hfutexam (4dec22)

git-svn-id: svn://tug.org/texlive/trunk@65192 c570f23f-e606-0410-a88d-b1316a301751

17 files changed, 1632 insertions, 1 deletions

diff --git a/Master/texmf-dist/doc/latex/hfutexam/CHANGELOG.md b/Master/texmf-dist/doc/latex/hfutexam/CHANGELOG.md
new file mode 100644
index 00000000000..c0174e3a9e4
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/CHANGELOG.md
@@ -0,0 +1,18 @@
+# 更新日志
+
+## [1.5] - 2022/12/04
++ 现在`\XiZhuRenQianMing`不设置或设置为空时即可不显示签名.
+
+## [1.4] - 2022/12/04
++ 修复了有时选择题选项未能正确计算行数的问题.
++ 修复了有时填空题答案会被横线截断的问题.
+
+## [1.3] - 2022/11/27
++ 移除了简易模式下学期的下划线.
++ 打分框的宽度减少了`10%`.
+
+## [1.2] - 2022/10/31
++ 选择题选项后增加了`5pt`的垂直间距.
+
+## [1.1] - 2022/10/30
++ 首次提交.
diff --git a/Master/texmf-dist/doc/latex/hfutexam/LICENSE b/Master/texmf-dist/doc/latex/hfutexam/LICENSE
new file mode 100644
index 00000000000..22443139013
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/LICENSE
@@ -0,0 +1,416 @@
+The LaTeX Project Public License
+=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
+
+LPPL Version 1.3c  2008-05-04
+
+Copyright 1999 2002-2008 LaTeX3 Project
+    Everyone is allowed to distribute verbatim copies of this
+    license document, but modification of it is not allowed.
+
+
+PREAMBLE
+========
+
+The LaTeX Project Public License (LPPL) is the primary license under
+which the LaTeX kernel and the base LaTeX packages are distributed.
+
+You may use this license for any work of which you hold the copyright
+and which you wish to distribute.  This license may be particularly
+suitable if your work is TeX-related (such as a LaTeX package), but 
+it is written in such a way that you can use it even if your work is 
+unrelated to TeX.
+
+The section `WHETHER AND HOW TO DISTRIBUTE WORKS UNDER THIS LICENSE',
+below, gives instructions, examples, and recommendations for authors
+who are considering distributing their works under this license.
+
+This license gives conditions under which a work may be distributed
+and modified, as well as conditions under which modified versions of
+that work may be distributed.
+
+We, the LaTeX3 Project, believe that the conditions below give you
+the freedom to make and distribute modified versions of your work
+that conform with whatever technical specifications you wish while
+maintaining the availability, integrity, and reliability of
+that work.  If you do not see how to achieve your goal while
+meeting these conditions, then read the document `cfgguide.tex'
+and `modguide.tex' in the base LaTeX distribution for suggestions.
+
+
+DEFINITIONS
+===========
+
+In this license document the following terms are used:
+
+   `Work'
+    Any work being distributed under this License.
+    
+   `Derived Work'
+    Any work that under any applicable law is derived from the Work.
+
+   `Modification' 
+    Any procedure that produces a Derived Work under any applicable
+    law -- for example, the production of a file containing an
+    original file associated with the Work or a significant portion of
+    such a file, either verbatim or with modifications and/or
+    translated into another language.
+
+   `Modify'
+    To apply any procedure that produces a Derived Work under any
+    applicable law.
+    
+   `Distribution'
+    Making copies of the Work available from one person to another, in
+    whole or in part.  Distribution includes (but is not limited to)
+    making any electronic components of the Work accessible by
+    file transfer protocols such as FTP or HTTP or by shared file
+    systems such as Sun's Network File System (NFS).
+
+   `Compiled Work'
+    A version of the Work that has been processed into a form where it
+    is directly usable on a computer system.  This processing may
+    include using installation facilities provided by the Work,
+    transformations of the Work, copying of components of the Work, or
+    other activities.  Note that modification of any installation
+    facilities provided by the Work constitutes modification of the Work.
+
+   `Current Maintainer'
+    A person or persons nominated as such within the Work.  If there is
+    no such explicit nomination then it is the `Copyright Holder' under
+    any applicable law.
+
+   `Base Interpreter' 
+    A program or process that is normally needed for running or
+    interpreting a part or the whole of the Work.    
+
+    A Base Interpreter may depend on external components but these
+    are not considered part of the Base Interpreter provided that each
+    external component clearly identifies itself whenever it is used
+    interactively.  Unless explicitly specified when applying the
+    license to the Work, the only applicable Base Interpreter is a
+    `LaTeX-Format' or in the case of files belonging to the 
+    `LaTeX-format' a program implementing the `TeX language'.
+
+
+
+CONDITIONS ON DISTRIBUTION AND MODIFICATION
+===========================================
+
+1.  Activities other than distribution and/or modification of the Work
+are not covered by this license; they are outside its scope.  In
+particular, the act of running the Work is not restricted and no
+requirements are made concerning any offers of support for the Work.
+
+2.  You may distribute a complete, unmodified copy of the Work as you
+received it.  Distribution of only part of the Work is considered
+modification of the Work, and no right to distribute such a Derived
+Work may be assumed under the terms of this clause.
+
+3.  You may distribute a Compiled Work that has been generated from a
+complete, unmodified copy of the Work as distributed under Clause 2
+above, as long as that Compiled Work is distributed in such a way that
+the recipients may install the Compiled Work on their system exactly
+as it would have been installed if they generated a Compiled Work
+directly from the Work.
+
+4.  If you are the Current Maintainer of the Work, you may, without
+restriction, modify the Work, thus creating a Derived Work.  You may
+also distribute the Derived Work without restriction, including
+Compiled Works generated from the Derived Work.  Derived Works
+distributed in this manner by the Current Maintainer are considered to
+be updated versions of the Work.
+
+5.  If you are not the Current Maintainer of the Work, you may modify
+your copy of the Work, thus creating a Derived Work based on the Work,
+and compile this Derived Work, thus creating a Compiled Work based on
+the Derived Work.
+
+6.  If you are not the Current Maintainer of the Work, you may
+distribute a Derived Work provided the following conditions are met
+for every component of the Work unless that component clearly states
+in the copyright notice that it is exempt from that condition.  Only
+the Current Maintainer is allowed to add such statements of exemption 
+to a component of the Work. 
+
+  a. If a component of this Derived Work can be a direct replacement
+     for a component of the Work when that component is used with the
+     Base Interpreter, then, wherever this component of the Work
+     identifies itself to the user when used interactively with that
+     Base Interpreter, the replacement component of this Derived Work
+     clearly and unambiguously identifies itself as a modified version
+     of this component to the user when used interactively with that
+     Base Interpreter.
+     
+  b. Every component of the Derived Work contains prominent notices
+     detailing the nature of the changes to that component, or a
+     prominent reference to another file that is distributed as part
+     of the Derived Work and that contains a complete and accurate log
+     of the changes.
+  
+  c. No information in the Derived Work implies that any persons,
+     including (but not limited to) the authors of the original version
+     of the Work, provide any support, including (but not limited to)
+     the reporting and handling of errors, to recipients of the
+     Derived Work unless those persons have stated explicitly that
+     they do provide such support for the Derived Work.
+
+  d. You distribute at least one of the following with the Derived Work:
+
+       1. A complete, unmodified copy of the Work; 
+          if your distribution of a modified component is made by
+          offering access to copy the modified component from a
+          designated place, then offering equivalent access to copy
+          the Work from the same or some similar place meets this
+          condition, even though third parties are not compelled to
+          copy the Work along with the modified component;
+
+       2. Information that is sufficient to obtain a complete,
+          unmodified copy of the Work.
+
+7.  If you are not the Current Maintainer of the Work, you may
+distribute a Compiled Work generated from a Derived Work, as long as
+the Derived Work is distributed to all recipients of the Compiled
+Work, and as long as the conditions of Clause 6, above, are met with
+regard to the Derived Work.
+
+8.  The conditions above are not intended to prohibit, and hence do not
+apply to, the modification, by any method, of any component so that it
+becomes identical to an updated version of that component of the Work as
+it is distributed by the Current Maintainer under Clause 4, above.
+
+9.  Distribution of the Work or any Derived Work in an alternative
+format, where the Work or that Derived Work (in whole or in part) is
+then produced by applying some process to that format, does not relax or
+nullify any sections of this license as they pertain to the results of
+applying that process.
+     
+10. a. A Derived Work may be distributed under a different license
+       provided that license itself honors the conditions listed in
+       Clause 6 above, in regard to the Work, though it does not have
+       to honor the rest of the conditions in this license.
+      
+    b. If a Derived Work is distributed under a different license, that
+       Derived Work must provide sufficient documentation as part of
+       itself to allow each recipient of that Derived Work to honor the 
+       restrictions in Clause 6 above, concerning changes from the Work.
+
+11. This license places no restrictions on works that are unrelated to
+the Work, nor does this license place any restrictions on aggregating
+such works with the Work by any means.
+
+12.  Nothing in this license is intended to, or may be used to, prevent
+complete compliance by all parties with all applicable laws.
+
+
+NO WARRANTY
+===========
+
+There is no warranty for the Work.  Except when otherwise stated in
+writing, the Copyright Holder provides the Work `as is', without
+warranty of any kind, either expressed or implied, including, but not
+limited to, the implied warranties of merchantability and fitness for a
+particular purpose.  The entire risk as to the quality and performance
+of the Work is with you.  Should the Work prove defective, you assume
+the cost of all necessary servicing, repair, or correction.
+
+In no event unless required by applicable law or agreed to in writing
+will The Copyright Holder, or any author named in the components of the
+Work, or any other party who may distribute and/or modify the Work as
+permitted above, be liable to you for damages, including any general,
+special, incidental or consequential damages arising out of any use of
+the Work or out of inability to use the Work (including, but not limited
+to, loss of data, data being rendered inaccurate, or losses sustained by
+anyone as a result of any failure of the Work to operate with any other
+programs), even if the Copyright Holder or said author or said other
+party has been advised of the possibility of such damages.
+
+
+MAINTENANCE OF THE WORK
+=======================
+
+The Work has the status `author-maintained' if the Copyright Holder
+explicitly and prominently states near the primary copyright notice in
+the Work that the Work can only be maintained by the Copyright Holder
+or simply that it is `author-maintained'.
+
+The Work has the status `maintained' if there is a Current Maintainer
+who has indicated in the Work that they are willing to receive error
+reports for the Work (for example, by supplying a valid e-mail
+address). It is not required for the Current Maintainer to acknowledge
+or act upon these error reports.
+
+The Work changes from status `maintained' to `unmaintained' if there
+is no Current Maintainer, or the person stated to be Current
+Maintainer of the work cannot be reached through the indicated means
+of communication for a period of six months, and there are no other
+significant signs of active maintenance.
+
+You can become the Current Maintainer of the Work by agreement with
+any existing Current Maintainer to take over this role.
+
+If the Work is unmaintained, you can become the Current Maintainer of
+the Work through the following steps:
+
+ 1.  Make a reasonable attempt to trace the Current Maintainer (and
+     the Copyright Holder, if the two differ) through the means of
+     an Internet or similar search.
+
+ 2.  If this search is successful, then enquire whether the Work
+     is still maintained.
+
+  a. If it is being maintained, then ask the Current Maintainer
+     to update their communication data within one month.
+     
+  b. If the search is unsuccessful or no action to resume active
+     maintenance is taken by the Current Maintainer, then announce
+     within the pertinent community your intention to take over
+     maintenance.  (If the Work is a LaTeX work, this could be
+     done, for example, by posting to comp.text.tex.)
