texlive[69089] Master/texmf-dist: jnuexam (11dec23)
commits+karl at tug.org
commits+karl at tug.org
Mon Dec 11 22:08:45 CET 2023
Revision: 69089
https://tug.org/svn/texlive?view=revision&revision=69089
Author: karl
Date: 2023-12-11 22:08:45 +0100 (Mon, 11 Dec 2023)
Log Message:
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jnuexam (11dec23)
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trunk/Master/texmf-dist/doc/latex/jnuexam/exam-a-empty.pdf
trunk/Master/texmf-dist/doc/latex/jnuexam/exam-a-empty.tex
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-% -*- coding: utf-8 -*-
-% !TEX program = xelatex
-
-% 直接包含 A4 试卷的 PDF 文件，生成双栏的 A3 试卷
-\documentclass[a3input]{jnuexam}
-\begin{document}
-\includepdf[pages=-,nup=2x1,offset=0 0,delta=0 0]{exam-a-empty}
-%\includepdf[pages=-,nup=2x1,offset=0 0,delta=0 0,frame]{exam-a}
-\end{document}
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-% -*- coding: utf-8 -*-
-% !TEX program = xelatex
-
-% 将原有的 A4 试卷改为 A3 试卷双栏排版
-\PassOptionsToClass{a3paper,noanswer}{jnuexam}
-\input{exam-a}
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@@ -0,0 +1,359 @@
+% -*- coding: utf-8 -*-
+% !TEX program = xelatex
+\documentclass{jnuexam}
+
+%\answerfalse %不显示答案
+
+\setexam{
+ binding = 2, % 装订线，1 仅空白试卷有，2 试卷和答案都有
+ scratch = 1, % 草稿纸数量，仅空白试卷有，A3 大小，双面印刷
+ seed = 19061116, % 随机数种子，用于改变 B 卷小题的随机顺序
+}
+
+\begin{document}
+
+\renewcommand{\niandu}{2017--2018}
+\renewcommand{\xueqi}{2}
+\renewcommand{\kecheng}{大学数学}
+\renewcommand{\zhuanye}{理工四学分} % 可以为空白
+\renewcommand{\jiaoshi}{张三，李四，王五} % 教师姓名
+\renewcommand{\shijian}{2018~年~06~月~28~日}
+\renewcommand{\bixiu}{1} % 1 为必修，0 为选修
+\renewcommand{\bijuan}{1} % 1 为闭卷，0 为开卷
+\renewcommand{\shijuan}{A} % A 或 B 或 C 卷
+\renewcommand{\neizhao}{1} % 1 打勾，0 不勾
+\renewcommand{\waizhao}{0} % 1 打勾，0 不勾
+
+\makehead % 生成试卷表头
+
+\makepart{填空题}{共~6~小题，每小题~3~分，共~18~分}
+
+\answertable[3em]{6}{3} % 生成答题栏：行高3em，总共6题，每行3题
+
+\begin{problem}
+设常数$k>0$，函数$f(x)=\ln x-\dfrac{x}{\e}+k$在$(0,+\infty)$内零点的个数为 \fillout{$2$}．
+\end{problem}
+
+\vfill
+
+\begin{problem}
+设$\va=(2,1,2)$，$\vb=(4,-1,10)$，$\vc=\vb-\lambda\va$，且$\va\bot\vc$，则$\lambda=$ \fillout{$3$}．
+\end{problem}
+
+\vfill
+
+\begin{problem}
+已知二阶行列式 $\left|\begin{array}{cc}
+ 1 & 2\\
+ - 3 & x
+\end{array}\right|=0$，则 $x=$ \fillout{$-6$}．
+\end{problem}
+
+\vfill
+
+\begin{problem}
+向量组 $\alpha_1=(1,1,0), \alpha_2=(0,1,1), \alpha_3=(1,0,1)$，
+则将向量 $\beta=(4, 5, 3)$ 表示为 $\alpha_1, \alpha_2, \alpha_3$
+的线性组合为 $\beta=$ \fillout{$3\alpha_1+2\alpha_2+\alpha_3$}．
+\end{problem}
