \ge 2\sqrt{ab}$,
+
+得$C \ge  2\sqrt{ab}+\sqrt{2ab}$
+\score{6}{8}由$S=\frac{1}{2}ab$，
+
+得$C \ge (2+2\sqrt{2})\sqrt{S}$，
+\score{4}{12}当且仅当$a=b=\sqrt{2S}$时取到最小值.
+\score{2}{14}
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [14] 已知$a\in\textbf{R}$,若关于$x$的方程$x^2+x+|a-\dfrac{1}{4}|+|a|=0$有实根，求$a$的取值范围.
+
+\end{minipage}
+\begin{solution}{6cm}
+\method
+方程即$|a-\frac{1}{4}|+|a|=-x^2-x\in[0,\frac{1}{4}]$,
+
+即解不等式$|a-\frac{1}{4}|+|a| \le \frac{1}{4}$.
+
+当$a\le 0$，化简得$\frac{1}{4}-2a \le \frac{1}{4}$，解得$a=0$;
+
+当$0 \lt a\le \frac{1}{4}$, 化简得$\frac{1}{4} \le \frac{1}{4}$，解得$a\in(0,\frac{1}{4}]$;
+
+当$a \gt \frac{1}{4}$, 化简得$2a-\frac{1}{4}\le\frac{1}{4}$，无解.
+
+综上所述，$a\in [0,\frac{1}{4}]$
+\score{14}{14}
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [14] 先阅读下列不等式的证法，再解决后面的问题:
+\[
+\text{证明:} (a_1b_1+a_2b_2)^2 \le  (a_1^2+a_2^2)(b_1^2+b_2^2)\]
+
+证: 令$A=\sqrt{a_1^2+a_2^2}$, $B=\sqrt{b_1^2+b_2^2}$
+
+\[
+\begin{array}{rl}
+\frac{a_1b_1+a_2b_2}{\sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}}& =\frac{a_1b_1}{AB}+\frac{a_2b_2}{AB}\\
+& = \frac{a_1}{A}\cdot\frac{b_1}{B}+\frac{a_2}{A}\cdot\frac{b_2}{B} \\
+& \le \frac{1}{2}(\frac{a_1^2}{A^2}+\frac{b_1^2}{B^2})+ \frac{1}{2}(\frac{a_2^2}{A^2}+\frac{b_2^2}{B^2})\\
+& =\frac{1}{2}(\frac{a_1^2+a_2^2}{A^2}+\frac{b_1^2+b_2^2}{B^2})\\
+& =1
+\end{array}
+\]
+故$ (a_1b_1+a_2b_2)^2 \le  (a_1^2+a_2^2)(b_1^2+b_2^2)$.
+\begin{subquestions}
+    \subquestion 若$x_1,y_1,x_2,y_2\in\symbf{R^+}$，利用上述结论，证明:\[
+(x_1+x_2)(y_1+y_2) \ge (\sqrt{x_1y_1}+\sqrt{x_2y_2})^2
+\]
+    \subquestion 若$x_1,y_1,x_2,y_2,z_1,z_2\in\symbf{R^+}$，模仿上述证法并结合(1)的证法，证明:\[
+(x_1+x_2)(y_1+y_2)(z_1+z_2) \ge (\sqrt[3]{x_1y_1z_1}+\sqrt[3]{x_2y_2z_2})^3
+\]
+(提示：若$a,b,c\in\symbf{R^+}$，有$\frac{a^3+b^3+c^3}{3}\ge abc$,)
+\end{subquestions}
+\end{minipage}
+\begin{solution}{3cm}
+\method
+(1)设$a_1=\sqrt{x_1}$，$a_2=\sqrt{x_2}$，$b_1=\sqrt{y_1}$，$b_2=\sqrt{y_2}$.
+
+由\[ (a_1b_1+a_2b_2)^2 \le  (a_1^2+a_2^2)(b_1^2+b_2^2)\]
+
+得\[ (\sqrt{x_1}\sqrt{y_1}+\sqrt{x_2}\sqrt{y_2})^2 \le  [((\sqrt{x_1})^2+(\sqrt{x_2})^2)((\sqrt{y_1})^2+(\sqrt{y_2})^2)]\]
+
+即\[
+(x_1+x_2)(y_1+y_2) \ge (\sqrt{x_1y_1}+\sqrt{x_2y_2})^2
+\]
+\score{6}{6}(2) 设$a_1=\sqrt[3]{x_1}, b_1=\sqrt[3]{y_1}, c_1=\sqrt[3]{z_1}$, $a_2=\sqrt[3]{x_2}, b_2=\sqrt[3]{y_2}, c_2=\sqrt[3]{z_2}$.
+
+要证
+\[
+(x_1+x_2)(y_1+y_2)(z_1+z_2) \ge (\sqrt[3]{x_1y_1z_1}+\sqrt[3]{x_2y_2z_2})^3
+\]
+
+即证
+\[
+(a_1b_1c_1+a_2b_2c_2)^3 \le  (a_1^3+a_2^3)(b_1^3+b_2^3)(c_1^3+c_2^3)
+\]
+
+(不换元，直接证不扣分)
+\score{2}{8}令$A=\sqrt[3]{a_1^3+a_2^3}$，$B=\sqrt[3]{b_1^3+b_2^3}$, $C=\sqrt[3]{c_1^3+c_2^3}$.
+
+\[
+\begin{array}{rl}
+& \frac{a_1b_1c_1+a_2b_2c_2}{\sqrt[3]{a_1^3+a_2^3}\sqrt[3]{b_1^3+b_2^3}\sqrt[3]{c_1^3+c_2^3}}\\
+= & \frac{a_1b_1c_1}{ABC}+\frac{a_2b_2c_2}{ABC}\\
+= & \frac{a_1}{A}\cdot\frac{b_1}{B}\cdot\frac{c_1}{C}+\frac{a_2}{A}\cdot\frac{b_2}{B}\cdot\frac{c_2}{C}\\
+\le & \frac{1}{3}(\frac{a_1^3}{A^3}+\frac{b_1^3}{B^3}+\frac{c_1^3}{C^3})+ \frac{1}{3}(\frac{a_2^3}{A^3}+\frac{b_2^3}{B^3}+\frac{c_2^3}{C^3})\\