+
+ 3a. If the Current Maintainer is reachable and agrees to pass
+     maintenance of the Work to you, then this takes effect
+     immediately upon announcement.
+     
+  b. If the Current Maintainer is not reachable and the Copyright
+     Holder agrees that maintenance of the Work be passed to you,
+     then this takes effect immediately upon announcement.  
+    
+ 4.  If you make an `intention announcement' as described in 2b. above
+     and after three months your intention is challenged neither by
+     the Current Maintainer nor by the Copyright Holder nor by other
+     people, then you may arrange for the Work to be changed so as
+     to name you as the (new) Current Maintainer.
+     
+ 5.  If the previously unreachable Current Maintainer becomes
+     reachable once more within three months of a change completed
+     under the terms of 3b) or 4), then that Current Maintainer must
+     become or remain the Current Maintainer upon request provided
+     they then update their communication data within one month.
+
+A change in the Current Maintainer does not, of itself, alter the fact
+that the Work is distributed under the LPPL license.
+
+If you become the Current Maintainer of the Work, you should
+immediately provide, within the Work, a prominent and unambiguous
+statement of your status as Current Maintainer.  You should also
+announce your new status to the same pertinent community as
+in 2b) above.
+
+
+WHETHER AND HOW TO DISTRIBUTE WORKS UNDER THIS LICENSE
+======================================================
+
+This section contains important instructions, examples, and
+recommendations for authors who are considering distributing their
+works under this license.  These authors are addressed as `you' in
+this section.
+
+Choosing This License or Another License
+----------------------------------------
+
+If for any part of your work you want or need to use *distribution*
+conditions that differ significantly from those in this license, then
+do not refer to this license anywhere in your work but, instead,
+distribute your work under a different license.  You may use the text
+of this license as a model for your own license, but your license
+should not refer to the LPPL or otherwise give the impression that
+your work is distributed under the LPPL.
+
+The document `modguide.tex' in the base LaTeX distribution explains
+the motivation behind the conditions of this license.  It explains,
+for example, why distributing LaTeX under the GNU General Public
+License (GPL) was considered inappropriate.  Even if your work is
+unrelated to LaTeX, the discussion in `modguide.tex' may still be
+relevant, and authors intending to distribute their works under any
+license are encouraged to read it.
+
+A Recommendation on Modification Without Distribution
+-----------------------------------------------------
+
+It is wise never to modify a component of the Work, even for your own
+personal use, without also meeting the above conditions for
+distributing the modified component.  While you might intend that such
+modifications will never be distributed, often this will happen by
+accident -- you may forget that you have modified that component; or
+it may not occur to you when allowing others to access the modified
+version that you are thus distributing it and violating the conditions
+of this license in ways that could have legal implications and, worse,
+cause problems for the community.  It is therefore usually in your
+best interest to keep your copy of the Work identical with the public
+one.  Many works provide ways to control the behavior of that work
+without altering any of its licensed components.
+
+How to Use This License
+-----------------------
+
+To use this license, place in each of the components of your work both
+an explicit copyright notice including your name and the year the work
+was authored and/or last substantially modified.  Include also a
+statement that the distribution and/or modification of that
+component is constrained by the conditions in this license.
+
+Here is an example of such a notice and statement:
+
+  %% pig.dtx
+  %% Copyright 2005 M. Y. Name
+  %
+  % 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 2005/12/01 or later.
+  %
+  % This work has the LPPL maintenance status `maintained'.
+  % 
+  % The Current Maintainer of this work is M. Y. Name.
+  %
+  % This work consists of the files pig.dtx and pig.ins
+  % and the derived file pig.sty.
+
+Given such a notice and statement in a file, the conditions
+given in this license document would apply, with the `Work' referring
+to the three files `pig.dtx', `pig.ins', and `pig.sty' (the last being
+generated from `pig.dtx' using `pig.ins'), the `Base Interpreter'
+referring to any `LaTeX-Format', and both `Copyright Holder' and
+`Current Maintainer' referring to the person `M. Y. Name'.
+
+If you do not want the Maintenance section of LPPL to apply to your
+Work, change `maintained' above into `author-maintained'.  
+However, we recommend that you use `maintained', as the Maintenance
+section was added in order to ensure that your Work remains useful to
+the community even when you can no longer maintain and support it
+yourself.
+
+Derived Works That Are Not Replacements
+---------------------------------------
+
+Several clauses of the LPPL specify means to provide reliability and
+stability for the user community. They therefore concern themselves
+with the case that a Derived Work is intended to be used as a
+(compatible or incompatible) replacement of the original Work. If
+this is not the case (e.g., if a few lines of code are reused for a
+completely different task), then clauses 6b and 6d shall not apply.
+
+
+Important Recommendations
+-------------------------
+
+ Defining What Constitutes the Work
+
+   The LPPL requires that distributions of the Work contain all the
+   files of the Work.  It is therefore important that you provide a
+   way for the licensee to determine which files constitute the Work.
+   This could, for example, be achieved by explicitly listing all the
+   files of the Work near the copyright notice of each file or by
+   using a line such as:
+
+    % This work consists of all files listed in manifest.txt.
+   
+   in that place.  In the absence of an unequivocal list it might be
+   impossible for the licensee to determine what is considered by you
+   to comprise the Work and, in such a case, the licensee would be
+   entitled to make reasonable conjectures as to which files comprise
+   the Work.
+
diff --git a/Master/texmf-dist/doc/latex/hfutexam/README.md b/Master/texmf-dist/doc/latex/hfutexam/README.md
new file mode 100644
index 00000000000..6162cde72d4
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/README.md
@@ -0,0 +1,7 @@
+# hfutexam: exam class for Hefei University of Technology (China)
+
+The package provides an exam class for Hefei University of Technology (China). 
+
++ Package: An exam class for Jinan University
++ Author:  Shenxing Zhang <zhangshenxing@hfut.edu.cn>
++ License: The LaTeX Project Public License 1.3c
diff --git a/Master/texmf-dist/doc/latex/hfutexam/hfutexam.pdf b/Master/texmf-dist/doc/latex/hfutexam/hfutexam.pdf
new file mode 100644
index 00000000000..47d4a72aaba
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/hfutexam.pdf
diff --git a/Master/texmf-dist/doc/latex/hfutexam/hfutexam.tex b/Master/texmf-dist/doc/latex/hfutexam/hfutexam.tex
new file mode 100644
index 00000000000..ffadfa29594
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/hfutexam.tex
@@ -0,0 +1,270 @@
+\documentclass{hfutexam}
+
+%% 示例所需的自定义命令
+\newcommand{\diff}{\,\mathrm{d}}
+\usetikzlibrary{arrows.meta, overlay-beamer-styles}
+\newfontfamily\couriernew{Courier New}
+\usepackage{enumitem}
+\usepackage{tcolorbox}
+\usepackage{listings}
+\makeatletter
+\definecolor{winered}{rgb}{0.5,0,0}
+\definecolor{lightgrey}{rgb}{0.9,0.9,0.9}
+\definecolor{frenchplum}{RGB}{190,20,83}
+\lstset{language=[LaTeX]TeX,
+	basicstyle=\couriernew,
+	texcsstyle=*\color{winered},
+	mathescape,
+	breaklines=true,
+	keywordstyle=\color{winered},
+	commentstyle=\color{green!70!black},
+	stringstyle=\color{green!50!blue},
+	frame=single,
+	tabsize=3,
+	framerule=0.5pt,
+	columns=flexible,
+	backgroundcolor=\color{black!5},
+	morekeywords={\diff, \maketitle, \titlesep, \BiaoTi, \XueNian, \XueQi, \KeChengDaiMa, \KeChengMingCheng, \XueFen, \KeChengXingZhi, \KaoShiXingShi, \ZhuanYeBanJi, \KaoShiRiQi, \MingTiJiaoShi, \XiZhuRenQianMing, \tigan, \scorebox, \score, \Score, \fillblank, \xx, \notice, \xuanzeti, \yihang, \erhang, \sihang, XeLaTeX},
+	keywordstyle=\color{winered},
+	morekeywords=[2]{hfutexam, shijuan, datizhi, cankaodaan, simple, nofangzheng, flalign, 5cm, enumerate, align},
+	keywordstyle=[2]\color{blue},
+}
+\tcbset{
+	colback=white,
+	colframe=blue,
+	boxrule=0.5pt,
+	arc=0pt,
+}
+\makeatother
+
+\begin{document}
+\BiaoTi{合肥工业大学试卷（A）}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年6月18日8:00-10:00}
+\MingTiJiaoShi{集体}
+\XiZhuRenQianMing{}
+
+
+\tigan{一、模板选项}
+
+\indent
+本模板 (2022/12/04 v1.5) 旨在为将合肥工业大学试卷的 word 格式转为\LaTeX{}格式.
+使用时, 只需在文档开头写上
+\begin{lstlisting}
+\documentclass[shijuan]{hfutexam}
+\end{lstlisting}
+即可使用.
+需要使用~{\color{blue}{\lstinline|UTF-8|}} 编码, 并使用 \lstinline|XeLaTeX| 至少编译两次, 以正确生成页码.
+
+\indent
+可使用的选项为: \lstinline|shijuan| (试卷), \lstinline|datizhi| (答题纸), \lstinline|cankaodaan| (参考答案) 和 \lstinline|simple| (简易模式). 如果留空则为默认值 \lstinline|shijuan| (试卷).
+\begin{enumerate}
+\item 试卷/答题纸/参考答案三个选项下页面会设置为 A3 大小, 三种情形的页眉页脚显示的内容以及标题的文字间隔有所不同.
+\item 简易模式选项下页面会设置为 A4 大小, 页眉页脚也较为简单. 此时需要使用命令 \lstinline|\maketitle| 来生成标题.
+一般用于保存(多张)试卷的内容，或者便于打印使用.
+\item 标题默认使用方正字体, 因此请在使用前先安装字体: {\bfseries\titlesongti 方正小标宋}和{\bfseries\titlefangsong 方正仿宋}(右键选择为所有用户安装), 否则请使用选项 \lstinline|nofangzheng| (采用新宋体和仿宋代替).
+\end{enumerate}
+
+\tigan{二、试卷信息}
+
+\indent
+通过下述命令来设置试卷信息.
+
+\textit{\color{blue}{试卷信息示例:}}
+\begin{lstlisting}
+\BiaoTi{合肥工业大学试卷（A）} % 试卷标题, 一般为: 合肥工业大学试卷（A）或（B)
+\XueNian{2021}{2022}                 % 学年起始和结束, 一般为相差 1 的 4 位数字
+\XueQi{二}                           % 学期, 一般为: 一, 二
+\KeChengDaiMa{034Y01}                % 课程代码
+\KeChengMingCheng{数学（下）} % 课程名称
+\XueFen{5}                            % 学分
+\KeChengXingZhi{必修}	             	% 课程性质, 只能为: 必修, 选修, 限修
+\KaoShiXingShi{闭卷}	                 	% 考试形式, 只能为: 开卷, 闭卷
+\ZhuanYeBanJi{少数民族预科班} 		% 专业班级, 一般不需要填写
+\KaoShiRiQi{2022年6月18日8:00-10:00} % 考试日期
+\MingTiJiaoShi{集体}                       % 命题教师
+\XiZhuRenQianMing{dengbing.png}  % 系主任签名
+\end{lstlisting}
+其中系主任签名处需要填写相应的图片名, 若不设置或设置为空则不显示.
+
+其它选项默认均为空, 可根据需要只填部分内容.
+
+\newpage
+\tigan{三、命令}
+\begin{enumerate}
+\item \lstinline|\tigan{三、命令}| 用于生成题干, 字体相对较大, 且为黑体. 小题建议使用~{\color{blue}\lstinline|enumerate|} 环境来生成.
