From 89213117b5298d9493446422708bf48384368825 Mon Sep 17 00:00:00 2001 From: Norbert Preining Date: Sat, 9 Nov 2019 03:01:38 +0000 Subject: CTAN sync 201911090301 --- macros/latex/contrib/jnuexam/exam-a-a3input.pdf | Bin 111416 -> 102199 bytes macros/latex/contrib/jnuexam/exam-a-a3split.pdf | Bin 103007 -> 94090 bytes macros/latex/contrib/jnuexam/exam-a-empty.pdf | Bin 100412 -> 91316 bytes macros/latex/contrib/jnuexam/exam-a.pdf | Bin 118234 -> 104945 bytes macros/latex/contrib/jnuexam/exam-a.tex | 68 ++++----- macros/latex/contrib/jnuexam/exam-b-a3input.pdf | Bin 111285 -> 102033 bytes macros/latex/contrib/jnuexam/exam-b-a3split.pdf | Bin 102968 -> 93887 bytes macros/latex/contrib/jnuexam/exam-b-empty.pdf | Bin 100270 -> 91140 bytes macros/latex/contrib/jnuexam/exam-b.pdf | Bin 118116 -> 104821 bytes macros/latex/contrib/jnuexam/jnuexam.cls | 178 ++++++++++++++++++------ macros/latex/contrib/jnuexam/jnuexam.pdf | Bin 65809 -> 74286 bytes macros/latex/contrib/jnuexam/jnuexam.tex | 107 +++++++++++--- 12 files changed, 259 insertions(+), 94 deletions(-) (limited to 'macros/latex/contrib/jnuexam') diff --git a/macros/latex/contrib/jnuexam/exam-a-a3input.pdf b/macros/latex/contrib/jnuexam/exam-a-a3input.pdf index 5c185d3fc7..c8ade76680 100644 Binary files a/macros/latex/contrib/jnuexam/exam-a-a3input.pdf and b/macros/latex/contrib/jnuexam/exam-a-a3input.pdf differ diff --git a/macros/latex/contrib/jnuexam/exam-a-a3split.pdf b/macros/latex/contrib/jnuexam/exam-a-a3split.pdf index f17c0ce2e3..de83638ce7 100644 Binary files a/macros/latex/contrib/jnuexam/exam-a-a3split.pdf and b/macros/latex/contrib/jnuexam/exam-a-a3split.pdf differ diff --git a/macros/latex/contrib/jnuexam/exam-a-empty.pdf b/macros/latex/contrib/jnuexam/exam-a-empty.pdf index 846854d86d..653325a652 100644 Binary files a/macros/latex/contrib/jnuexam/exam-a-empty.pdf and b/macros/latex/contrib/jnuexam/exam-a-empty.pdf differ diff --git a/macros/latex/contrib/jnuexam/exam-a.pdf b/macros/latex/contrib/jnuexam/exam-a.pdf index c57592e675..5eb115847a 100644 Binary files a/macros/latex/contrib/jnuexam/exam-a.pdf and b/macros/latex/contrib/jnuexam/exam-a.pdf differ diff --git a/macros/latex/contrib/jnuexam/exam-a.tex b/macros/latex/contrib/jnuexam/exam-a.tex index 44bc600fae..54b2d1ac40 100644 --- a/macros/latex/contrib/jnuexam/exam-a.tex +++ b/macros/latex/contrib/jnuexam/exam-a.tex @@ -39,10 +39,10 @@ \vfill \begin{problem} -已知二阶行列式 $\text{$\left|\begin{array}{cc} +已知二阶行列式 $\left|\begin{array}{cc} 1 & 2\\ - 3 & x -\end{array}\right|$=0}$,则 $x=$ \fillout{$-6$}. +\end{array}\right|=0$,则 $x=$ \fillout{$-6$}. \end{problem} \vfill @@ -178,10 +178,10 @@ \begin{solution} \everymath{\displaystyle}% -原式$=\int\e^{2x}\,\sec^2 x\dx+2\int\e^{2x}\,\tan x\dx$ \score{2} -\hspace{5em}${}=\int\e^{2x}\,\d(\tan x)+ 2\int\e^{2x}\,\tan x\dx$ \score{4} -\hspace{5em}${}=\e^{2x}\,\tan x - 2\int\e^{2x}\,\tan x\dx+ 2\int\e^{2x}\,\tan x\dx$ \score{6} -\hspace{5em}${}=\e^{2x}\,\tan x + C$ \score{8} +原式 \? $=\int\e^{2x}\,\sec^2 x\dx+2\int\e^{2x}\,\tan x\dx$ \score{2} +\+ $=\int\e^{2x}\,\d(\tan x)+ 2\int\e^{2x}\,\tan x\dx$ \score{4} +\+ $=\e^{2x}\,\tan x - 2\int\e^{2x}\,\tan x\dx+ 2\int\e^{2x}\,\tan x\dx$ \score{6} +\+ $=\e^{2x}\,\tan x + C$ \score{8} \end{solution} \vfill @@ -220,7 +220,7 @@ \bigskip \begin{solution} -$A = \left|\begin{array}{cccc} +$A \? = \left|\begin{array}{cccc} 0 & 1 & 2 & 3\\ 1 & 2 & 3 & 0\\ 2 & 3 & 0 & 1\\ @@ -235,7 +235,7 @@ $A = \left|\begin{array}{cccc} - 1 & - 6 & 1\\ - 6 & - 8 & 2 \end{array}\right|$ \score{4} -\qquad $= -\left|\begin{array}{ccc} +\+ $= -\left|\begin{array}{ccc} 1 & 2 & 3\\ 0 & - 4 & 4\\ 0 & 4 & 20 @@ -255,11 +255,11 @@ x_2 x_3 + 9 x^2_3$ 化为标准形 $f = d_1 y^2_1 + d_2 y^2_2 + d_3 y^2_3$ . \bigskip \begin{solution} -$f = x_1^2 + 2 x_1 x_2 - 6 x_1 x_3 + 2 x_2^2 - 12 x_2 x_3 + 9 x^2_3$ \par -\qquad$= x_1^2 + 2 x_1 (x_2 - 3 x_3) + (x_2 - 3 x_3)^2 + x_2^2 - 6 x_2 x_3 $ \par -\qquad$= (x_1 + x_2 - 3 x_3)^2 + x_2^2 - 6 x_2 x_3$ \score{3} -\qquad$= (x_1 + x_2 - 3 x_3)^2 + x_2^2 - 2 x_2 \cdot 3 x_3 + (3 x_3)^2 - 9x_3^2$ \par -\qquad$= (x_1 + x_2 - 3 x_3)^2 + (x_2 - 3 x_3)^2 - 9 x_3^2$ \score{6} +$f \? = x_1^2 + 2 x_1 x_2 - 6 x_1 x_3 + 2 x_2^2 - 12 x_2 x_3 + 9 x^2_3$ \par + \+ $= x_1^2 + 2 x_1 (x_2 - 3 x_3) + (x_2 - 3 x_3)^2 + x_2^2 - 6 x_2 x_3 $ \par + \+ $= (x_1 + x_2 - 3 x_3)^2 + x_2^2 - 6 x_2 x_3$ \score{3} + \+ $= (x_1 + x_2 - 3 x_3)^2 + x_2^2 - 2 x_2 \cdot 3 x_3 + (3 x_3)^2 - 9x_3^2$ \par + \+ $= (x_1 + x_2 - 3 x_3)^2 + (x_2 - 3 x_3)^2 - 9 x_3^2$ \score{6} 令$y_1 = x_1 + x_2 - 3 x_3, y_2 = x_2 - 3 x_3, y_3 = x_3$, \newline 则$f = y_1^2 + y_2^2 - 9y_3^2$为标准形.\score{8} \end{solution} @@ -270,19 +270,19 @@ $f = x_1^2 + 2 x_1 x_2 - 6 x_1 x_3 + 2 x_2^2 - 12 x_2 x_3 + 9 x^2_3$ \par \begin{problem} 设每发炮弹命中飞机的概率是0.2且相互独立,现在发射100发炮弹.\par -\step 用切贝谢夫不等式估计命中数目$\xi$在10发到30发之间的概率.\par -\step 用中心极限定理估计命中数目$\xi$在10发到30发之间的概率. +(1) 用切贝谢夫不等式估计命中数目$\xi$在10发到30发之间的概率.\par +(2) 用中心极限定理估计命中数目$\xi$在10发到30发之间的概率. \end{problem} \bigskip \begin{solution} $E\xi = n p = 100 \cdot 0.2 = 20, D\xi = n p q = 100 \cdot 0.2 \cdot 0.8 = 16$. \score{2} -\step $P (10 < \xi < 30) = P (| \xi - E \xi | < 10) \ge 1 - \frac{D\xi}{10^2} +(1) $P (10 < \xi < 30) = P (|\xi - E\xi| < 10) \ge 1 - \frac{D\xi}{10^2} = 1 - \frac{16}{100} = 0.84$. \score{4} -\step $P (10 < \xi < 30) \approx \Phi_0 \left( \frac{30 - 20}{\sqrt{16}}\right) - - \Phi_0 \left( \frac{10 - 20}{\sqrt{16}} \right)$ \score{6} -\qquad $= 2 \Phi_0 (2.5) - 1 = 2 \cdot 0.9938 - 1 =0.9876$ \score{8} +(2) $P (10 < \xi < 30) \? \approx \Phi_0\left(\frac{30 - 20}{\sqrt{16}}\right) + - \Phi_0\left(\frac{10 - 20}{\sqrt{16}}\right)$ \score{6} + \+ $= 2 \Phi_0(2.5) - 1 = 2 \cdot 0.9938 - 1 =0.9876$ \score{8} \end{solution} \vfill @@ -295,11 +295,11 @@ $E\xi = n p = 100 \cdot 0.2 = 20, D\xi = n p q = 100 \cdot 0.2 \cdot 0.8 = 16$. \bigskip \begin{solution} -\step 待检假设 $H_0 : \mu = 3140$. \score{1} -\step 选取统计量 $T = \frac{\bar{X}-\mu}{S / \sqrt{n}} \sim t(n-1)$. \score{3} -\step 查表得到 $t_{\alpha} = t_{\alpha} (n - 1) = t_{0.01} (15) =2.947$. \score{5} -\step 计算统计值 $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} =\frac{3160-3140}{100/4} = 0.8$.