+= & \frac{1}{3}(\frac{a_1^3+a_2^3}{A^3}+\frac{b_1^3+b_2^3}{B^3}++\frac{c_1^3+c_2^3}{C^3})\\
+= & 1
+\end{array}
+\]
+
+故\[(a_1b_1c_1+a_2b_2c_2)^3 \le  (a_1^3+a_2^3)(b_1^3+b_2^3)(c_1^3+c_2^3)\]
+
+得证.
+\score{6}{14}
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [16] 公元2222年，有一种高危传染疾病在全球范围内蔓延，被感染者的潜伏期可以长达10年，期间会有约$0.05\%$的概率传染给他人，一旦发病三天内即死亡。某城市总人口约200万人，专家分析其中约有$1000$名感染者，为了防止疾病继续扩散，疾病预防控制中心现决定对全市人口进行血液检测以筛选出被感染者。由于检测试剂十分昂贵且数量有限，需要将血样混合后一起检测以节约试剂。已知感染者的检测结果为阳性, 未被感染者则为阴性. 阳性血样与阴性血样混合后的检测结果为阳性, 同为阴性或阳性的血样混合后结果不发生改变.
+\begin{subquestions}
+    \subquestion 若对全市人口进行平均分组，同一分组的血样将被混合到一起检测，若发现结果为阳性，则再在该分组内逐个检测排查. 设每个组$x$个人，那么最坏情况下，需要进行约多少次检测可以找到所有的被感染者? 在当前方案下，若要使检测的次数尽可能少，每个分组的最优人数是？
+    \subquestion 在(1)的检测方案中，对于检测结果为阳性的组采取逐一检测排查的方法并不是很好，或可将这些组的血样再进行一次分组混合血样检测，然后再进行逐一排查。仍然考虑最坏的情况，请问两次要如何分组，使检测总次数尽可能少？
+    \subquestion 在(2)的检测方案中，进行了两次分组混合血样检测。仍然考虑最坏情况，若再进行若干次分组混合血样检测，是否会使检测次数更少？请给出最优的检测方案.
+\end{subquestions}
+\end{minipage}
+\begin{solution}{6cm}
+\method
+设需要进行$y$次检测. 由每个组$x$个人，则有$\frac{2000000}{x}$个小组, 最坏情况是1000人被分配到了不同的组. 
+
+\[
+y=\begin{cases}
+\frac{2\cdot 10^6}{x}+2\cdot 10^6, & x \ge 2000 \\
+\frac{2\cdot 10^6}{x}+1000\cdot x, & x \lt 2000 \\
+\end{cases}
+\]
+
+(题目要求近似计算，可以不考虑整除，考虑整除不扣分)
+\score{2}{2}分组数小于1000的情况比逐个检测还要多，可以不用考虑，
+
+$y=\frac{2\times 10^6}{x}+1000\cdot x \ge 2\sqrt{2\cdot 10^9}$.
+
+当$x=\sqrt{2\times10^3}\approx45$人，检测次数最少.
+
+(结果为44-45都给分)
+\score{2}{4}(2) 设第二次分组每个组$w$人.
+
+\[
+\begin{array}{rl}
+y & = \frac{2\times10^6}{x}+\frac{1000x}{w}+1000w\\
+& \ge 3\sqrt[3]{2\times10^{12}}\\
+\end{array}
+\]
+
+当$\frac{2\times10^6}{x}=\frac{1000x}{w}=1000w$时，
+即$x=2000^\frac{2}{3}\approx159$,$w=2000^\frac{1}{3}\approx13$，
+检测次数最少.
+
+所以第一次分组159人一组，第二次分组13人一组.
+
+(也可以用两次基本不等式求得，允许$10\%$的误差.)
+\score{6}{10}(3) 若进行$k$次分组，设第$i$次分组时每组人数为$x_i$. 则总的检测次数为:
+\[
+\begin{array}{rl}
+y=&\dfrac{2\times10^6}{x_1}+\dfrac{1000x_1}{x_2}+\cdots\\
+&+\dfrac{1000x_{k-1}}{x_k}+1000x_k\\
+\ge & (k+1)\sqrt[k+1]{2\times10^{3k+2}}
+\end{array}
+\]
+
+利用计算器可得，
+
+1次分组， $y \ge 2\sqrt{2\cdot 10^9}\approx8.94\times10^4$;
+
+2次分组， $y \ge 3\sqrt[3]{2\cdot 10^{12}}\approx3.78\times10^4$;
+
+3次分组， $y \ge 4\sqrt[4]{2\cdot 10^{15}}\approx2.68\times10^4$;
+
+4次分组， $y \ge 5\sqrt[5]{2\cdot 10^{18}}\approx2.29\times10^4$;
+
+5次分组， $y \ge 6\sqrt[6]{2\cdot 10^{21}}\approx2.13\times10^4$;
+
+6次分组， $y \ge 7\sqrt[7]{2\cdot 10^{24}}\approx2.073\times10^4$;
+
+7次分组， $y \ge 8\sqrt[8]{2\cdot 10^{27}}\approx2.068\times10^4$;
+
+8次分组， $y \ge 9\sqrt[9]{2\cdot 10^{30}}\approx2.094\times10^4$;
+
+9次分组， $y \ge 10\sqrt[10]{2\cdot 10^{33}}\approx2.138\times10^4$;
+
+可见进行7次分组混合血样检测最优.
+
+其中第k次分组时$(2000)^{\frac{8-k}{8}}$人一组.
+\score{6}{16}
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [18] 已知函数$f(x)=a{{x}^{2}}-\frac{1}{2}x+c$（$a$、$c\in R$），满足$f(1)=0$，且$f(x)\ge 0$在$x\in R$时恒成立．
+\begin{subquestions}
+    \subquestion 求$a$、$c$的值；
+    \subquestion 若$h(x)=\frac{3}{4}{{x}^{2}}-bx+\frac{b}{2}-\frac{1}{4}$，解不等式$f(x)+h(x) \lt 0$；
+    \subquestion 是否存在实数$m$，使函数$g(x)=f(x)-mx$在区间$[m,m+2]$上有最小值$-5$？若存在，请求出$m$的值；若不存在，请说明理由．