+\item \hspace{-8mm}\scorebox\hspace{8mm}\lstinline|\scorebox| 用于生成打分框, 请放置在答题纸一行的开头使用.
+\vspace{-2mm}
+\item \lstinline|\notice| 用于生成答题纸提示信息, 请放置在答题纸的正文开始处.
+\item 答题纸中可能需要设置一定高度的空白, 使用命令 \lstinline|\hspace{5cm}| 之类的命令即可. 也可以使用 \lstinline|\newpage| 换到新的一页(或分栏).
+\end{enumerate}
+
+\tigan{填空题相关}
+\begin{enumerate}[resume]
+\item \lstinline|\fillblank[长度][最低高度]{内容}| 用于生成填空题的空白, 内容可以为空. 其中长度默认值是~{\color{blue}{\lstinline|3.5cm|}}, 最低高度默认值是~{\color{blue}{\lstinline|1cm|}} (答题纸和参考答案)或~{\color{blue}{\lstinline|0.5cm|}} (其它).
+\end{enumerate}
+
+\textit{\color{blue}{填空题示例:}}
+\begin{lstlisting}
+\textbf{请将你的答案对应填在横线上：}
+
+\textbf{1.} \fillblank{}, 
+\textbf{2.} \fillblank[5cm]{}, 
+\textbf{3.} \fillblank{}.
+\end{lstlisting}
+
+\begin{tcolorbox}
+\textbf{请将你的答案对应填在横线上：}
+
+\textbf{1.} \fillblank[3.5cm][1cm]{}, 
+\textbf{2.} \fillblank[5cm][1cm]{}, 
+\textbf{3.} \fillblank[3.5cm][1cm]{}.
+\end{tcolorbox}
+
+\tigan{选择题相关}
+\begin{enumerate}[resume]
+\item \lstinline|\xx{选项}{选项}{选项}{选项}| 用于生成选择题的选项, 直接在选择题题干后使用即可. 该命令会自动根据选项长度设置行数. 只支持四个选项, 选项会自动带上 ABCD.
+\item 如果想要手动改变每行显示的选项数, 可使用命令 \lstinline|\xx[每行显示的选项数]{选项}{选项}{选项}{选项}|, 每行只能显示 1, 2 或 4 个选项.
+\item \lstinline|\xuanzeti{题号}{答案}| 用于生成答题纸选择题的答题区域, 或参考答案选择题的答案区域.
+\end{enumerate}
+
+\textit{\color{blue}{选择题示例:}}
+\begin{lstlisting}
+\begin{enumerate}
+\item 柳宗元的《江雪》包含下面哪一句? (~~~~)
+\xx[2]{一山鸟飞绝}{百山鸟飞绝}{千山鸟飞绝}{亿山鸟飞绝}
+\item 张志和的《渔歌子》是(~~~~).
+\xx{东塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+{南塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+{西塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+{北塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+\end{enumerate}
+\end{lstlisting}
+
+\begin{tcolorbox}
+\begin{enumerate}
+\item 柳宗元的《江雪》包含下面哪一句? (~~~~).
+\xx[2]{一山鸟飞绝}{百山鸟飞绝}{千山鸟飞绝}{亿山鸟飞绝}
+\item 张志和的《渔歌子》是(~~~~).
+\xx{东塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+{南塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+{西塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+{北塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+\end{enumerate}
+\end{tcolorbox}
+
+%\newpage
+\textit{\color{blue}{选择题示例:}}
+\begin{lstlisting}
+\textbf{请将你所选择的字母 A, B, C, D 之一对应填在下列表格里：}
+
+\xuanzeti{\textbf{题号}}{\textbf{答案}}%
+\xuanzeti{1}{}\xuanzeti{2}{}\xuanzeti{3}{}\xuanzeti{4}{}
+\end{lstlisting}
+
+\begin{tcolorbox}
+\textbf{请将你所选择的字母 A, B, C, D 之一对应填在下列表格里：}
+
+\xuanzeti{\textbf{题号}}{\textbf{答案}}%
+\xuanzeti{1}{}\xuanzeti{2}{}\xuanzeti{3}{}\xuanzeti{4}{}
+\end{tcolorbox}
+
+\newpage
+\tigan{得分点相关}
+\begin{enumerate}[resume]
+\item \lstinline|\score{数值}| 用于在参考答案一行结尾处生成得分点的虚线.\score2
+\item \lstinline|\Score{(2分, 缺少常数得1分)}| 用于自定义得分说明.\Score{(2分, 缺少常数得1分)}
+\item 在公式中也可使用, 但是需要编译两次才会正常计算出虚线长度.
+\end{enumerate}
+\textit{\color{blue}{得分点示例:}}
+\begin{lstlisting}
+\[\int e^x\diff x=e^x+C. \Score{(4分, 缺少常数得2分)}\]
+\begin{align*}
+\int\sin x\diff x&=-\cos x+C, \Score{(4分, 缺少常数得2分)}\\
+\int_0^\pi(1+\sin x)\diff x&=\pi+2. \score5
+\end{align*}
+\end{lstlisting}
+
+\begin{tcolorbox}
+\[\int e^x\diff x=e^x+C. \Score{(4分, 缺少常数得2分)}\]
+\begin{align*}
+\int\sin x\diff x&=-\cos x+C, \Score{(4分, 缺少常数得2分)}\\
+\int_0^\pi(1+\sin x)\diff x&=\pi+2. \score5
+\end{align*}
+\end{tcolorbox}
+
+\textbf{如有疑问或建议, 欢迎联系我: {\color{red}{zhangshenxing@hfut.edu.cn}} 或 {\color{blue}{QQ362037052}}.}
+
+\textbf{CTAN: \color{blue}https://www.ctan.org/pkg/hfutexam}
+
+\newpage
+\tigan{一、填空题（每题3分，共18分）}
+\begin{enumerate}
+\item 如果 $f(x)>0$ 且 $\displaystyle\lim_{x\to\infty}f(x)=0$, 则 $\displaystyle\lim_{x\to\infty}\bigl[1+f(x)\bigr]^{1/f(x)}=$\fillblank{}.
+\item 设 $y=\sin(x^2+1)$, 则 $\diff y=$\fillblank{}.
+\item 极限 $\displaystyle\lim_{n\to\infty}\left(\frac1{n^2-1}+\frac2{n^2-2}+\cdots+\frac n{n^2-n}\right)=$\fillblank{}.
+\item 曲线 $y=2\ln(x+1)$ 在点 $(1,2\ln2)$ 处的切线方程为\fillblank{}.
+\item 若 $e^{y-1}=1+xy$, 则 $\dfrac{\diff y}{\diff x}\bigg|_{x=0}=$\fillblank{}.
+\item 如果函数 $f(x)$ 的定义域是 $(0,+\infty)$, 且 $x=0$ 是曲线 $y=f(x)$ 的垂直渐近线, 那么 $\displaystyle\lim_{x\to0^+}\frac1{f(x)}=$\fillblank{}.
+\end{enumerate}
+
+\tigan{二、选择题（每题3分，共18分）}
+\begin{enumerate}
+\item 当 $x\to+\infty$ 时, $\dfrac1x$ 和(~~~~)是等价无穷小.
+% 自动根据选项长度设置行数
+\xx{$\sin\dfrac1x$}{$\sin x$}{$e^{-x}$}{$e^{1/x}$}
+\item 若当 $x\to0$ 时, $\arctan(e^x-1)\cdot(\cos x-1)$ 和 $x^n$ 是同阶无穷小, 则 $n=$(~~~~).
+\xx{$0$}{$1$}{$2$}{$3$}
+\item 设 $f(x)=\arctan\dfrac1{x(x-1)^2}$, 则 $x=0$ 是 $f(x)$ 的(~~~~).
+\xx{可去间断点}{跳跃间断点}{第二类间断点}{连续点}
+\item
+\begin{tikzpicture}[overlay,xshift=12.5cm,yshift=-3cm]
+\draw[-Stealth,thick](-3,0)--(3,0);
+\draw[-Stealth,thick](0,-0.8)--(0,2.5);
+\draw[very thick,smooth,domain=-55:55] plot ({\x/50-1.3}, {tan(\x)*tan(\x)});
+\draw[very thick,smooth,domain=0.15:2] plot ({\x}, {-ln(\x)});
+\draw
+	(-0.3,-0.3) node {$O$}
+	(2.8,-0.3) node {$x$}
+	(-0.3,2.3) node {$y$};
+\end{tikzpicture}
+设 $f(x)$ 是定义在 $(-\infty,+\infty)$ 上的连续函数, 且 $f'(x)$ 的图像如下图所示, 则 $f(x)$ 有(~~~~).
+% 手动设置为每行1个
+\xx[1]{一个极大值点，没有极小值点}{没有极大值点，一个极小值点}{一个极大值点和一个极小值点}{一个极大值点和两个极小值点}
+\newpage
+\item 设函数 $f(x)$ 在点 $x=0$ 处可导, 且 $f(0)=0$, 则 $\displaystyle\lim_{x\to0}\frac{f(x^{2022})+x^{2021}f(x)}{x^{2022}}=$(~~~~).
+\xx{$0$}{$f'(0)$}{$2f'(0)$}{$2022f'(0)$}
+\item 如果点 $(x_0,y_0)$ 是曲线 $y=f(x)$ 的拐点, 则 $f''(x_0)=$(~~~~).
+\xx{$0$}{$\infty$}{不存在}{$0$ 或不存在}
+\end{enumerate}
+
+\tigan{三、解答题（每题8分，共64分）}
+\begin{enumerate}
+\item 求极限 $\displaystyle\lim_{x\to-1}\frac{x^2-1}{x^2+3x+2}$.
+\item 求极限 $\displaystyle\lim_{x\to0}\frac{e^x-1-x}{\arcsin x^2}$.
+\item 设 $\begin{cases}x=t^2+t&\\y=t^3+t&\end{cases}$, 求 $\dfrac{\diff y}{\diff x}$ 和 $\dfrac{\diff^2 y}{\diff x^2}$.
+\item 设 $f(x)=\begin{cases}x\arctan\dfrac1x,&x<0,\\x^2+ax+b,&x\ge0.\end{cases}$
+求常数 $a,b$ 使得函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导, 并求出此时曲线 $y=f(x)$ 的渐近线.
+\item	求函数 $f(x)=x^3-x^2-x$ 在区间 $[-2,2]$ 上的最大值和最小值.
+\item 证明: 当 $-\dfrac\pi2<x_1<x_2<\dfrac\pi2$ 时, $\tan x_2-\tan x_1\ge x_2-x_1$.
+\item 设函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导, 且 $f(1)=0$.
+证明: 存在 $\xi\in(0,1)$ 使得 $\xi f'(\xi)+2022f(\xi)=0$.
+\item 设函数 $f(x)=\ln x+\dfrac2{x^2}, x\in(0,+\infty)$. 求
+\begin{enumerate}
+\item[(1)] 函数 $f(x)$ 的增减区间及极值;
+\item[(2)] 曲线 $y=f(x)$ 的凹凸区间及拐点.