\score{7} -\step 由于 $| t | < t_{\alpha}$, 故接受 $H_0$, 即假设成立. \score{8} +(1) 待检假设 $H_0 : \mu = 3140$. \score{1} +(2) 选取统计量 $T = \frac{\widebar{X}-\mu}{S / \sqrt{n}} \sim t(n-1)$. \score{3} +(3) 查表得到 $t_{\alpha} = t_{\alpha} (n - 1) = t_{0.01} (15) =2.947$. \score{5} +(4) 计算统计值 $t = \frac{\widebar{x} - \mu_0}{s/\sqrt{n}} =\frac{3160-3140}{100/4} = 0.8$.\score{7} +(5) 由于 $| t | < t_{\alpha}$, 故接受 $H_0$, 即假设成立. \score{8} \end{solution} \vfill @@ -308,6 +308,8 @@ $E\xi = n p = 100 \cdot 0.2 = 20, D\xi = n p q = 100 \cdot 0.2 \cdot 0.8 = 16$. \makepart{证明题}{共~2~小题,每小题~8~分,共~16~分} +\renewcommand{\solutionname}{证} % 将“解”字改为“证”字 + \begin{problem} 设数列$\{x_n\}$满足$x_1=\sqrt2$,$x_{n+1}=\sqrt{2+x_n}$.证明数列收敛,并求出极限. \end{problem} @@ -315,13 +317,13 @@ $E\xi = n p = 100 \cdot 0.2 = 20, D\xi = n p q = 100 \cdot 0.2 \cdot 0.8 = 16$. \bigskip \begin{solution} -\step 事实上,由于$x_1<2$,且$x_k<2$时 +(1) 事实上,由于$x_1<2$,且$x_k<2$时 $$x_{k+1}=\sqrt{2+x_k}<\sqrt{2+2}=2,$$ 由数学归纳法知对所有$n$都有$x_n<2$,即数列有上界. 又由于 $$\frac{x_{n+1}}{x_n}=\sqrt{\frac{2}{x_n^2}+\frac{1}{x_n}}>\sqrt{\frac{2}{2^2}+\frac{1}{2}}=1,$$ 所以数列单调增加.由极限存在准则II,数列必定收敛.\score{4} -\step 设数列的极限为$A$,对递推公式两边同时取极限得到 +(2) 设数列的极限为$A$,对递推公式两边同时取极限得到 $$A=\sqrt{2+A}.$$ 解得$A=2$,即数列$\{x_n\}$的极限为$2$.\score{8} \end{solution} @@ -329,29 +331,29 @@ $$A=\sqrt{2+A}.$$ \vfill \begin{problem} -设事件$A$和$B$相互独立,证明$A$和$\bar{B}$相互独立. +设事件$A$和$B$相互独立,证明$A$和$\widebar{B}$相互独立. \end{problem} \bigskip \begin{solution} -$P (A \cdot \bar{B}) = P (A - B) = P (A - A B)$ \score{2} -\qquad $= P (A) - P (A B) = P (A) - P (A) P (B)$ \score{4} -\qquad $= P (A) (1 - P (B)) = P (A) P (\bar{B})$ \score{6} -所以$A$和$\bar{B}$相互独立.\score{8} +\? $P (A \cdot \widebar{B}) = P (A - B) = P (A - A B)$ \score{2} +\< $= P (A) - P (A B) = P (A) - P (A) P (B)$ \score{4} +\< $= P (A) (1 - P (B)) = P (A) P (\widebar{B})$ \score{6} +所以$A$和$\widebar{B}$相互独立.\score{8} \end{solution} \vfill \makedata{一些可能用到的数据} %附录数据 -\begin{tabu}{*{4}{X[l,$]}} +\begin{tabularx}{\linewidth}{*{4}{>{$}X<{$}}} \hline \Phi_0(0.5)=0.6915 & \Phi_0(1)=0.8413 & \Phi_0(2)=0.9773 & \Phi_0(2.5)=0.9938 \\ t_{0.01}(8)=3.355 & t_{0.01}(9)=3.250 & t_{0.01}(15)=2.947 & t_{0.01}(16)=2.921 \\ \chi_{0.005}^2(8)=22.0 & \chi_{0.005}^2(9)=23.6 & \chi_{0.005}^2(15)=32.8 & \chi_{0.005}^2(16)=34.3 \\ \chi_{0.995}^2(8)=1.34 & \chi_{0.995}^2(9)=1.73 & \chi_{0.995}^2(15)=4.60 & \chi_{0.995}^2(16)=5.14 \\ \hline -\end{tabu} +\end{tabularx} \end{document} diff --git a/macros/latex/contrib/jnuexam/exam-b-a3input.pdf b/macros/latex/contrib/jnuexam/exam-b-a3input.pdf index 9b0fd8d18d..7351872d94 100644 Binary files a/macros/latex/contrib/jnuexam/exam-b-a3input.pdf and b/macros/latex/contrib/jnuexam/exam-b-a3input.pdf differ diff --git a/macros/latex/contrib/jnuexam/exam-b-a3split.pdf b/macros/latex/contrib/jnuexam/exam-b-a3split.pdf index c224b9ab33..ff8a2e40ba 100644 Binary files a/macros/latex/contrib/jnuexam/exam-b-a3split.pdf and b/macros/latex/contrib/jnuexam/exam-b-a3split.pdf differ diff --git a/macros/latex/contrib/jnuexam/exam-b-empty.pdf