+\end{subquestions}
+\end{minipage}
+\begin{solution}{6cm}
+\method
+由$f(1)=0$，得$a+c=\frac{1}{2}$,
+\score{1}{1}因为$f(x)\ge 0$在$x\in R$时恒成立，所以$a \gt 0$且△$=\frac{1}{4}-4ac\le 0$，$ac\ge \frac{1}{16}$，
+\score{1}{2}即$a\left( \frac{1}{2}-a \right)\ge \frac{1}{16}$，${{a}^{2}}-\frac{1}{2}a+\frac{1}{16}\le 0$，${{\left( a-\frac{1}{4} \right)}^{2}}\le 0$，所以$a=c=\frac{1}{4}$．
+\score{2}{4}(2) 由（1）得$f(x)=\frac{1}{4}{{x}^{2}}-\frac{1}{2}x+\frac{1}{4}$，由$f(x)+h(x) \lt 0$，得
+${{x}^{2}}-\left( b+\frac{1}{2} \right)x+\frac{b}{2} \lt 0$，即$(x-b)\left( x-\frac{1}{2} \right) \lt 0$，
+\score{3}{7}所以，当$b \lt \frac{1}{2}$时，原不等式解集为$(b,\frac{1}{2})$；
+当$b \gt \frac{1}{2}$时，原不等式解集为$(\frac{1}{2},b)$；
+当$b=\frac{1}{2}$时，原不等式解集为空集 ．
+\score{3}{10}(3) $g(x)=\frac{1}{4}{{x}^{2}}-\left( \frac{1}{2}+m \right)x+\frac{1}{4}$，
+\score{1}{11}$g(x)$的图像是开口向上的抛物线，对称轴为直线$x=2m+1$．
+假设存在实数$m$，使函数$g(x)$在区间$[m,m+2]$上有最小值$-5$．
+\ding{192}	当$2m+1 \lt m$，即$m \lt -1$时，函数$g(x)$在区间$[m,m+2]$上是增函数，所以$g(m)=-5$，即$\frac{1}{4}{{m}^{2}}-\left( \frac{1}{2}+m \right)m+\frac{1}{4}=-5$，解得$m=-3$或$m=\frac{7}{3}$，
+因为$m \lt -1$，所以$m=-3$；
+\score{2}{13}\ding{193}当$m\le 2m+1\le m+2$，即$-1\le m\le 1$时，函数$g(x)$的最小值为$g(2m+1)=-5$，即
+$\frac{1}{4}{{(2m+1)}^{2}}-\left( \frac{1}{2}+m \right)(2m+1)+\frac{1}{4}=-5$，解得$m=-\frac{1}{2}-\frac{\sqrt{21}}{2}$或$m=-\frac{1}{2}+\frac{\sqrt{21}}{2}$，均舍去；
+\score{2}{15}\ding{194}当$2m+1 \gt m+2$，即$m \gt 1$时，$g(x)$在区间$[m,m+2]$上是减函数，所以$g(m+2)=-5$，即$\frac{1}{4}{{(m+2)}^{2}}-\left( \frac{1}{2}+m \right)(m+2)+\frac{1}{4}=-5$，解得$m=-1-2\sqrt{2}$或$m=-1+2\sqrt{2}$，因$m \gt 1$，所以$m=-1+2\sqrt{2}$．
+\score{2}{17}综上，存在实数$m$，$m=-3$或$m=-1+2\sqrt{2}$时，函数$g(x)$在区间$[m,m+2]$上有最小值$-5$．
+\score{1}{18}
+\end{solution}
+\end{questions}
+\end{groups}
+\label{lastpage}
+\end{document}
diff --git a/macros/xetex/latex/bhcexam/examples/logo.png b/macros/xetex/latex/bhcexam/examples/logo.png
new file mode 100644
index 0000000000..ccbb38ad13
--- /dev/null
+++ b/macros/xetex/latex/bhcexam/examples/logo.png
diff --git a/macros/xetex/latex/bhcexam/examples/naive.pdf b/macros/xetex/latex/bhcexam/examples/naive.pdf
new file mode 100644
index 0000000000..e6846ff3b0
--- /dev/null
+++ b/macros/xetex/latex/bhcexam/examples/naive.pdf
diff --git a/macros/xetex/latex/bhcexam/examples/naive.tex b/macros/xetex/latex/bhcexam/examples/naive.tex
new file mode 100644
index 0000000000..369d23ee6e
--- /dev/null
+++ b/macros/xetex/latex/bhcexam/examples/naive.tex
@@ -0,0 +1,86 @@
+% 使用 BHCexam 文档类，并传递选项
+\documentclass[answers]{BHCexam}
+\usepackage{hyperref}
+
+\begin{document}
+
+% 第一行主标题
+\title{BHCexam试卷排版宏包}
+
+% 第二行主标题
+\subtitle{样例}
+
+% 考试说明
+\notice{满分100分, 10分钟完成.}
+
+% 命题人信息
+\author{微信关注公众号：橘子数学}
+
+% 考试日期
+\date{2019.12.1}
+
+% 生成试卷头
+\maketitle
+
+\begin{groups}
+
+% 第一个题组，显示分值，不预留空间
+\group{填空}{本题组共1小题，共30.0分}
+\begin{questions}[s]
+
+% 填空题，两个空
+\question[30] 橘子数学的网址是\key{www.mathcrowd.cn}, 微信服务号\key{橘子数学}.
+\question[30] 橘子数学趣味挑战的网址是\key{qa.mathcrowd.cn},微信订阅号是\key{试题工坊}, \key{橘子数学题库}.