+\end{enumerate}
+\end{enumerate}
+
+\end{document}
+
+
+
+
diff --git a/Master/texmf-dist/doc/latex/hfutexam/hfutexam_cankaodaan.pdf b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_cankaodaan.pdf
new file mode 100644
index 00000000000..3776c11ac84
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_cankaodaan.pdf
diff --git a/Master/texmf-dist/doc/latex/hfutexam/hfutexam_cankaodaan.tex b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_cankaodaan.tex
new file mode 100644
index 00000000000..f9f20890d59
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_cankaodaan.tex
@@ -0,0 +1,158 @@
+\documentclass[cankaodaan]{hfutexam}
+\usepackage{extarrows}
+\newcommand{\diff}{\,\mathrm{d}}
+
+\begin{document}
+\BiaoTi{合肥工业大学试卷参考答案（A）}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年6月18日8:00-10:00}
+\MingTiJiaoShi{集体}
+%\XiZhuRenQianMing{dengbing.png}
+
+\tigan{一、填空题（每小题3分，共18分）}
+
+\textbf{请将你的答案对应填在横线上：}
+
+\textbf{1.} \fillblank{$e$}, 
+\textbf{2.} \fillblank{$2x\cos(x^2+1)\diff x$}, 
+\textbf{3.} \fillblank{$\dfrac12$}, 
+
+\textbf{4.} \fillblank{$y=x-1+2\ln 2$},
+\textbf{5.} \fillblank{$1$},
+\textbf{6.} \fillblank{$0$}.
+
+\tigan{二、选择题（每小题3分，共18分）}
+
+\textbf{请将你所选择的字母 A, B, C, D 之一对应填在下列表格里：}
+
+\xuanzeti{\textbf{题号}}{\textbf{答案}}%
+\xuanzeti{1}{A}%
+\xuanzeti{2}{D}%
+\xuanzeti{3}{B}%
+\xuanzeti{4}{A}%
+\xuanzeti{5}{C}%
+\xuanzeti{6}{D}
+
+\tigan{三、解答题（每小题8分，共64分）}
+% 得分点命令 \score1, 得分点长度是自动的
+% \Score{(2分, 缺少常数得1分)} 用于自定义得分说明
+
+\textbf{1. （8分）【解】}
+\begin{align*}
+\lim_{x\to-1}\frac{x^2-1}{x^2+3x+2}&=\lim_{x\to-1}\frac{(x-1)(x+1)}{(x+2)(x+1)} \score3\\
+&=\lim_{x\to-1}\frac{x-1}{x+2} \score3\\
+&=\frac{-2}1=-2. \score2
+\end{align*}
+
+\textbf{2. （8分）【解】}
+\begin{align*}
+\lim_{x\to0}\frac{e^x-1-x}{\arcsin x^2}&=\lim_{x\to0}\frac{e^x-1-x}{x^2} \score3\\
+&\xlongequal{\text{洛必达}}\lim_{x\to0}\frac{e^x-1}{2x} \score3\\
+&=\lim_{x\to0}\frac{x}{2x}=\frac12. \score2
+\end{align*}
+
+\newpage
+
+\textbf{3. （8分）【解】}
+\begin{align*}
+\frac{\diff y}{\diff x}&=\frac{\diff y/\diff t}{\diff x/\diff t} \score2\\
+&=\frac{3t^2+1}{2t+1}, \score2\\
+\frac{\diff^2 y}{\diff x^2}&=\frac{\diff y'/\diff t}{\diff x/\diff t} \score2\\
+&=\frac{6t(2t+1)-(3t^2+1)2}{(2t+1)^3}=\frac{6t^2+6t-2}{(2t+1)^3}. \score2
+\end{align*}
+
+\textbf{4. （8分）【解】}
+
+\indent 由于 $f(x)$ 在 $x=0$ 处连续, 因此
+\begin{align*}
+f(0)&=f(0^+) \score1\\
+		&=b=\lim_{x\to0^-}x\arctan\frac1x=0\times\left(-\frac\pi2\right)=0. \score1
+\end{align*}
+
+\indent 由于 $f(x)$ 在 $x=0$ 处可导, 因此
+\begin{align*}
+f'_-(0)&=f'_+(0), \score1\\
+f'_-(0)&=\lim_{x\to0^-}\frac{x\arctan\frac1x}x=\lim_{x\to0^-}\arctan\frac1x=-\frac\pi2 \score1\\
+f'_+(0)&=(2x+a)|_{x=0}=a, \score1
+\end{align*}
+因此 $a=-\dfrac\pi2$. \score1
+
+\indent 由于
+\begin{align*}
+\lim_{x\to+\infty}\frac yx&=\lim_{x\to+\infty}\left(x-\frac\pi2\right)=+\infty, \score1\\
+\lim_{x\to-\infty}\frac yx&=\lim_{x\to-\infty}\arctan\frac1x=0,\\
+\lim_{x\to-\infty}y&=\lim_{x\to-\infty}x\arctan\frac1x=\lim_{t\to0^-}\frac{\arctan t}t=1,
+\end{align*}
+因此曲线 $y=f(x)$ 的渐近线只有 $y=1$. \score1
+
+\newpage
+
+\textbf{5. （8分）【解】}
+
+\indent 由
+\[f'(x)=3x^2-2x-1=(3x+1)(x-1)=0 \score2\]
+可得驻点 $x=-\dfrac13,1$. \score2
+
+\indent 由于
+\[f(-2)=-10,\quad f(2)=2,\quad f\left(-\frac13\right)=\frac5{27},\quad f(1)=-1, \score2\]
+因此最大值为 $2$, 最小值为 $-10$. \score2
+
+\textbf{6. （8分）【证明】}
+
+\textbf{证法一}: 设 $f(x)=\tan x-x$, 则 \score2
+\begin{align*}
+f'(x)=\frac1{\cos^2x}-1=\tan^2x\ge0. \score2
+\end{align*}
+因此 $f(x)$ 在 $\left(-\dfrac\pi2,\dfrac\pi2\right)$ 上单调递增, 从而 \score2
+\begin{align*}
+f(x_2)\ge f(x_1),\quad\tan x_2-\tan x_1\ge x_2-x_1. \score2
+\end{align*}
+
+\textbf{证法二}: 设 $f(x)=\tan x$, 则 $f(x)$ 在 $[x_1,x_2]$ 上连续, $(x_1,x_2)$ 内可导. \score2
+
+\indent 由拉格朗日中值定理, 存在 $\xi\in(x_1,x_2)$ 使得
+\begin{align*}
+\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(\xi), \score2
+\end{align*}
+即
+\begin{align*}
+\frac{\tan x_2-\tan x_1}{x_2-x_1}=\frac1{\cos^2\xi}\ge1. \score2
+\end{align*}
+所以 $\tan x_2-\tan x_1\ge x_2-x_1$. \score2
+\newpage
+
+\textbf{7. （8分）【证明】}
+
+\indent 设 $F(x)=x^{2022}f(x)$, \score2\\
+则 $F(x)$ 在 $[0,1]$ 上连续, $(0,1)$ 内可导, \score1\\
+且 $F(0)=0,F(1)=f(1)=0$. \score1
+
+\indent 由罗尔中值定理, 存在 $\xi\in(0,1)$ 使得 $F'(\xi)=0$. \score2\\
+由于 $F'(x)=x^{2022}f'(x)+2022x^{2021}f(x)$ 且 $\xi\neq0$, \score1\\
+所以 $\xi f'(\xi)+2022f(\xi)=1$. \score1
+
+
+\textbf{8. （8分）【解】}
+
+(1) 
+\[f'(x)=\frac1x-\frac4{x^3}=\frac{x^2-4}{x^3}=\frac{(x+2)(x-2)}{x^3}. \score1\]
+当 $0<x<2$ 时, $f'(x)<0$. 当 $x>2$ 时, $f'(x)>0$. \score1\\
+因此 $(0,2]$ 是 $f(x)$ 的单减区间, $[2,+\infty)$ 是 $f(x)$ 的单增区间. \Score{(1分, 写成开区间不扣分)}\\
+所以 $f(x)$ 只有唯一的极小值 $f(2)=\ln2+\dfrac12$. \score1
+
+(2) 
+\[f''(x)=-\frac1{x^2}+\frac{12}{x^4}=-\frac{x^2-12}{x^4}=-\frac{(x-2\sqrt3)(x+2\sqrt3)}{x^4}. \score1\]
+当 $0<x<2\sqrt3$ 时, $f''(x)>0$. 当 $x>2\sqrt3$ 时, $f''(x)<0$. \score1\\
+因此 $(0,2\sqrt3]$ 是曲线 $y=f(x)$ 的凹区间, \\
+$[2\sqrt3,+\infty)$ 是曲线 $y=f(x)$ 的凸区间, \Score{(1分, 写成开区间不扣分)}\\
+拐点为 $\left(2\sqrt3,\ln(2\sqrt3)+\dfrac16\right)$. \score1
+
+\end{document}
+
diff --git a/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.pdf b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.pdf
new file mode 100644
index 00000000000..ad9dc41adef
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.pdf
diff --git a/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.tex b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.tex
new file mode 100644
index 00000000000..d5ad8907e45
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.tex
@@ -0,0 +1,73 @@
+\documentclass[datizhi]{hfutexam}
+\newcommand{\diff}{\,\mathrm{d}}
+
+\begin{document}
+\BiaoTi{合肥工业大学考试专用答卷纸（A）}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年6月18日8:00-10:00}
+\MingTiJiaoShi{集体}
+%\XiZhuRenQianMing{dengbing.png}
+
+
+\notice
+
+%\scorebox 为打分框, 必须放在行首
+\scorebox\tigan{一、填空题（每小题3分，共18分）}
+
+\textbf{请将你的答案对应填在横线上：}
+
+\textbf{1.} \fillblank{}, 
+\textbf{2.} \fillblank{}, 
+\textbf{3.} \fillblank{}, 
+
+\textbf{4.} \fillblank{}, 
+\textbf{5.} \fillblank{}, 
+\textbf{6.} \fillblank{}.
+
+\scorebox\tigan{二、选择题（每小题3分，共18分）}
+
+\textbf{请将你所选择的字母 A, B, C, D 之一对应填在下列表格里：}
+
+\xuanzeti{\textbf{题号}}{\textbf{答案}}%
+\xuanzeti{1}{}%
+\xuanzeti{2}{}%
+\xuanzeti{3}{}%
+\xuanzeti{4}{}%
+\xuanzeti{5}{}%
+\xuanzeti{6}{}
+
+\tigan{三、解答题（每小题8分，共64分）}
+
+\scorebox\textbf{1. （8分）【解】}
+% \vspace 用于生成一定高度的空白, \newpage 直接换页
+\vspace{3cm}
+
+\scorebox\textbf{2. （8分）【解】}
+\newpage
+
+\scorebox\textbf{3. （8分）【解】}
+\vspace{7cm}
+
+\scorebox\textbf{4. （8分）【解】}
+\newpage
+
+\scorebox\textbf{5. （8分）【解】}
+\vspace{7cm}
+
+\scorebox\textbf{6. （8分）【证明】}
+\newpage
+
+\scorebox\textbf{7. （8分）【证明】}
+\vspace{7cm}
+
+\scorebox\textbf{8. （8分）【解】}
+
+\end{document}
+
diff --git a/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.pdf b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.pdf
new file mode 100644
index 00000000000..97f57ce4027
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.pdf
diff --git a/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.tex b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.tex
new file mode 100644
index 00000000000..e4a5617329f
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.tex
@@ -0,0 +1,82 @@
+\documentclass{hfutexam}
+\newcommand{\diff}{\,\mathrm{d}}
+\usetikzlibrary{arrows.meta, overlay-beamer-styles}
+
+\begin{document}
+\BiaoTi{合肥工业大学试卷（A）}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年6月18日8:00-10:00}
+\MingTiJiaoShi{集体}
+\XiZhuRenQianMing{}
+% \XiZhuRenQianMing{dengbing.png}
+
+\tigan{一、填空题（每题3分，共18分）}
+\begin{enumerate}
+\item 如果 $f(x)>0$ 且 $\displaystyle\lim_{x\to\infty}f(x)=0$, 则 $\displaystyle\lim_{x\to\infty}\bigl[1+f(x)\bigr]^{1/f(x)}=$\fillblank{}.