b/macros/latex/contrib/jnuexam/exam-b-empty.pdf index 1b27aa32c8..9809fb34cc 100644 Binary files a/macros/latex/contrib/jnuexam/exam-b-empty.pdf and b/macros/latex/contrib/jnuexam/exam-b-empty.pdf differ diff --git a/macros/latex/contrib/jnuexam/exam-b.pdf b/macros/latex/contrib/jnuexam/exam-b.pdf index 8a378a7e3d..69dd16f187 100644 Binary files a/macros/latex/contrib/jnuexam/exam-b.pdf and b/macros/latex/contrib/jnuexam/exam-b.pdf differ diff --git a/macros/latex/contrib/jnuexam/jnuexam.cls b/macros/latex/contrib/jnuexam/jnuexam.cls index e839799d0e..84e36954a5 100644 --- a/macros/latex/contrib/jnuexam/jnuexam.cls +++ b/macros/latex/contrib/jnuexam/jnuexam.cls @@ -6,18 +6,22 @@ % ---------------------------------------------------------------------------- \NeedsTeXFormat{LaTeX2e} -\ProvidesClass{jnuexam}[2018/11/21 v0.5 An exam class for Jinan University] +\ProvidesClass{jnuexam}[2019/11/08 v0.7 An exam class for Jinan University] \newif\ifsidebyside \sidebysidefalse % 是否 A3 纸张 \newif\ifreverse \reversefalse % 是否逆序出题 \newif\ifanswer \answertrue % 是否显示答案 +\newif\ifamsfonts \amsfontsfalse % 切换数学字体 \newif\ifsourcehan \sourcehanfalse % 切换思源字体 +\newif\ifdisplay \displayfalse % 切换展示公式 \newif\ifcollection \collectionfalse % 用于试卷题库 \DeclareOption{a3paper}{\sidebysidetrue} \DeclareOption{reverse}{\reversetrue} \DeclareOption{noanswer}{\answerfalse} +\DeclareOption{amsfonts}{\amsfontstrue} \DeclareOption{sourcehan}{\sourcehantrue} +\DeclareOption{display}{\displaytrue} \DeclareOption{collection}{\collectiontrue} \DeclareOption*{\PassOptionsToClass{\CurrentOption}{ctexart}} %其它选项 @@ -28,14 +32,22 @@ % 14bp 12bp 10.5bp 9bp \LoadClass[cs4size,UTF8,noindent]{ctexart} +\ifamsfonts + \RequirePackage{amssymb} +\else + \RequirePackage[utopia]{mathdesign} % charter, utopia + \renewcommand\bfdefault{bx} + \let\oldoiint\oiint\renewcommand{\oiint}{\oldoiint\nolimits} + \DeclareTextCommandDefault{\nobreakspace}{\leavevmode\nobreak\ } +\fi + \ifsidebyside \RequirePackage[a3paper,landscape,twocolumn,columnsep=40mm,left=50mm,right=30mm,top=25mm,bottom=25mm]{geometry} \else \RequirePackage[a4paper,left=30mm,right=30mm,top=25mm,bottom=25mm]{geometry} \fi -\RequirePackage{tabu} -\RequirePackage{amssymb} +\RequirePackage{tabularx} \RequirePackage{lastpage} \RequirePackage{fancyhdr} \RequirePackage{xcolor} @@ -44,9 +56,12 @@ \RequirePackage{etoolbox} \RequirePackage{calc} +\newcolumntype{Y}{>{\centering\arraybackslash}X} +\newcolumntype{n}[1]{>{\centering\arraybackslash}m{#1}} + \setlength{\parindent}{0em} -\setlength{\lineskiplimit}{3pt} -\setlength{\lineskip}{3pt} +\setlength{\lineskiplimit}{4pt} +\setlength{\lineskip}{4pt} %% --------------------------------------------------------------------------- %% 密封线命令 \mifengxian @@ -116,10 +131,10 @@ \newcommand{\head@table@a}{% \begin{tabular}{l} - \underbox{11em}{\niandu}学年度第\underbox{5.5em}{\xueqi}学期 \\ - 课程名称:\underbox{17.5em}{\kecheng\ifx\zhuanye\my@empty\else\kern0pt(\zhuanye)\fi} \\ - 授课教师:\underparbox{17.5em}{\centering\rule{0pt}{3ex}\jiaoshi} \\ - 考试时间:\underbox{17.5em}{\shijian} \\ + \underbox{11\ccwd}{\niandu}学年度第\underbox{5.5\ccwd}{\xueqi}学期 \\ + 课程名称:\underbox{17.5\ccwd}{\kecheng\ifx\zhuanye\my@empty\else\kern0pt(\zhuanye)\fi} \\ + 授课教师:\underparbox{17.5\ccwd}{\centering\rule{0pt}{3ex}\jiaoshi} \\ + 考试时间:\underbox{17.5\ccwd}{\shijian} \\ \end{tabular} } @@ -136,9 +151,9 @@ \newcommand{\head@table@c}{% \begin{tabular}{l} - \underspace{10em}学院\underspace{8em}专业\underspace{7.5em}班\kern0pt(\kern0pt{}级\kern0pt) \\[1em] - 姓名\underspace{8em}学号\underspace{10em}\hfill - \bfseries 内招~[\ischeck{\neizhao}] 外招~[\ischeck{\waizhao}] %\\[0.8em] + \underspace{10\ccwd}学院\underspace{8\ccwd}专业\underspace{7.5\ccwd}班\kern0pt(\kern0pt{}级\kern0pt) \\[1em] + 姓名\underspace{8\ccwd}学号\underspace{10\ccwd}\hfill + \bfseries 内招~[\ischeck{\neizhao}]~~外招~[\ischeck{\waizhao}] %\\[0.8em] \end{tabular} } @@ -151,25 +166,25 @@ \ifx\shijuan\my@temp@c\renewcommand{\shijuan}{D}\fi \fi \noindent - \begin{tabu}{|@{}X@{}|} + \begin{tabularx}{\linewidth}{|@{}X@{}|} \hline\renewcommand{\arraystretch}{1.5}% - \begin{tabu}{@{}>{\bfseries}l@{}|@{}X@{}|@{}l@{}} + {\begin{tabularx}{\linewidth}{@{}>{\bfseries}l@{}|@{}X@{}|@{}l@{}} \begin{tabular}{l}教\\ 师\\ 填\\ 写\end{tabular} & \head@table@a & \head@table@b - \end{tabu}\\ + \end{tabularx}}\\ \hline\renewcommand{\arraystretch}{0.9}% - \begin{tabu}{@{}>{\bfseries}l@{}|@{}X@{}} + {\begin{tabularx}{\linewidth}{@{}>{\bfseries}l@{}|@{}X@{}} \begin{tabular}{l}考\\ 生\\ 填\\ 写\end{tabular} & \head@table@c - \end{tabu}\\ + \end{tabularx}}\\ \hline - \end{tabu} - \vspace{1em} - \noindent\begin{tabu}{|*{8}{X[c]|}} + \end{tabularx} + \par\vspace{1em} + \noindent\begin{tabularx}{\linewidth}{|*{8}{Y|}} \hline - \bfseries{题 号} & 一 & 二 & 三 & 四 & 五 & 六 & 总 分\\ + \textbf{题号} & 一 & 二 & 三 & 四 & 五 & 六 & 总分\\ \hline - \bfseries{得 分} & & & & & & & \\ + \parbox[c][2em][c]{2.2em}{\bfseries 得分} & & & & & & & \\ \hline - \end{tabu} + \end{tabularx} } %% --------------------------------------------------------------------------- @@ -218,6 +233,7 @@ \xdef\allproblems{} \xdef\lastproblem{} \newcounter{problem} +\newcommand{\solutionname}{解} \newcounter{choice} % 后面选择题的 abcd 环境要用到 \newcounter{step} % 后面解答题的 \step 命令要用到 @@ -228,14 +244,14 @@ \setcounter{problem}{0}% \stepcounter{section}% \vspace{1em}% - \noindent\begin{tabu}{|X[c]|X[c]|X[6]} + \noindent\begin{tabularx}{\linewidth}{|n{1.7cm}|n{1.7cm}|X} \cline{1-2} 得分 & 评阅人 & \textbf{\Chinese{section}、#1}\\ \cline{1-2} & & (#2) \\ \cline{1-2} - \end{tabu} - \vspace{1em} + \end{tabularx} + \par\vspace{1em} } \newcommand{\makedata}[1]{% @@ -249,11 +265,11 @@ \newenvironment{problemreal}{% \stepcounter{problem}\setcounter{choice}{0}\setcounter{step}{0}% - \textsf{\color{blue}\arabic{problem}}.\;\,\ignorespaces + \textbf{\textsf{{\color{blue}\arabic{problem}}.}}\;\,\ignorespaces }{\par} \newenvironment{solutionreal}{% \setcounter{step}{0}% - \textsf{\color{blue}解答}\quad\ignorepars + \textbf{\textsf{{\color{blue}\solutionname}.}}\;\,\ignorepars }{\par} \let \oldnewpage = \newpage @@ -373,15 +389,16 @@ \newcommand{\answertable}[3][1em]{% 答题须知:本题答案必须写在如下表格中,否则不给分.