+
+\end{questions}
+
+% 第二个题组，显示分值，不预留空间
+\group{选择}{本题组共2小题，共40.0分}
+\begin{questions}[ps]
+
+% 选择题，四个选项
+\question[30] 以下哪一项不是橘子数学社区的宗旨\key{C}.
+\fourchoices{开放}{高效}{无视版权}{合作}
+
+% 解答，4cm 参数被忽略
+\begin{solution}{4cm}
+\method 橘子数学社区的宗旨是开放、高效、合作、变革.
+\method 见 \url{http://docs.mathcrowd.cn/zh_CN/latest/community/principles.html}
+\end{solution}
+
+% 选择题，五个选项
+\question[40] 以下数学公式显示有明显瑕疵的是\key{D}.
+\fivechoices{$\sin A$}{$2+3\mathrm{i}$}{$x^2$}{$\ln x$}{$\mathrm{e}^{\mathrm{i}\theta}$}
+
+\begin{solution}{4cm}
+\methodonly D 中正确的公式显示效果为$\ln{x}$.
+\end{solution}
+\end{questions}
+
+% 第三个题组，显示分值，预留空间
+\group{主观题}{本题组共1小题，共30.0分}
+\begin{questions}[st]
+% 简答题，两个小问
+\question[30] 请回答以下问题：
+\begin{subquestions}
+    \subquestion 你觉得有必要创建这样一个试题社区吗? 为什么?
+    \subquestion 你对社区的建设有什么建议.
+\end{subquestions}
+
+% 解答，学生版会预留8cm的答题空间.
+\begin{solution}{4cm}
+	\methodonly 欢迎加入用户群组发言讨论. 
+	
+telegram 交流群组: https://t.me/mathcrowd
+
+QQ 群: 319701002
+
+Github项目页: \url{https://github.com/mathedu4all/bhcexam}
+
+\score{30}{30}
+	
+\end{solution}
+\end{questions}
+
+\end{groups}
+\end{document}
diff --git a/macros/xetex/latex/bhcexam/examples/qrcode.png b/macros/xetex/latex/bhcexam/examples/qrcode.png
new file mode 100644
index 0000000000..b15dd4d96b
--- /dev/null
+++ b/macros/xetex/latex/bhcexam/examples/qrcode.png
diff --git a/systems/texlive/tlnet/tlpkg/installer/ctan-mirrors.pl b/systems/texlive/tlnet/tlpkg/installer/ctan-mirrors.pl
index b23aaefccc..729c304174 100644
--- a/systems/texlive/tlnet/tlpkg/installer/ctan-mirrors.pl
+++ b/systems/texlive/tlnet/tlpkg/installer/ctan-mirrors.pl
@@ -114,7 +114,6 @@ $mirrors = {
       'https://ftp.tu-chemnitz.de/pub/tex/' => 1,
       'https://mirror.clientvps.com/CTAN/' => 1,
       'https://mirror.dogado.de/tex-archive/' => 1,
-      'https://mirror.informatik.hs-fulda.de/tex-archive/' => 1,
       'https://mirror.physik.tu-berlin.de/pub/CTAN/' => 1,
       'https://packages.oth-regensburg.de/ctan/' => 1,
     },
@@ -122,6 +121,9 @@ $mirrors = {
       'http://ftp.ntua.gr/mirror/ctan/' => 1,
       'https://ftp.cc.uoc.gr/mirrors/CTAN/' => 1,
     },
+    'Hungary' => {
+      'https://mirror.szerverem.hu/ctan/' => 1,
+    },
     'Italy' => {
       'https://ctan.mirror.garr.it/mirrors/ctan/' => 1,
     },
@@ -187,7 +189,6 @@ $mirrors = {
     },
     'USA' => {
       'http://bay.uchicago.edu/CTAN/' => 1,
-      'http://mirrors.ibiblio.org/pub/mirrors/CTAN/' => 1,
       'https://ctan.math.illinois.edu/' => 1,
       'https://ctan.math.utah.edu/ctan/tex-archive/' => 1,
       'https://ctan.math.washington.edu/tex-archive/' => 1,
diff --git a/systems/texlive/tlnet/tlpkg/texlive.tlpdb b/systems/texlive/tlnet/tlpkg/texlive.tlpdb
index d02f10ae0a..830c14c138 100644
--- a/systems/texlive/tlnet/tlpkg/texlive.tlpdb
+++ b/systems/texlive/tlnet/tlpkg/texlive.tlpdb
@@ -26,7 +26,7 @@ depend container_split_src_files/1
 depend frozen/0
 depend minrelease/2016
 depend release/2022
-depend revision/64086
+depend revision/64093
 