+\item 设 $y=\sin(x^2+1)$, 则 $\diff y=$\fillblank{}.
+\item 极限 $\displaystyle\lim_{n\to\infty}\left(\frac1{n^2-1}+\frac2{n^2-2}+\cdots+\frac n{n^2-n}\right)=$\fillblank{}.
+\item 曲线 $y=2\ln(x+1)$ 在点 $(1,2\ln2)$ 处的切线方程为\fillblank{}.
+\item 若 $e^{y-1}=1+xy$, 则 $\dfrac{\diff y}{\diff x}\bigg|_{x=0}=$\fillblank{}.
+\item 如果函数 $f(x)$ 的定义域是 $(0,+\infty)$, 且 $x=0$ 是曲线 $y=f(x)$ 的垂直渐近线, 那么 $\displaystyle\lim_{x\to0^+}\frac1{f(x)}=$\fillblank{}.
+\end{enumerate}
+
+\tigan{二、选择题（每题3分，共18分）}
+\begin{enumerate}
+\item 当 $x\to+\infty$ 时, $\dfrac1x$ 和(~~~~)是等价无穷小.
+% 自动根据选项长度设置行数
+\xx{$\sin\dfrac1x$}{$\sin x$}{$e^{-x}$}{$e^{1/x}$}
+\item 若当 $x\to0$ 时, $\arctan(e^x-1)\cdot(\cos x-1)$ 和 $x^n$ 是同阶无穷小, 则 $n=$(~~~~).
+\xx{$0$}{$1$}{$2$}{$3$}
+\item 设 $f(x)=\arctan\dfrac1{x(x-1)^2}$, 则 $x=0$ 是 $f(x)$ 的(~~~~).
+\xx{可去间断点}{跳跃间断点}{第二类间断点}{连续点}
+\item
+\begin{tikzpicture}[overlay,xshift=13cm,yshift=-3.5cm]
+\draw[-Stealth,thick](-3,0)--(3,0);
+\draw[-Stealth,thick](0,-1)--(0,3);
+\draw[very thick,smooth,domain=-55:55] plot ({\x/50-1.3}, {tan(\x)*tan(\x)});
+\draw[very thick,smooth,domain=0.15:2] plot ({\x}, {-ln(\x)});
+\draw
+	(-0.3,-0.3) node {$O$}
+	(2.8,-0.3) node {$x$}
+	(-0.3,2.8) node {$y$};
+\end{tikzpicture}
+设 $f(x)$ 是定义在 $(-\infty,+\infty)$ 上的连续函数, 且 $f'(x)$ 的图像如下图所示, 则 $f(x)$ 有(~~~~).
+% 手动设置为每行1个
+\xx[1]{一个极大值点，没有极小值点}{没有极大值点，一个极小值点}{一个极大值点和一个极小值点}{一个极大值点和两个极小值点}
+\newpage
+\item 设函数 $f(x)$ 在点 $x=0$ 处可导, 且 $f(0)=0$, 则 $\displaystyle\lim_{x\to0}\frac{f(x^{2022})+x^{2021}f(x)}{x^{2022}}=$(~~~~).
+\xx{$0$}{$f'(0)$}{$2f'(0)$}{$2022f'(0)$}
+\item 如果点 $(x_0,y_0)$ 是曲线 $y=f(x)$ 的拐点, 则 $f''(x_0)=$(~~~~).
+\xx{$0$}{$\infty$}{不存在}{$0$ 或不存在}
+\end{enumerate}
+
+\tigan{三、解答题（每题8分，共64分）}
+\begin{enumerate}
+\item 求极限 $\displaystyle\lim_{x\to-1}\frac{x^2-1}{x^2+3x+2}$.
+\item 求极限 $\displaystyle\lim_{x\to0}\frac{e^x-1-x}{\arcsin x^2}$.
+\item 设 $\begin{cases}x=t^2+t&\\y=t^3+t&\end{cases}$, 求 $\dfrac{\diff y}{\diff x}$ 和 $\dfrac{\diff^2 y}{\diff x^2}$.
+\item 设 $f(x)=\begin{cases}x\arctan\dfrac1x,&x<0,\\x^2+ax+b,&x\ge0.\end{cases}$
+求常数 $a,b$ 使得函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导, 并求出此时曲线 $y=f(x)$ 的渐近线.
+\item	求函数 $f(x)=x^3-x^2-x$ 在区间 $[-2,2]$ 上的最大值和最小值.
+\item 证明: 当 $-\dfrac\pi2<x_1<x_2<\dfrac\pi2$ 时, $\tan x_2-\tan x_1\ge x_2-x_1$.
+\item 设函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导, 且 $f(1)=0$.
+证明: 存在 $\xi\in(0,1)$ 使得 $\xi f'(\xi)+2022f(\xi)=0$.
+\item 设函数 $f(x)=\ln x+\dfrac2{x^2}, x\in(0,+\infty)$. 
+\begin{enumerate}
+\item[(1)] 函数 $f(x)$ 的增减区间及极值;
+\item[(2)] 曲线 $y=f(x)$ 的凹凸区间及拐点.
+\end{enumerate}
+\end{enumerate}
+
+\end{document}
+
+
+
+
diff --git a/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.pdf b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.pdf
new file mode 100644
index 00000000000..18b857b946c
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.pdf
diff --git a/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.tex b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.tex
new file mode 100644
index 00000000000..54175ece65f
--- /dev/null
+++ b/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.tex
@@ -0,0 +1,260 @@
+\documentclass[simple]{hfutexam}
+\newcommand{\diff}{\,\mathrm{d}}
+\usetikzlibrary{arrows.meta, overlay-beamer-styles}
+\RequirePackage{extarrows}
+
+\begin{document}
+\BiaoTi{合肥工业大学期中试卷}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年5月13日8:00-10:00}
+\MingTiJiaoShi{集体}
+\maketitle
+
+\begin{enumerate}
+\item \textbf{(10分)} 求函数 $\displaystyle f(x)=\ln\frac1{\sqrt{x^2-1}}+\arctan\frac1x$ 的定义域.
+\item \textbf{(5分)} 求函数 $\displaystyle y=\begin{cases}
+1/x,&x<0,\\1,&x=0,\\1+e^{-x},&x>0\end{cases}$ 的反函数.
+\item \textbf{(10分)} 求极限 $\displaystyle\lim_{x\to0^-}(1-x)^{1/x}$.
+\item \textbf{(5分)} 求极限 $\displaystyle\lim_{x\to-2}\frac{x^2-4}{x^3+8}$.
+\item \textbf{(5分)} 求极限 $\displaystyle\lim_{x\to0}\frac{\sin(e^{-x}-1)}{\arctan(1-\cos x)}$.
+\item \textbf{(5分)} 求极限 $\displaystyle\lim_{x\to0}\frac{\sqrt{1+2x-x^2}-\sqrt{1-2x+x^2}}x$. 
+\item \textbf{(5分)} 求极限 $\displaystyle\lim_{x\to\infty}\left(\cos\frac1x\right)^{\frac1{\ln(1+x^2)-2\ln x}}$.
+\item \textbf{(5分)} 求极限 $\displaystyle\lim_{x\to\infty}\left(\frac\pi{e^x-1}-\arctan\frac x2\right)$. 
+\item \textbf{(5分)} 求极限 $\displaystyle\lim_{n\to\infty}\left(\frac1{n^2+2}+\frac2{n^2+4}+\cdots+\frac n{n^2+2n}\right)$.
+\item \textbf{(5分)} 设 $a_1=4,a_{n+1}=\sqrt{a_n+6}$, 证明 $\displaystyle\lim_{n\to\infty}a_n$ 存在并求之.
+\item \textbf{(10分)} 证明 $e^x+x=4$ 在 $(0,+\infty)$ 内有零点.
+\item \textbf{(5分)} 设函数 $f(x)$ 在 $[-1,1]$ 上连续, 且 $f(-1)\le1\le f(1)$. 证明存在 $\xi\in[-1,1]$, 使得 $f(\xi)=\xi^2$.
+\item \textbf{(10分)} 求 $y=e^{x+1}\sin x-e^2\sin1$ 的导数.
+\item \textbf{(5分)} 求 $y=\arctan e^x$ 的导数.
+\item \textbf{(5分)} 求曲线 $y=\tan x$ 在点 $\left(-\dfrac\pi4,-1\right)$ 处的切线方程和法线方程.
+\item \textbf{(5分)} 设
+$\displaystyle f(x)=\begin{cases}\dfrac{e^{3x}-1}{\arctan x},&x<0,\\2x+a,&x\ge0\end{cases}$
+在 $x=0$ 处连续, 求常数 $a$.
+\end{enumerate}
+\newpage
+
+\BiaoTi{合肥工业大学试卷（A）}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年6月18日8:00-10:00}
+\MingTiJiaoShi{集体}
+\maketitle
+
+\tigan{一、填空题（每题3分，共18分）}
+\begin{enumerate}
+\item 如果 $f(x)>0$ 且 $\displaystyle\lim_{x\to\infty}f(x)=0$, 则 $\displaystyle\lim_{x\to\infty}\bigl[1+f(x)\bigr]^{1/f(x)}=$\fillblank{}.
+\item 设 $y=\sin(x^2+1)$, 则 $\diff y=$\fillblank{}.
+\item 极限 $\displaystyle\lim_{n\to\infty}\left(\frac1{n^2-1}+\frac2{n^2-2}+\cdots+\frac n{n^2-n}\right)=$\fillblank{}.
+\item 曲线 $y=2\ln(x+1)$ 在点 $(1,2\ln2)$ 处的切线方程为\fillblank{}.
+\item 若 $e^{y-1}=1+xy$, 则 $\dfrac{\diff y}{\diff x}\bigg|_{x=0}=$\fillblank{}.
+\item 如果函数 $f(x)$ 的定义域是 $(0,+\infty)$, 且 $x=0$ 是曲线 $y=f(x)$ 的垂直渐近线, 那么 $\displaystyle\lim_{x\to0^+}\frac1{f(x)}=$\fillblank{}.
+\end{enumerate}
+
+\tigan{二、选择题（每题3分，共18分）}
+\begin{enumerate}
+\item 当 $x\to+\infty$ 时, $\dfrac1x$ 和(~~~~)是等价无穷小.
+\xx{$\sin\dfrac1x$}{$\sin x$}{$e^{-x}$}{$e^{1/x}$}
+\item 若当 $x\to0$ 时, $\arctan(e^x-1)\cdot(\cos x-1)$ 和 $x^n$ 是同阶无穷小, 则 $n=$(~~~~).
+\xx{$0$}{$1$}{$2$}{$3$}
+\item 设 $f(x)=\arctan\dfrac1{x(x-1)^2}$, 则 $x=0$ 是 $f(x)$ 的(~~~~).
+\xx{可去间断点}{跳跃间断点}{第二类间断点}{连续点}
+\item 
+\begin{tikzpicture}[overlay,xshift=12cm,yshift=-3cm]
+\draw[-Stealth,thick](-3,0)--(3,0);
+\draw[-Stealth,thick](0,-0.8)--(0,2.5);
+\draw[very thick,smooth,domain=-55:55] plot ({\x/50-1.3}, {tan(\x)*tan(\x)});
+\draw[very thick,smooth,domain=0.15:2] plot ({\x}, {-ln(\x)});
+\draw
+	(-0.3,-0.3) node {$O$}
+	(2.8,-0.3) node {$x$}
+	(-0.3,2.3) node {$y$};
+\end{tikzpicture}
+设 $f(x)$ 是定义在 $(-\infty,+\infty)$ 上的连续函数, 且 $f'(x)$ 的图像如下图所示, 则 $f(x)$ 有(~~~~).