\par - \begin{tabu}{|c|*{#3}{X[c]|}} + \begin{tabularx}{\linewidth}{|c|*{#3}{Y|}} \hline \answer@lines{#1}{#2}{#3} - \end{tabu}% + \end{tabularx}% \par\vspace{0.8em}% } %% --------------------------------------------------------------------------- %% 答案切换命令 \answer +%% 判断命令 \true 和 \false %% 填空命令 \fillin 和 \fillout %% 选择命令 \pickin 和 \pickout %% 四个选项排版环境 abcd,根据四个选项的长度自动排成一行、两行或四行 @@ -389,11 +406,14 @@ \newcommand{\answer}[1]{\ifanswer#1\else\phantom{#1}\fi} +\newcommand{\cdotfill}{\leavevmode\xleaders\hbox to 0.5em{\hss$\cdot$\hss}\hfill\kern0pt\relax} +\newcommand{\true}{\unskip\nobreak\cdotfill(\makebox[1.5em]{\color{blue}\answer{$\checkmark$}})} +\newcommand{\false}{\unskip\nobreak\cdotfill(\makebox[1.5em]{\color{blue}\answer{\sffamily x}})} + \newcommand{\ulinefill}[1]{\xleaders\hbox{\underline{\vphantom{#1}\kern1pt}}\hfill\kern0pt} \newcommand{\fillout}[1]{\allowbreak\hbox{}\nobreak\ulinefill{#1}\underline{\color{blue}\answer{#1}}\ulinefill{#1}} \newcommand{\fillin}[1]{\underline{\hspace{1em}\color{blue}\answer{#1}\hspace{1em}}} -\newcommand{\cdotfill}{\leavevmode\xleaders\hbox to 0.5em{\hss$\cdot$\hss}\hfill\kern0pt\relax} \newcommand{\pickout}[1]{\unskip\nobreak\cdotfill(\makebox[1.5em]{\color{blue}\answer{#1}})} \newcommand{\pickin}[1]{\unskip\nobreak\hspace{0.3em}(\makebox[1.5em]{\color{blue}\answer{#1}})\hspace{0.3em}\ignorespaces} @@ -431,20 +451,85 @@ \fi } -%\newcommand{\my@item}{\ifnum\value{choice}=0\par\fi\stepcounter{choice}} -%\newcommand{\fullitem}[1]{\my@item\parbox{\linewidth}{(\Alph{choice})\ #1\rule[-0.5em]{0pt}{0.5em}}\hfill\ignorespaces} -%\newcommand{\halfitem}[2][0.5]{\my@item\makebox[#1\linewidth][l]{(\Alph{choice})\ #2}\hfill\ignorespaces} -%\newcommand{\quaritem}[2][0.25]{\my@item\makebox[#1\linewidth][l]{(\Alph{choice})\ #2}\hfill\ignorespaces} - %% --------------------------------------------------------------------------- %% 解答题步骤命令 \step %% --------------------------------------------------------------------------- \newcommand{\step}{% \stepcounter{step}% - \makebox[2em][l]{\ttfamily(\arabic{step})}% + \textsf{(\arabic{step})}\;\,% } +%% --------------------------------------------------------------------------- +%% 自由对齐命令 \tabpoint, \tabto, \tableft +%% 命令 \tabpoint 记录当前的水平位置,也可以简写为 \? +%% 命令 \tabto 跳到之前记录的位置,也可以简写为 \+ +%% 命令 \tableft 跳到之前记录的位置的左侧,也可以简写为 \< +%% 这些自由对齐命令需要编译两次才能生效 +%% --------------------------------------------------------------------------- + +\usepackage{zref-savepos} + +\@ifundefined{zsaveposx}{\let\zsaveposx\zsavepos}{} % 旧版本无 \zsaveposx 命令 + +\newcounter{saveposcnt} +\newcounter{useposcnt} +\renewcommand*{\thesaveposcnt}{savepos\number\value{saveposcnt}} +\renewcommand*{\theuseposcnt}{usepos\number\value{useposcnt}} + +\def\my@alignment@offset{} + +\def\my@alignment@list{} +\forcsvlist{\listadd\my@alignment@list}{=,<,>,\le,\ge,\leq,\geq,\approx} + +\newlength{\my@alignment@kern} + +\newcommand*{\my@alignment@check}[1]{% + \ifx\my@let@token #1% + \def\my@alignment@offset{5}% + \listbreak + \fi +} + +\newcommand{\my@alignment@next}{% + \ifdefempty{\my@alignment@offset}{% + \def\my@alignment@offset{0}% + \forlistloop{\my@alignment@check}{\my@alignment@list}% + }{}% + \settowidth{\my@alignment@kern}{$\mkern\my@alignment@offset mu$}% + \stepcounter{saveposcnt}% + \rlap{\kern\my@alignment@kern\zsaveposx{\thesaveposcnt}}% +} + +\newcommand*{\tabpoint}[1][]{% + \leavevmode + \def\my@alignment@offset{#1}% + \futurelet\my@let@token\my@alignment@next +} +\let \? = \tabpoint + +\newcommand*{\tabto}{% + \stepcounter{useposcnt}% + \zsaveposx{\theuseposcnt}% + \noindent + \hskip\zposx{\thesaveposcnt}sp\relax + \hskip-\zposx{\theuseposcnt}sp\relax + \ignorespaces +} +\let \+ = \tabto + +\newcommand*{\tableft}{% + \settowidth{\my@alignment@kern}{$=\mkern5mu$}% + \stepcounter{useposcnt}% + \zsaveposx{\theuseposcnt}% + \noindent + \hskip\zposx{\thesaveposcnt}sp\relax + \hskip-\zposx{\theuseposcnt}sp\relax + \hskip-\my@alignment@kern + \ignorespaces +} +\let \< = \tableft + %% --------------------------------------------------------------------------- %% 评分命令 \score %% --------------------------------------------------------------------------- @@ -466,6 +551,14 @@ }% } +%% --------------------------------------------------------------------------- +%% 文档选项 display 将全部公式都设为展示公式 +%% 命令 \display 将当前环境的公式都设为展示公式 +%% --------------------------------------------------------------------------- + +\newcommand{\display}{\everymath\expandafter{\the\everymath\displaystyle}} +\ifbool{display}{\display}{} + %% --------------------------------------------------------------------------- %% 载入个人定制文件 jnuexam.cfg %% 中文字体切换选项 sourcehan @@ -526,6 +619,7 @@ \RequirePackage{CJKfntef} \RequirePackage{multirow} \RequirePackage{diagbox} +\RequirePackage{tabu} \RequirePackage{relsize} \newcommand{\Int}{\mathop{\mathlarger{\int}}} @@ -557,9 +651,15 @@ \renewrobustcmd{\d}{\ifbool{mmode}{\diff}{\oldd}}% } -\newcommand{\va}{\vec{a\vphantom{b}}} +% from mathabx package +\DeclareFontFamily{U}{mathx}{\hyphenchar\font45} +\DeclareFontShape{U}{mathx}{m}{n}{<-> mathx10}{} +\DeclareSymbolFont{mathx}{U}{mathx}{m}{n} +\DeclareMathAccent{\widebar}{0}{mathx}{"73} + +\newcommand{\va}{\vec{a}} \newcommand{\vb}{\vec{b}} -\newcommand{\vc}{\vec{c\vphantom{b}}} +\newcommand{\vc}{\vec{c}} \newcommand{\vd}{\vec{d}} \newcommand{\ve}{\vec{e}} \newcommand{\vi}{\vec{i}} diff --git a/macros/latex/contrib/jnuexam/jnuexam.pdf b/macros/latex/contrib/jnuexam/jnuexam.pdf index 79517e4623..dac435686d 100644 Binary files a/macros/latex/contrib/jnuexam/jnuexam.pdf and b/macros/latex/contrib/jnuexam/jnuexam.pdf differ diff --git a/macros/latex/contrib/jnuexam/jnuexam.tex b/macros/latex/contrib/jnuexam/jnuexam.tex index f41fb784f6..bfc98e0834 100644 --- a/macros/latex/contrib/jnuexam/jnuexam.tex +++ b/macros/latex/contrib/jnuexam/jnuexam.tex @@ -5,9 +5,7 @@ \usepackage[UTF8,noindent]{ctex} \usepackage{arev} -\usepackage{ragged2e} -\usepackage{listings} -\usepackage{tabu} +\usefonttheme{professionalfonts} \makeatletter @@ -28,39 +26,45 @@ \setlength{\parskip}{7pt plus 1pt minus 1pt} -\justifying -\let\oldraggedright\raggedright -\let\raggedright\justifying - -\lstset{ - basicstyle=\ttfamily\color{blue!50!red} -} - -\lstnewenvironment{code}{}{} - \setbeamersize{text margin left=8mm,text margin right=8mm} -\newenvironment{framex}{\begin{frame}[fragile=singleslide,environment=framex]}{\end{frame}} +\setbeamercolor{normal text}{bg=gray!20} \setbeamertemplate{frametitle}{\strut\insertframetitle\strut\par} \setbeamertemplate{navigation symbols}{} +\newcommand{\cdotfill}{\leavevmode\xleaders\hbox to 0.5em{\hss$\cdot$\hss}\hfill\kern0pt\relax} + +\usepackage{tabularx} + \newcommand{\ulinefill}[1]{\xleaders\hbox{\underline{\vphantom{#1}\kern1pt}}\hfill\kern0pt} \newcommand{\fillbox}[1]{\ulinefill{#1}\underline{#1}\ulinefill{#1}} -\setbeamercolor{normal text}{bg=gray!20} - \setbeamertemplate{title page}{% \renewcommand{\arraystretch}{2}% \usebeamerfont{title} - \begin{tabu}{|X|} + \begin{tabularx}{\linewidth}{|X|} \hline 模板名称:\fillbox{\usebeamercolor[fg]{title}\inserttitle} \\ 模板作者:\fillbox{\insertauthor} \\ 所在单位:\fillbox{\insertinstitute} \\ 更新日期:\fillbox{\the\year}年\fillbox{\the\month}月\fillbox{\the\day}日\\ \hline - \end{tabu}% + \end{tabularx}% +} + +\usepackage{ragged2e} + +\justifying +\let\oldraggedright\raggedright +\let\raggedright\justifying + +\usepackage{fancyvrb} + +\newenvironment{framex}{\begin{frame}[fragile=singleslide,environment=framex]}{\end{frame}} + +\DefineVerbatimEnvironment{code}{Verbatim}{% + formatcom=\color{blue!50!red}% } \begin{document} @@ -135,6 +139,22 @@ 其中 \verb!\zhuanye! 和 \verb!\shijian! 命令的内容可以为空。 \end{framex} +\begin{framex} +\frametitle{判断题目} +\begin{code} +\makepart{判断题}{题数分值} + +\begin{problem} +第一道判断题描述。\true +\end{problem} + +\begin{problem} +第二道判断题描述。\false +\end{problem} +\end{code} +其中 \verb!\true! 和 \verb!\false! 命令分别表示正确和错误。 +\end{framex} + \begin{framex} \frametitle{填空题目} \begin{code} @@ -242,6 +262,15 @@ \end{code} \end{framex} +\begin{framex} +\frametitle{解答名称} +通过重新定义 \verb!\solutionname! 命令,可以改变 \verb!solution! 环境的名称。 +比如下面例子将“解答”二字改为“证明”: +\begin{code} +\renewcommand{\solutionname}{证明} +\end{code} +\end{framex} + \begin{framex} \frametitle{评分命令} 计算题和证明题等主观题的排版方法是完全一样的。在编写这些主观题的解答时, @@ -255,6 +284,40 @@ $2+2=4$ \score{8} 评分命令 \verb!\score! 也可在 \verb!align*! 等数学环境中使用,此时评分显示在公式编号位置。 \end{framex} +\begin{framex} +\frametitle{对齐命令} +此文档类提供几个对齐命令,用于在不同行之间对齐。比如 +\vskip1em\hrule +我们有$(a+b)^2 = (a+b)(a+b)$ \par +\leavevmode\phantom{我们有$(a+b)^2$}${}= a^2 + 2ab + b^2$ \cdotfill 2分 +\vskip0.6em\hrule\vskip1em +\begin{code} +我们有$(a+b)^2 \? = (a+b)(a+b)$ \\ + \+$= a^2+2ab+b^2$ \score{2} +\end{code} +第一个公式内部的 \verb!\?! 保存当前水平位置, +而第二个公式前面的 \verb!\+! 表示跳到之前保存的位置。 +\par +这两个对齐命令 \verb!\?! 和 \verb!\+! 需要编译两次才能生效。 +\end{framex} + +\begin{framex} +\frametitle{对齐命令} +此文档类提供几个对齐命令,用于在不同行的对齐。比如 +\vskip1em\hrule +我们有$(a+b)^2 = (a+b)(a+b)$ \par +\leavevmode\phantom{我们\,}${}= a^2 + 2ab + b^2$ \cdotfill 2分 +\vskip0.6em\hrule\vskip1em +\begin{code} +我们有 \? $(a+b)^2 = (a+b)(a+b)$ \\ + \< $= a^2+2ab+b^2$ \score{2} +\end{code} +第一行公式前面的 \verb!\?! 保存当前水平位置, +而第二行公式前面的 \verb!\