 name 00texlive.image
 category TLCore
@@ -4609,7 +4609,7 @@ depend setting_available_architectures:aarch64-linux amd64-freebsd amd64-netbsd
 
 name 00texlive.installer
 category TLCore
-revision 64074
+revision 64088
 shortdesc TeX Live standalone installer package
 longdesc This package defines the files to go into the installer
 longdesc archives (install-tl-unx.tar.gz, install-tl.zip) built by the
@@ -35837,35 +35837,45 @@ catalogue-version 0.4
 
 name bhcexam
 category Package
-revision 39041
-shortdesc An exam class designed for Mathematics Teachers in China
-relocated 1
-longdesc The class based on the exam class, and is specially designed
-longdesc for High School Mathematics Teachers in China.
-containersize 3832
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-docfiles size=10
- RELOC/doc/latex/bhcexam/Makefile
- RELOC/doc/latex/bhcexam/README details="Readme"
- RELOC/doc/latex/bhcexam/test1.tex
- RELOC/doc/latex/bhcexam/test2.tex
- RELOC/doc/latex/bhcexam/test3.tex
- RELOC/doc/latex/bhcexam/test4.tex
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-srcfiles size=7
- RELOC/source/latex/bhcexam/BHCexam.dtx
- RELOC/source/latex/bhcexam/BHCexam.ins
+revision 64093
+shortdesc An exam class for mathematics teachers in China
+relocated 1
+longdesc This exam class is specially designed for mathematics teachers
+longdesc in China. It is used by mathcrowd.cn (an opensource math exam
+longdesc database) as the default class for exporting exam papers to
+longdesc pdf. Using BHCexam you can separate the format and the content
+longdesc very well; export both teacher paper and student paper; typeset
+longdesc multiple choice questions with 3-6 options keeping adaptively
+longdesc neat alignment; typeset cloze questions with a customizable
+longdesc underline; typeset questions with subquestions in lists; group
+longdesc questions in a list to control whether to show score, leave
+longdesc spacing, initialize question number; and more (see BHCexam
+longdesc Documentation).
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+ RELOC/doc/xelatex/bhcexam/examples/example_teacher_paper.pdf details="Example of use (teacher paper)" language="zh"
+ RELOC/doc/xelatex/bhcexam/examples/example_teacher_paper.tex
+ RELOC/doc/xelatex/bhcexam/examples/logo.png
+ RELOC/doc/xelatex/bhcexam/examples/naive.pdf
+ RELOC/doc/xelatex/bhcexam/examples/naive.tex
+ RELOC/doc/xelatex/bhcexam/examples/qrcode.png
 runfiles size=4
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- RELOC/tex/latex/bhcexam/BHCexam.cls
-catalogue-also exam
-catalogue-ctan /macros/latex/contrib/bhcexam
-catalogue-license lppl
-catalogue-topics exam
-catalogue-version 0.4
+ RELOC/tex/xelatex/bhcexam/BHCexam.cls
+catalogue-contact-announce https://github.com/mathedu4all/bhcexam/wiki
+catalogue-contact-bugs https://github.com/mathedu4all/bhcexam/issues
+catalogue-contact-repository https://github.com/mathedu4all/bhcexam
+catalogue-ctan /macros/xetex/latex/bhcexam
+catalogue-license lppl1.3
+catalogue-topics exam class chinese
+catalogue-version 1.7
 