+\xx[1]{一个极大值点，没有极小值点}{没有极大值点，一个极小值点}{一个极大值点和一个极小值点}{一个极大值点和两个极小值点}
+\item 设 $f(x)$ 在点 $x=0$ 处可导, 且 $f(0)=0$, 则 $\displaystyle\lim_{x\to0}\frac{f(x^{2022})+x^{2021}f(x)}{x^{2022}}=$(~~~~).
+\xx{$0$}{$f'(0)$}{$2f'(0)$}{$2022f'(0)$}
+\item 如果点 $(x_0,y_0)$ 是曲线 $y=f(x)$ 的拐点, 则 $f''(x_0)=$(~~~~).
+\xx{$0$}{$\infty$}{不存在}{$0$ 或不存在}
+\end{enumerate}
+
+\tigan{三、解答题（每题8分，共64分）}
+\begin{enumerate}
+\item 求极限 $\displaystyle\lim_{x\to-1}\frac{x^2-1}{x^2+3x+2}$.
+\item 求极限 $\displaystyle\lim_{x\to0}\frac{e^x-1-x}{\arcsin x^2}$.
+\item 设 $\begin{cases}x=t^2+t&\\y=t^3+t&\end{cases}$, 求 $\dfrac{\diff y}{\diff x}$ 和 $\dfrac{\diff^2 y}{\diff x^2}$.
+\item 设 $f(x)=\begin{cases}x\arctan\dfrac1x,&x<0,\\x^2+ax+b,&x\ge0.\end{cases}$
+求常数 $a,b$ 使得函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导, 并求出此时曲线 $y=f(x)$ 的渐近线.
+\item	求函数 $f(x)=x^3-x^2-x$ 在区间 $[-2,2]$ 上的最大值和最小值.
+\item 证明: 当 $-\dfrac\pi2<x_1<x_2<\dfrac\pi2$ 时, $\tan x_2-\tan x_1\ge x_2-x_1$.
+\item 设函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导, 且 $f(1)=0$.
+证明: 存在 $\xi\in(0,1)$ 使得 $\xi f'(\xi)+2022f(\xi)=0$.
+\item 设函数 $f(x)=\ln x+\dfrac2{x^2}, x\in(0,+\infty)$. 求
+\begin{enumerate}
+\item[(1)] 函数 $f(x)$ 的增减区间及极值；
+\item[(2)] 曲线 $y=f(x)$ 的凹凸区间及拐点.
+\end{enumerate}
+\end{enumerate}
+
+\newpage
+\BiaoTi{合肥工业大学试卷参考答案（A）}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年6月18日8:00-10:00}
+\MingTiJiaoShi{集体}
+\maketitle
+
+\tigan{一、填空题（每小题3分，共18分）}
+
+\textbf{请将你的答案对应填在横线上：}
+
+\textbf{1.} \fillblank{$e$}, 
+\textbf{2.} \fillblank{$2x\cos(x^2+1)\diff x$}, 
+\textbf{3.} \fillblank{$\dfrac12$}, \\
+\textbf{4.} \fillblank{$y=x-1+2\ln 2$},
+\textbf{5.} \fillblank{$1$},
+\textbf{6.} \fillblank{$0$}.
+
+\tigan{二、选择题（每小题3分，共18分）}
+
+\textbf{请将你所选择的字母 A, B, C, D 之一对应填在下列表格里：}
+
+\xuanzeti{\textbf{题号}}{\textbf{答案}}%
+\xuanzeti{1}{A}%
+\xuanzeti{2}{D}%
+\xuanzeti{3}{B}%
+\xuanzeti{4}{A}%
+\xuanzeti{5}{C}%
+\xuanzeti{6}{D}
+
+\tigan{三、解答题（每小题8分，共64分）}
+
+\textbf{1. （8分）【解】}
+\vspace{-\baselineskip}
+
+\begin{align*}
+\lim_{x\to-1}\frac{x^2-1}{x^2+3x+2}&=\lim_{x\to-1}\frac{(x-1)(x+1)}{(x+2)(x+1)} \score3\\
+&=\lim_{x\to-1}\frac{x-1}{x+2} \score3\\
+&=\frac{-2}1=-2. \score2
+\end{align*}
+
+\textbf{2. （8分）【解】}
+\vspace{-\baselineskip}
+
+\begin{align*}
+\lim_{x\to0}\frac{e^x-1-x}{\arcsin x^2}&=\lim_{x\to0}\frac{e^x-1-x}{x^2} \score3\\
+&\xlongequal[]{\text{洛必达}}\lim_{x\to0}\frac{e^x-1}{2x} \score3\\
+&=\lim_{x\to0}\frac{x}{2x}=\frac12. \score2
+\end{align*}
+
+\textbf{3. （8分）【解】}
+\vspace{-\baselineskip}
+
+\begin{align*}
+\frac{\diff y}{\diff x}&=\frac{\diff y/\diff t}{\diff x/\diff t} \score2\\
+&=\frac{3t^2+1}{2t+1}, \score2\\
+\frac{\diff^2 y}{\diff x^2}&=\frac{\diff y'/\diff t}{\diff x/\diff t} \score2\\
+&=\frac{6t(2t+1)-(3t^2+1)2}{(2t+1)^3}=\frac{6t^2+6t-2}{(2t+1)^3}. \score2
+\end{align*}
+
+\newpage
+\textbf{4. （8分）【解】}
+
+\indent 由于 $f(x)$ 在 $x=0$ 处连续, 因此
+\begin{align*}
+f(0)&=f(0^+) \score1\\
+&=b=\lim_{x\to0^-}x\arctan\frac1x=0\times\left(-\frac\pi2\right)=0. \score1
+\end{align*}
+
+\indent 由于 $f(x)$ 在 $x=0$ 处可导, 因此
+\begin{align*}
+f'_-(0)&=f'_+(0), \score1\\
+f'_-(0)&=\lim_{x\to0^-}\frac{x\arctan\frac1x}x=\lim_{x\to0^-}\arctan\frac1x=-\frac\pi2 \score1\\
+f'_+(0)&=(2x+a)|_{x=0}=a, \score1
+\end{align*}
+因此 $a=-\dfrac\pi2$. \score1
+
+\indent 由于
+\begin{align*}
+\lim_{x\to+\infty}\frac yx&=\lim_{x\to+\infty}\left(x-\frac\pi2\right)=+\infty, \score1\\
+\lim_{x\to-\infty}\frac yx&=\lim_{x\to-\infty}\arctan\frac1x=0,\\
+\lim_{x\to-\infty}y&=\lim_{x\to-\infty}x\arctan\frac1x=\lim_{t\to0^-}\frac{\arctan t}t=1,
+\end{align*}
+因此曲线 $y=f(x)$ 的渐近线只有 $y=1$. \score1
+
+\textbf{5. （8分）【解】}
+
+\indent 由
+\[f'(x)=3x^2-2x-1=(3x+1)(x-1)=0 \score2\]
+可得驻点 $x=-\dfrac13,1$. \score2
+
+\indent 由于
+\[f(-2)=-10,\quad f(2)=2,\quad f\left(-\frac13\right)=\frac5{27},\quad f(1)=-1, \score2\]
+因此最大值为 $2$, 最小值为 $-10$. \score2
+
+\textbf{6. （8分）【证明】}
+
+\textbf{证法一}: 设 $f(x)=\tan x-x$, 则 \score2
+\[f'(x)=\frac1{\cos^2x}-1=\tan^2x\ge0. \score2\]
+因此 $f(x)$ 在 $\left(-\dfrac\pi2,\dfrac\pi2\right)$ 上单调递增, 从而 \score2
+\[f(x_2)\ge f(x_1),\quad\tan x_2-\tan x_1\ge x_2-x_1. \score2\]
+
+\newpage
+\textbf{证法二}: 设 $f(x)=\tan x$, 则 $f(x)$ 在 $[x_1,x_2]$ 上连续, $(x_1,x_2)$ 内可导. \score2
+
+\indent 由拉格朗日中值定理, 存在 $\xi\in(x_1,x_2)$ 使得
+\[\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(\xi), \score2\]
+即
+\[\frac{\tan x_2-\tan x_1}{x_2-x_1}=\frac1{\cos^2\xi}\ge1. \score2\]
+所以 $\tan x_2-\tan x_1\ge x_2-x_1$. \score2
+
+\textbf{7. （8分）【证明】}
+
+\indent 设 $F(x)=x^{2022}f(x)$, \score2\\
+则 $F(x)$ 在 $[0,1]$ 上连续, $(0,1)$ 内可导, \score1\\
+且 $F(0)=0,F(1)=f(1)=0$. \score1
+
+\indent 由罗尔中值定理, 存在 $\xi\in(0,1)$ 使得 $F'(\xi)=0$. \score2\\
+由于 $F'(x)=x^{2022}f'(x)+2022x^{2021}f(x)$ 且 $\xi\neq0$, \score1\\
+所以 $\xi f'(\xi)+2022f(\xi)=1$. \score1
+
+
+\textbf{8. （8分）【解】}
+
+(1) 
+\[f'(x)=\frac1x-\frac4{x^3}=\frac{x^2-4}{x^3}=\frac{(x+2)(x-2)}{x^3}. \score1\]
+当 $0<x<2$ 时, $f'(x)<0$. 当 $x>2$ 时, $f'(x)>0$. \score1\\
+因此 $(0,2]$ 是 $f(x)$ 的单减区间,\\
+$[2,+\infty)$ 是 $f(x)$ 的单增区间. \Score{(1分, 写成开区间不扣分)}\\
+所以 $f(x)$ 只有唯一的极小值 $f(2)=\ln2+\dfrac12$. \score1
+
+(2) 
+\[f''(x)=-\frac1{x^2}+\frac{12}{x^4}=-\frac{x^2-12}{x^4}=-\frac{(x-2\sqrt3)(x+2\sqrt3)}{x^4}. \score1\]
+当 $0<x<2\sqrt3$ 时, $f''(x)>0$. 当 $x>2\sqrt3$ 时, $f''(x)<0$. \score1\\
+因此 $(0,2\sqrt3]$ 是曲线 $y=f(x)$ 的凹区间,\\
+$[2\sqrt3,+\infty)$ 是曲线 $y=f(x)$ 的凸区间,\Score{(1分, 写成开区间不扣分)}\\
+拐点为 $\left(2\sqrt3,\ln(2\sqrt3)+\dfrac16\right)$. \score1
+
+\end{document}
+
+
+
+
diff --git a/Master/texmf-dist/tex/latex/hfutexam/hfutexam.cls b/Master/texmf-dist/tex/latex/hfutexam/hfutexam.cls
new file mode 100644
index 00000000000..4d560ba7b4b
--- /dev/null
+++ b/Master/texmf-dist/tex/latex/hfutexam/hfutexam.cls
@@ -0,0 +1,346 @@
+% 文件 `hfutexam.cls'
+% !TEX TS-program = xelatex
+% !TEX encoding = UTF-8 Unicode
+% 合肥工业大学试卷模板
+% 作者: 张神星
+% 使用前请先安装字体: 方正小标宋、方正仿宋, 否则请使用选项 nofangzheng (采用新宋体和仿宋代替)
+% 编译模式： XeLaTeX
+% 你可以任意修改或再次分发该文件
+\NeedsTeXFormat{LaTeX2e}
+\ProvidesClass{hfutexam}[2022/12/04 v1.5 HFUTExam document class by Zhang Shenxing]
+% 文档选项 shijuan, datizhi, cankaodaan, simple
+\newif\ifHFUT@ShiJuan\HFUT@ShiJuantrue
+\newif\ifHFUT@DaTiZhi\HFUT@DaTiZhifalse
+\newif\ifHFUT@CanKaoDaAn\HFUT@CanKaoDaAnfalse
+\newif\ifHFUT@Simple\HFUT@Simplefalse
+\DeclareOption{datizhi}{\HFUT@ShiJuanfalse\HFUT@DaTiZhitrue}
+\DeclareOption{cankaodaan}{\HFUT@ShiJuanfalse\HFUT@CanKaoDaAntrue}
+\DeclareOption{simple}{\HFUT@ShiJuanfalse\HFUT@Simpletrue}
+% 文档选项 nofangzheng
+\newif\ifHFUT@Fandol\HFUT@Fandoltrue
+\DeclareOption{nofangzheng}{\HFUT@Fandolfalse}
+% 基于 ctexart 文档类