 name bib-fr
 category Package
@@ -86809,7 +86819,7 @@ catalogue-version 4.33
 
 name customdice
 category Package
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 shortdesc Simple commands for drawing customisable dice
 relocated 1
 longdesc The customdice package for LaTeX, LuaLaTeX and XeTeX that
@@ -86817,26 +86827,26 @@ longdesc provides functionality for drawing dice. The aim is to provide
 longdesc highly-customisable but simple-to-use commands, allowing:
 longdesc adding custom text to dice faces; control over colouring;
 longdesc control over sizing.
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 catalogue-contact-bugs https://github.com/prowlett/customdice/issues
 catalogue-contact-repository https://github.com/prowlett/customdice
 catalogue-ctan /graphics/pgf/contrib/customdice
 catalogue-license cc-by-sa-4
 catalogue-topics graphics-use
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 name cutwin
 category Package
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 name luatexja
 category Package
-revision 64072
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 shortdesc Typeset Japanese with Lua(La)TeX
 relocated 1
 longdesc The package offers support for typesetting Japanese documents
 longdesc with LuaTeX. Either of the Plain and LaTeX2e formats may be
 longdesc used with the package.
 depend luatexbase
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 docfiles size=979
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  RELOC/doc/luatex/luatexja/README details="Readme"
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  RELOC/doc/luatex/luatexja/luatexja.ins
 srccontainersize 83300
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 catalogue-ctan /macros/luatex/generic/luatexja
 catalogue-license bsd
 catalogue-topics japanese luatex class macro-gen
-catalogue-version 20220807.0
+catalogue-version 20220810.0
 
 name luatexko
 category Package
@@ -251803,7 +251813,7 @@ catalogue-version 1.0
 
 name proflycee
 category Package
-revision 64024
+revision 64090
 shortdesc A LaTeX package for French maths teachers in high school
 relocated 1
 longdesc This package provides some commands to help French mathematics
@@ -251812,14 +251822,17 @@ longdesc create splines with "derivative control"; \paramCF and \ligneCF
 longdesc in order to create an xcas-windows-like; \envconsolepythontex
 longdesc and \envcodepythontex to create code presentation and code
 longdesc execution with pythontex.
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+docfiles size=230
  RELOC/doc/latex/proflycee/ProfLycee-doc.pdf details="Package documentation" language="fr"
  RELOC/doc/latex/proflycee/ProfLycee-doc.tex
  RELOC/doc/latex/proflycee/README.md details="Readme"
+ RELOC/doc/latex/proflycee/basique.csv
+ RELOC/doc/latex/proflycee/cap.csv
+ RELOC/doc/latex/proflycee/perroquet.csv
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  RELOC/doc/latex/proflycee/pl-doc-stats_b.png
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@@ -251834,7 +251847,7 @@ catalogue-contact-repository https://github.com/cpierquet/ProfLycee
 catalogue-ctan /macros/latex/contrib/proflycee
 catalogue-license lppl1.3c
 catalogue-topics maths french teaching
-catalogue-version 1.1.8
+catalogue-version 1.2
 
 name program
 category Package
@@ -278260,7 +278273,7 @@ catalogue-topics dissertation class
 
 name simplebnf
 category Package
-revision 63263
+revision 64091
 shortdesc A simple package to format Backus-Naur form (BNF)
 relocated 1
 longdesc This package provides a simple way to format Backus-Naur form
@@ -278268,11 +278281,11 @@ longdesc (BNF). The included bnfgrammar environment parses BNF
 longdesc expressions (possibly annotated), so users can write readable
 longdesc BNF expressions in their documents. The package requires expl3,
 longdesc xparse, and mathtools.
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  RELOC/doc/latex/simplebnf/README.md details="Readme"
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@@ -278285,7 +278298,7 @@ catalogue-contact-support https://github.com/Zeta611/simplebnf/issues
 catalogue-ctan /macros/latex/contrib/simplebnf
 catalogue-license mit
 catalogue-topics expl3 formal-spec
-catalogue-version 0.3.0
+catalogue-version 0.3.1
 
 name simplecd
 category Package
@@ -306184,7 +306197,7 @@ docfiles size=376
 
 name texlive-scripts
 category TLCore
-revision 64074
+revision 64088
 shortdesc TeX Live infrastructure programs
 longdesc Includes install-tl, tl-portable, rungs, etc.; not needed for
 longdesc tlmgr to run but still ours. Not included in tlcritical.
@@ -306192,10 +306205,10 @@ depend texlive-scripts.ARCH
 depend texlive.infra
 execute addMap mathpple.map
 postaction shortcut type=menu name="TeX Live command-line"   cmd=TEXDIR/tlpkg/installer/tl-cmd.bat
-containersize 113528
-containerchecksum f2c7868ac1ad92b7947a5dc22ff61c559ccb7b408b4d662c0b9d5b109b44caaa2cf61591f8102a7c545e9e72e4510d41677132ea72ce6dc9b47a266043223820
+containersize 113532
+containerchecksum 7d4456dcadb368c084460b6498935ded333b53bbdc6943bbf421cd7ac9d2115098d1194100755e7a40febb88b00842946066cb60c5b858a8f407bd6fd43ebc3b
 doccontainersize 419224
-doccontainerchecksum d32ffdece9c23d795ca3975ec1734048324ce14ed4831ab631524576c36d1d3a3655af6821b174889a8d90f2746d12707dca7c1981746691f292501a0eae9482
+doccontainerchecksum 0c35ddff511d9ac03939c456501d2248514f31b15136476c6aaec741c456ca26533475651629efb03eee4532b52df329820e85fc81cfd33a34b2dfcf0875303c
 docfiles size=516
  doc.html
  texmf-dist/doc/man/man1/fmtutil-sys.1
diff --git a/systems/texlive/tlnet/tlpkg/texlive.tlpdb.md5 b/systems/texlive/tlnet/tlpkg/texlive.tlpdb.md5
index 3e26d466aa..a5cf1e4dd8 100644
--- a/systems/texlive/tlnet/tlpkg/texlive.tlpdb.md5
+++ b/systems/texlive/tlnet/tlpkg/texlive.tlpdb.md5
@@ -1 +1 @@
-fde4811ca40a2be6ef67215c92f889c3  texlive.tlpdb
+ddd602c9ce8f9098e27fcd9b3f2084b2  texlive.tlpdb
diff --git a/systems/texlive/tlnet/tlpkg/texlive.tlpdb.sha512 b/systems/texlive/tlnet/tlpkg/texlive.tlpdb.sha512
index f6ea1cf901..0ea4433ad9 100644
--- a/systems/texlive/tlnet/tlpkg/texlive.tlpdb.sha512
+++ b/systems/texlive/tlnet/tlpkg/texlive.tlpdb.sha512
@@ -1 +1 @@
-f04b8f79496a1d197503ab02c18321d114a8438a213b62611b8c9a7915425220fe1e14aade757bdc63c00480d515e62f4254235629898eb05e38a8de7a7aa90f  texlive.tlpdb
+786899849ff54db49803b47f081f54d760c403c79cf1f419114996f11199fb600a41196ec370fe07f5ba41121a6de0b263306ee298fea1215a76a5a4118b9fcb  texlive.tlpdb
diff --git a/systems/texlive/tlnet/tlpkg/texlive.tlpdb.sha512.asc b/systems/texlive/tlnet/tlpkg/texlive.tlpdb.sha512.asc
index 2b4d7553f1..5fc5416ece 100644
--- a/systems/texlive/tlnet/tlpkg/texlive.tlpdb.sha512.asc
+++ b/systems/texlive/tlnet/tlpkg/texlive.tlpdb.sha512.asc
@@ -1,10 +1,10 @@
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diff --git a/systems/texlive/tlnet/tlpkg/texlive.tlpdb.xz b/systems/texlive/tlnet/tlpkg/texlive.tlpdb.xz
index 2f95f2bd9f..9f093cd8de 100644
--- a/systems/texlive/tlnet/tlpkg/texlive.tlpdb.xz
+++ b/systems/texlive/tlnet/tlpkg/texlive.tlpdb.xz
diff --git a/systems/win32/miktex/tm/packages/next/pr.ini b/systems/win32/miktex/tm/packages/next/pr.ini
index be4ea7fc7e..05cd4f651a 100644
--- a/systems/win32/miktex/tm/packages/next/pr.ini
+++ b/systems/win32/miktex/tm/packages/next/pr.ini
@@ -1,13 +1,13 @@
 

 [repository]

-date=1660076266

+date=1660162672

 lastupd=miktex-arctrl-bin-x64-2.9 miktex-asymptote-bin-x64-2.9 miktex-autosp-bin-x64-2.9 miktex-axohelp-bin-x64-2.9 miktex-bibarts-bin-x64-2.9 miktex-bibtex-bin-x64-2.9 miktex-bibtex8bit-bin-x64-2.9 miktex-bzip2-bin-x64-2.9 miktex-cairo-bin-x64-2.9 miktex-chktex-bin-x64-2.9 miktex-cjkutils-bin-x64-2.9 miktex-console-bin-x64-2.9 miktex-curl-bin-x64-2.9 miktex-cweb-bin-x64-2.9 miktex-devnag-bin-x64-2.9 miktex-doc miktex-dvicopy-bin-x64-2.9 miktex-dvipdfmx-bin-x64-2.9 miktex-dvipng-bin-x64-2.9 miktex-dvips-bin-x64-2.9

-lstdigest=ae03e78ad2073d2af5440f4a917e106c

+lstdigest=34719537d8d983aac441854c0e3440d9

 numpkg=5848

 relstate=next

-version=8256

+version=8257

 

 ;;;;This configuration file is signed by a MiKTeX maintainer. The signature follows.