+\DeclareOption*{\PassOptionsToClass{\CurrentOption}{ctexart}}
+\ProcessOptions\relax
+\ifHFUT@Simple\else\PassOptionsToClass{twocolumn}{ctexart}\fi
+\ProcessOptions\relax
+\LoadClass[12pt,oneside]{ctexart}
+% 页面设置
+\RequirePackage{amsmath,amssymb,amsthm}
+\RequirePackage{graphicx}
+\RequirePackage{enumitem}
+\RequirePackage{geometry}
+\ifHFUT@Simple
+	\geometry{a4paper,scale=0.8}
+\else
+	\geometry{
+		paperheight=29.7cm,
+		paperwidth=42cm,
+		left=3.17cm,
+		right=3.17cm,
+		voffset=2.2cm,
+		headheight=77pt,
+		headsep=12pt,
+		footskip=1cm
+	}
+\fi
+% 试卷信息
+\newcommand{\BiaoTi}[1]{\gdef\HFUT@BiaoTi{#1}}
+\newcommand{\XueNian}[2]{\gdef\HFUT@XueNians{#1}\gdef\HFUT@XueNiane{#2}}
+\newcommand{\XueQi}[1]{\gdef\HFUT@XueQi{#1}}
+\newcommand{\KeChengDaiMa}[1]{\gdef\HFUT@KeChengDaiMa{#1}}
+\newcommand{\KeChengMingCheng}[1]{\gdef\HFUT@KeChengMingCheng{#1}}
+\newcommand{\XueFen}[1]{\gdef\HFUT@XueFen{#1}}
+\newcommand{\KeChengXingZhi}[1]{\gdef\HFUT@KeChengXingZhi{#1}}
+\newcommand{\KaoShiXingShi}[1]{\gdef\HFUT@KaoShiXingShi{#1}}
+\newcommand{\ZhuanYeBanJi}[1]{\gdef\HFUT@ZhuanYeBanJi{#1}}
+\newcommand{\KaoShiRiQi}[1]{\gdef\HFUT@KaoShiRiQi{#1}}
+\newcommand{\MingTiJiaoShi}[1]{\gdef\HFUT@MingTiJiaoShi{#1}}
+\newcommand{\XiZhuRenQianMing}[1]{\gdef\HFUT@XiZhuRenQianMing{#1}}
+\BiaoTi{}
+\XueNian{}{}
+\XueQi{}
+\KeChengDaiMa{}
+\KeChengMingCheng{}
+\XueFen{}
+\KeChengXingZhi{}
+\KaoShiXingShi{}
+\ZhuanYeBanJi{}
+\KaoShiRiQi{}
+\MingTiJiaoShi{}
+\XiZhuRenQianMing{}
+% 粗字体设定
+\newfontfamily\timesnewroman[AutoFakeBold={1.5}]{Times New Roman} % 粗罗马
+\ifHFUT@Fandol
+	\setCJKfamilyfont{titlesongti}[AutoFakeBold={1.5}]{FZXiaoBiaoSong-B05S} % 粗方正小标宋
+	\newfontfamily\entitlesongti[AutoFakeBold={1.5}]{FZXiaoBiaoSong-B05S}
+	\setCJKfamilyfont{titlefangsong}[AutoFakeBold={1.5}]{FZFangSong-Z02S} % 粗方正仿宋
+	\newfontfamily\entitlefangsong[AutoFakeBold={1.5}]{FZFangSong-Z02S}
+\else
+	\setCJKfamilyfont{titlesongti}[AutoFakeBold={4}]{NSimSun} % 粗新宋体
+	\newfontfamily\entitlesongti[AutoFakeBold={4}]{NSimSun}
+	\setCJKfamilyfont{titlefangsong}[AutoFakeBold={1.5}]{FangSong} % 粗仿宋
+	\newfontfamily\entitlefangsong[AutoFakeBold={1.5}]{FangSong}
+\fi
+\newcommand{\titlesongti}{\entitlesongti\CJKfamily{titlesongti}}
+\newcommand{\titlefangsong}{\entitlefangsong\CJKfamily{titlefangsong}}
+\setCJKfamilyfont{xinsongti}[AutoFakeBold={1.5}]{NSimSun} % 粗新宋体
+\newfontfamily\enxinsongti[AutoFakeBold={1.5}]{NSimSun}
+\newcommand{\xinsongti}{\enxinsongti\CJKfamily{xinsongti}}
+\setCJKfamilyfont{cusongti}[AutoFakeBold={1.5}]{SimSun} % 粗宋体
+\newcommand{\cusongti}{\CJKfamily{cusongti}}
+% 重新设定字体大小
+\renewcommand{\LARGE}{\fontsize{21}{21}}
+\renewcommand{\large}{\fontsize{14}{14}}
+\renewcommand{\normalsize}{\fontsize{12}{12}}
+\renewcommand{\small}{\fontsize{10.5}{10.5}}
+% 缩进
+\setlength{\parindent}{0em}
+\renewcommand{\indent}{\hspace*{2em}}
+% 分栏
+\setlength\columnsep{0.8cm} % 分栏间距
+\columnseprule=0.5pt % 分栏线宽度
+% 填空题
+\newlength{\ltemp}
+\RequirePackage[normalem]{ulem}
+\ifHFUT@ShiJuan%
+	\NewDocumentCommand\fillblank{O{3.5cm} O{0.5cm} m}{\uline{\makebox[#1]{\rule{0pt}{#2}#3}}}%
+	\else\ifHFUT@DaTiZhi%
+	\NewDocumentCommand\fillblank{O{3.5cm} O{1cm} m}{\uline{\makebox[#1]{\rule{0pt}{#2}#3}}}%
+	\else%
+	\NewDocumentCommand\fillblank{O{3.5cm} O{1cm} m}{%
+		\settodepth{\ltemp}{#3}%
+		\uline{\makebox[#1]{\rule{0pt}{#2}\raisebox{\ltemp}{#3}}}%
+	}%
+\fi\fi
+% 页眉页脚
+\RequirePackage{fancyhdr}
+\RequirePackage{lastpage}
+\RequirePackage{etoolbox}
+\renewcommand\headrulewidth{0.5pt} % 页眉线宽度
+\renewcommand\footrulewidth{0.5pt} % 页脚线宽度
+\pagestyle{fancy}
+\fancyhf{}
+\newcommand{\boxyes}{{\raisebox{-0.5mm}{\LARGE$\checkmark\hspace{-1.2em}\square$}}}
+\newcommand{\boxno}{{\raisebox{-0.5mm}{\LARGE$\square$}}}
+\newcommand{\filltitle}[2]{\uline{\makebox[#1]{#2}}}
+\newlength{\ltitle}
+\settowidth{\ltitle}{\HFUT@BiaoTi}
+\ifHFUT@ShiJuan%试卷页眉页脚
+	\fancyhead[C]{%
+		\hfill\bfseries\LARGE\titlesongti\ifdimcomp\ltitle>{0.5\linewidth}{\HFUT@BiaoTi}{\makebox[0.5\linewidth][s]{\HFUT@BiaoTi}}\hfill%
+		\large\cusongti 共\filltitle{1.5em}{\timesnewroman\pageref{LastPage}}页第\filltitle{1.5em}{\timesnewroman\thepage}页\\\vspace{5pt}%
+		\timesnewroman\HFUT@XueNians\titlefangsong～\timesnewroman\HFUT@XueNiane\titlefangsong 学年第\filltitle{2em}{\HFUT@XueQi}学期\hfill%
+		课程代码\filltitle{5em}{\HFUT@KeChengDaiMa}\hfill%
+		课程名称\filltitle{12em}{\HFUT@KeChengMingCheng}\hfill%
+		学分\filltitle{2.5em}{\HFUT@XueFen}\hfill%
+		课程性质: 必修\ifdefstring{\HFUT@KeChengXingZhi}{必修}{\boxyes}{\boxno}%
+             选修\ifdefstring{\HFUT@KeChengXingZhi}{选修}{\boxyes}{\boxno}%
+             限修\ifdefstring{\HFUT@KeChengXingZhi}{限修}{\boxyes}{\boxno}\hfill%
+		考试形式: 开卷\ifdefstring{\HFUT@KaoShiXingShi}{开卷}{\boxyes}{\boxno}%
+             闭卷\ifdefstring{\HFUT@KaoShiXingShi}{闭卷}{\boxyes}{\boxno}\\\vspace{2pt}%
+		专业班级（教学班）\filltitle{11em}{\HFUT@ZhuanYeBanJi}\hfill%
+		考试日期\filltitle{16em}{\HFUT@KaoShiRiQi}\hfill%
+		命题教师\fillblank[5.5em][0.6cm]{\HFUT@MingTiJiaoShi}\hfill%
+		系（所或教研室）主任审批签名%
+		\begin{tikzpicture}[overlay,xshift=3.25em,yshift=0.15cm]%
+			\node at (0,0) {\ifx\HFUT@XiZhuRenQianMing\@empty\else\includegraphics[height=0.9cm]{\HFUT@XiZhuRenQianMing}\fi};%
+		\end{tikzpicture}%
+		\filltitle{6.5em}{}\vspace{4pt}%
+	}
+	\fancyfoot[C]{\small\vspace{0.5\baselineskip}命题教师注意事项:
+		1. 主考教师必须于考试一周前将“试卷A”、“试卷B”经教研室主任审批签字后送教务科印刷。\hspace{1em}%
+		2. 请命题教师用黑色水笔工整地书写题目或用A4纸横式打印贴在试卷版芯中。
+	}
+\fi
+\ifHFUT@DaTiZhi%答题纸页眉页脚
+	\fancyhead[C]{%
+		\bfseries\LARGE\xinsongti\scalebox{2.0}[1.0]{%
+			\ifdimcomp\ltitle>{0.35\linewidth}{\HFUT@BiaoTi}{\makebox[0.35\linewidth][s]{\HFUT@BiaoTi}}%
+		}\\\vspace{8pt}%
+		\large\timesnewroman\HFUT@XueNians\titlefangsong～\timesnewroman\HFUT@XueNiane\titlefangsong 学年第\filltitle{2em}{\HFUT@XueQi}学期\hfill%
+		课程代码\filltitle{6.5em}{\HFUT@KeChengDaiMa}\hfill%
+		课程名称\filltitle{13em}{\HFUT@KeChengMingCheng}\hfill%
+		命题教师\filltitle{7em}{\HFUT@MingTiJiaoShi}\hfill%
+		系主任审批%
+		\begin{tikzpicture}[overlay,xshift=3.5em,yshift=0.2cm]%
+			\node at (0,0) {\ifx\HFUT@XiZhuRenQianMing\@empty\else\includegraphics[height=0.9cm]{\HFUT@XiZhuRenQianMing}\fi};%
+		\end{tikzpicture}%
+		\filltitle{7em}{}\\\vspace{2pt}%
+		教学班级\filltitle{10.5em}{}\hfill%
+		学生姓名\fillblank[8em][0.6cm]{}\hfill%
+		学号\filltitle{8em}{}\hfill%
+		考试日期\filltitle{16em}{\HFUT@KaoShiRiQi}\hfill%
+		成绩\filltitle{6.3em}{}\vspace{4pt}%
+	}
+	\fancyfoot[C]{\small\vspace{0.5\baselineskip}
+		第 \timesnewroman\thepage 页~共 \timesnewroman\pageref{LastPage} 页
+	}
+\fi
+\ifHFUT@CanKaoDaAn%参考答案页眉页脚
+	\fancyhead[C]{%