 ;;;;-----BEGIN MIKTEX SIGNATURE-----

-;;;; signature/miktex: VxevTL1+/QxR3ajZURcbvh+k8qG0ir6daMnexAvJv0QQwUadZvwnibN5QDxjiftS5XD6a9AbZGtkpTW5EwTCH6UYjnnG5WUL6Q1i/UPS4m+g8DqnEE4icIcLBzN9k1hI/hA6Ll4Uky8DNqZ2i5N3+QZtrt7KB/yE2010UAAzwbdSIKdH02u902POo7XfbTpPqzxEqBVVA7zm5xuHapkVMSk9O35vUDkA7mx14v1ZASddpl8WqxtZkMne7xPISqnBxk3GkMDm2jrxFP4NAGhEaviYnOJKSJX0vz9WgYQdO9rzpfpPPkmmeJWfXAufktec2IzcNSqAAUejuKqeLRYwSw==

+;;;; signature/miktex: Gc6zSyy6um6GOMQ0KwMumvMg21Y1YVcnHyOHWkbKcIWGvPrVI4SS5ilkw3CPbqT3fTEI1qOBhW/4WUf8Z3jNlQIK5zrpN1idGUaLEB9nI0HkT2X5RXhJt15PhOBKYt/U10RoZ+YyjuH6T3YNiMS42CPNfGts2uqOtqXPr4Nh5WP4LJGkrwtslUvd60NFGT+nSit8Ss/ECQvVf3sOxxZ2kcckzBkdKypzR8nATxcvVBteIt6R+c6bSZvkLhki/IBbY5BPKf3ARO/msHD4DFWpI4+qp9V6zaaW2aSoEgwvfXd/tVE4XyGnxmMH5cV8BBiw14s1rC59PrBvYuNeyG7KQQ==

 ;;;;-----END MIKTEX SIGNATURE-----

diff --git a/systems/win32/miktex/tm/packages/pr.ini b/systems/win32/miktex/tm/packages/pr.ini
index c7120352c3..23a7ac9e55 100644
--- a/systems/win32/miktex/tm/packages/pr.ini
+++ b/systems/win32/miktex/tm/packages/pr.ini
@@ -1,13 +1,13 @@
 

 [repository]

-date=1660075897

+date=1660162289

 lastupd=wrapstuff wrapstuff__doc wrapstuff__source miktex-dvips-bin-x64-2.9 miktex-runtime-bin-x64-2.9 miktex-ucrt-bin miktex-vc140-bin apa7 apa7__doc apa7__source buctthesis buctthesis__doc buctthesis__source csassignments csassignments__doc csassignments__source exam-zh exam-zh__doc exam-zh__source fixdif

-lstdigest=6dfc095593e3aaa1c206ce8dc5b4a478

+lstdigest=0c022d56551a080ec684acce3dc32d6d

 numpkg=5848

 relstate=stable

-version=8256

+version=8257

 

 ;;;;This configuration file is signed by a MiKTeX maintainer. The signature follows.

 ;;;;-----BEGIN MIKTEX SIGNATURE-----

-;;;; signature/miktex: tsRCvYdbfXCm+ns9l/BORUjw99tsN3FlJg89n3+jpTua400qwhur0DgxRxhB7NgJfpzcUEYlu1K0Qx6yNKtR8ZQq9Gz92ue2d2dOXv3htQVhGjTTT0yWnbHy34w3SZjMF9pOPDkMTIn6giv9Skf7mWP44mW6Yg/gTnQZLU3iYx4BEeqKM6vkH2C2kG/TC54sqIos32mqIP5fTnvDYlm3l/F7O5BkERYN9Mp5SWXZAD+h9m3FEjsdrLTspgJFAjHd18RS0m1h/17NABs5MwBuQev1xdMMPa5iq0obYE9+CLLx7Dy3UuBmi/xcdNHamleNSlUxKHfmBAN5RF1svo6U6w==

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 ;;;;-----END MIKTEX SIGNATURE-----

diff --git a/timestamp b/timestamp
index 5fd345ce1e..bf3cd05c8e 100644
--- a/timestamp
+++ b/timestamp
@@ -2,4 +2,4 @@
 #   The source CTAN of this site's material:
 dante.ctan.org
 #   The year-month-day-hour-minute of this site's material:
-2022-08-10-02-02
+2022-08-11-02-02