+		\bfseries\LARGE\titlesongti\ifdimcomp\ltitle>{0.6\linewidth}{\HFUT@BiaoTi}{\makebox[0.6\linewidth][s]{\HFUT@BiaoTi}}\\\vspace{5pt}%
+		\large\timesnewroman\HFUT@XueNians\titlefangsong～\timesnewroman\HFUT@XueNiane\titlefangsong 学年第\filltitle{2em}{\HFUT@XueQi}学期\hfill%
+		课程代码\filltitle{6.5em}{\HFUT@KeChengDaiMa}\hfill%
+		课程名称\filltitle{13em}{\HFUT@KeChengMingCheng}\hfill%
+		命题教师\filltitle{7em}{\HFUT@MingTiJiaoShi}\hfill%
+		系主任审批%
+		\begin{tikzpicture}[overlay,xshift=3.5em,yshift=0.2cm]%
+			\node at (0,0) {\ifx\HFUT@XiZhuRenQianMing\@empty\else\includegraphics[height=0.9cm]{\HFUT@XiZhuRenQianMing}\fi};%
+		\end{tikzpicture}%
+		\filltitle{7em}{}\\\vspace{2pt}%
+		教学班级\filltitle{10.5em}{}\hfill%
+		学生姓名\fillblank[8em][0.6cm]{}\hfill%
+		学号\filltitle{8em}{}\hfill%
+		考试日期\filltitle{16em}{\HFUT@KaoShiRiQi}\hfill%
+		成绩\filltitle{6.3em}{}\vspace{4pt}%
+	}
+	\fancyfoot[C]{\small\vspace{0.5\baselineskip}
+		第 \timesnewroman\thepage 页~共 \timesnewroman\pageref{LastPage} 页
+	}
+\fi
+\ifHFUT@Simple%简单模式页眉页脚
+	\renewcommand\headrulewidth{0pt}
+	\renewcommand\footrulewidth{0pt}
+	\renewcommand{\maketitle}{%
+		\begin{center}
+		\bfseries\LARGE\titlesongti%
+		\ifdimcomp\ltitle>{0.9\linewidth}{\HFUT@BiaoTi}{\makebox[0.9\linewidth][s]{\HFUT@BiaoTi}}\\%
+		\makebox[0.75\linewidth]{\large\timesnewroman\HFUT@XueNians\titlefangsong～\timesnewroman\HFUT@XueNiane\titlefangsong 学年%
+			第{\HFUT@XueQi}学期\hfill\HFUT@KeChengMingCheng(\HFUT@KeChengDaiMa)}
+		\end{center}
+	}
+	\fancyfoot[C]{\small\vspace{0.5\baselineskip}
+		第 \timesnewroman\thepage 页~共 \timesnewroman\pageref{LastPage} 页
+	}
+\fi
+% 题号加粗
+\renewcommand{\labelenumi}{{\bfseries \theenumi.}}
+% 题干
+\newcommand\tigan[1]{\noindent{\large\textbf{#1}}}
+% 打分栏
+\RequirePackage{tikz}
+\usetikzlibrary{overlay-beamer-styles}
+\newcommand\scorebox{%
+	\vspace{0.5\baselineskip}\noindent%
+	\begin{tikzpicture}[overlay,xshift=13.8cm,yshift=-1.6cm]%
+		\draw (0,0) rectangle (3.6,2);%
+		\draw (1.8,0)--(1.8,2);%
+		\draw (0,1)--(3.6,1);%
+		\draw (0.9,1.5) node {\textbf{\normalsize 得分}};%
+		\draw (2.7,1.5) node {\textbf{\normalsize 阅卷人}};%
+	\end{tikzpicture}%
+}
+% 答题纸提示信息
+\newcommand\notice{%
+	\noindent\textbf{\small 考生注意事项：\\%
+	\indent 1. 本试卷分试题与答卷两部分；\\%
+	\indent 2. 所有试题的解答（包括选择、填空）必须写在专用答卷纸上，在试题上直接作答一律无效；\\%
+	\indent 3. 考试结束后，必须将试题、答卷整理上交，不得将试题带离考场；\\%
+	\indent 4. 考生务必认真填写班级、姓名、学号等信息。}\par%
+	{\leavevmode\xleaders\hbox{\rule[4pt]{8pt}{0.5pt}\,}\hfill\null}%
+}
+% 选择题, 根据选项内容长度自动排版
+\newlength{\lxxmax}
+\newlength{\lquar}
+\newlength{\lhalf}
+\newlength{\lfull}
+\newcounter{lxxtype}
+\NewDocumentCommand\xx{O{0} m m m m}{%
+	\setlength{\lfull}{\columnwidth}%
+	\addtolength{\lfull}{-\leftmargin}%
+	\setlength{\lhalf}{0.5\lfull}%
+	\setlength{\lquar}{0.25\lfull}%
+	\setcounter{lxxtype}{0}%
+	\ifnum#1=1\setcounter{lxxtype}{1}\fi%
+	\ifnum#1=2\setcounter{lxxtype}{2}\fi%
+	\ifnum#1=4\setcounter{lxxtype}{4}\fi%
+	\settowidth{\lxxmax}{A.~#2~}% 获取最长选项长度
+	\settowidth{\ltemp}{B.~#3~}%
+	\ifdimcomp\ltemp>\lxxmax{\setlength{\lxxmax}{\ltemp}}{}%
+	\settowidth{\ltemp}{C.~#4~}%
+	\ifdimcomp\ltemp>\lxxmax{\setlength{\lxxmax}{\ltemp}}{}%
+	\settowidth{\ltemp}{D.~#5~}%
+	\ifdimcomp\ltemp>\lxxmax{\setlength{\lxxmax}{\ltemp}}{}%
+	\ifnum\value{lxxtype}=0%
+		\setcounter{lxxtype}{4}%
+		\ifdimcomp\lxxmax>\lquar{\setcounter{lxxtype}{2}}{}%
+		\ifnum\value{lxxtype}=2%
+			\ifdimcomp\lxxmax>\lhalf{\setcounter{lxxtype}{1}}{}%
+		\fi%
+	\fi%
+	\vspace{5pt}%
+	\ifnum\value{lxxtype}=1%
+		\\\makebox[\lfull][l]{A.~#2}%
+		\\\makebox[\lfull][l]{B.~#3}%
+		\\\makebox[\lfull][l]{C.~#4}%
+		\\\makebox[\lfull][l]{D.~#5}%
+	\fi%
+	\ifnum\value{lxxtype}=2%
+		\\\makebox[\lhalf][l]{A.~#2}%
+			\makebox[\lhalf][l]{B.~#3}%
+		\\\makebox[\lhalf][l]{C.~#4}%
+			\makebox[\lhalf][l]{D.~#5}%
+	\fi%
+	\ifnum\value{lxxtype}=4%
+		\\\makebox[\lquar][l]{A.~#2}% 
+			\makebox[\lquar][l]{B.~#3}%
+			\makebox[\lquar][l]{C.~#4}%
+			\makebox[\lquar][l]{D.~#5}%
+	\fi%
+}
+% 选择题答题框
+\newcommand\xuanzeti[2]{%
+	\begin{tikzpicture}%
+		\draw (0,0) rectangle (1.8,2);%
+		\draw (0,1)--(1.8,1);%
+		\draw (0.9,0.5) node {#2} (0.9,1.5) node {\textbf{#1}};%
+	\end{tikzpicture}%
+}
+% 得分点命令
+% https://ask.latexstudio.net/ask/question/7557.html
+\RequirePackage{zref-savepos}
+\makeatletter
+\ExplSyntaxOn
+\zref@require@unique
+\NewDocumentCommand{\examscore}{O{} m}{
+	\mode_if_math:TF
+		{ \@@_math_cdotfill:n {\text{#2}} }
+		{ \__examzh_cdotfill: #2 }
+	\mode_if_math:F{
+		\ignorespaces
+	}
+}
+\cs_new:Npn \__examzh_cdotfill:
+{
+	\mode_leave_vertical:
+	\cleaders \hb@xt@ .44em {\hss $\cdot$ \hss} \hfill
+	\kern\z@
+}
+\cs_new_protected:Npn \@@_math_cdotfill:n #1
+{
+	\stepcounter { zref@unique }
+	\hbox_overlap_right:n
+	{
+		\zsaveposx { \thezref@unique L }
+		\zref@ifrefundefined { \thezref@unique R }
+		{ }
+		{
+			\cleaders
+			\hbox_to_wd:nn { .44em } { \hss $\cdot$ \hss }
+			\skip_horizontal:n
+			{
+				\zposx { \thezref@unique R } sp
+				- \zposx { \thezref@unique L } sp
+			}
+		}
+	}
+	\tag * { \zsaveposx { \thezref@unique R } #1 }
+}
+\ExplSyntaxOff
+\makeatother
+\newcommand\Score[1]{~~~\examscore{#1}}
+\newcommand\score[1]{~~~\examscore{(#1分)}}
+\renewcommand\le{\leqslant}
+\renewcommand\ge{\geqslant}
+
diff --git a/Master/tlpkg/bin/tlpkg-ctan-check b/Master/tlpkg/bin/tlpkg-ctan-check
index a85dbe3b6d3..92cd36d19c2 100755
--- a/Master/tlpkg/bin/tlpkg-ctan-check
+++ b/Master/tlpkg/bin/tlpkg-ctan-check
@@ -396,7 +396,7 @@ my @TLP_working = qw(
     hep-math hep-math-font hep-paper hep-reference hep-text hep-title hepnames
     hepparticles hepthesis hepunits here hereapplies heros-otf hershey-mp
     heuristica hexboard hexgame
-    hf-tikz hfbright hfoldsty hfutthesis
+    hf-tikz hfbright hfoldsty hfutexam hfutthesis
     hhtensor hideanswer highlightlatex hindawi-latex-template hindmadurai
     histogr historische-zeitschrift hitec hitreport
     hitszthesis hitszbeamer hithesis
diff --git a/Master/tlpkg/tlpsrc/collection-publishers.tlpsrc b/Master/tlpkg/tlpsrc/collection-publishers.tlpsrc
index fb211a273d5..3d7d50809b4 100644
--- a/Master/tlpkg/tlpsrc/collection-publishers.tlpsrc
+++ b/Master/tlpkg/tlpsrc/collection-publishers.tlpsrc
@@ -89,6 +89,7 @@ depend hagenberg-thesis
 depend har2nat
 depend hecthese
 depend hep-paper
+depend hfutexam
 depend hfutthesis
 depend hithesis
 depend hitszthesis
diff --git a/Master/tlpkg/tlpsrc/hfutexam.tlpsrc b/Master/tlpkg/tlpsrc/hfutexam.tlpsrc
new file mode 100644
index 00000000000..e69de29bb2d
--- /dev/null
+++ b/Master/tlpkg/tlpsrc/hfutexam.tlpsrc