+
+\vfill
+
+\begin{problem}
+已知随机变量$\xi$的期望和方差各为$E\xi=3, D\xi=2$, 则$E\xi^2=$ \fillout{$11$}．
+\end{problem}
+
+\vfill
+
+\begin{problem}
+已知$\xi$和$\eta$相互独立且$\xi\sim N(1,4), \eta\sim N(2,5)$，则$\xi-2\eta\sim$ \fillout{$N(-3,24)$}．
+\end{problem}
+
+\vfill
+
+\newpage
+
+\makepart{单选题}{共~6~小题，每小题~3~分，共~18~分}
+
+\answertable{6}{6} % 生成答题栏：默认行高，总共8题，每行8题
+
+\begin{problem}
+在下列等式中，正确的结果是\pickout{C}
+\begin{abcd}
+\item $\int f'(x)\dx=f(x)$
+\item $\int \d f(x)=f(x)$
+\item $\frac{\d}{\dx}\big(\int f(x)\dx\big)=f(x)$
+\item $\d\big(\int f(x)\dx\big)=f(x)$
+\end{abcd}
+\end{problem}
+
+\bigskip
+
+\begin{problem}
+假设$F(x)$是连续函数$f(x)$的一个原函数，则必有\pickout{A}
+\begin{abcd}
+\item $F(x)$是偶函数 $\Leftrightarrow$ $f(x)$是奇函数
+\item $F(x)$是奇函数 $\Leftrightarrow$ $f(x)$是偶函数
+\item $F(x)$是周期函数 $\Leftrightarrow$ $f(x)$是周期函数
+\item $F(x)$是单调函数 $\Leftrightarrow$ $f(x)$是单调函数
+\end{abcd}
+\end{problem}
+
+\bigskip
+
+\begin{problem}
+设矩阵 $A = \left(\begin{array}{ccc}
+ 1 & 1 & 0\\
+ 1 & x & 0\\
+ 0 & 0 & 1
+\end{array}\right)$ 其中两个特征值为 $\lambda_1 = 1$ 和 $\lambda_2
+= 2$，则 $x=$ \pickout{B}
+\begin{abcd}
+\item $2$
+\item $1$
+\item $0$
+\item $-1$
+\end{abcd}
+\end{problem}
+
+\bigskip
+
+\begin{problem}
+二次型 $f = 4 x_1^2 - 2 x_1 x_2 + 6 x_2^2$ 对应的矩阵等于 \pickout{C}
+\begin{abcd}
+\item $\left(\begin{array}{cc}
+ 4 & - 2\\
+ - 2 & 6
+\end{array}\right)$
+\item $\left(\begin{array}{cc}
+ 2 & - 2\\
+ - 2 & 3
+\end{array}\right)$
+\item $\left(\begin{array}{cc}
+ 4 & - 1\\
+ - 1 & 6
+\end{array}\right)$
+\item $\left(\begin{array}{cc}
+ 2 & - 1\\
+ - 1 & 3
+\end{array}\right)$
+\end{abcd}
+\end{problem}
+
+\bigskip
+
+\begin{problem}
+下列说法\CJKunderline{不正确}的是\pickout{B}
+\begin{abcd}
+\item 大数定律说明了大量相互独立且同分布的随机变量的均值的稳定性
+\item 大数定律说明大量相互独立且同分布的随机变量的均值近似于正态分布
+\item 中心极限定理说明了大量相互独立且同分布的随机变量的和的稳定性
+\item 中心极限定理说明大量相互独立且同分布的随机变量的和近似于正态分布
+\end{abcd}
+\end{problem}
+
+\bigskip
+
+\begin{problem}
+对总体$X$和样本$(X_1,\cdots,X_n)$的说法哪个是\CJKunderline{不正确}的\pickout{D}
+\begin{abcd}
+\item 总体是随机变量
+\item 样本是$n$元随机变量
+\item $X_1, \cdots, X_n$相互独立
+\item $X_1 = X_2 =\cdots = X_n$
+\end{abcd}
+\end{problem}
+
+\bigskip
+
+\newpage
+
+\makepart{计算题}{共~6~小题，每小题~8~分，共~48~分}
+
+\begin{problem}
+求不定积分$\displaystyle\int\e^{2x}\,(\tan x+1)^2\dx$。
+\end{problem}
+
+\smallskip
+
+\begin{solution}
+\everymath{\displaystyle}%
+原式 \? $=\int\e^{2x}\,\sec^2 x\dx+2\int\e^{2x}\,\tan x\dx$ \points{2}
+\+ $=\int\e^{2x}\,\d(\tan x)+ 2\int\e^{2x}\,\tan x\dx$ \points{4}
+\+ $=\e^{2x}\,\tan x - 2\int\e^{2x}\,\tan x\dx+ 2\int\e^{2x}\,\tan x\dx$ \points{6}
+\+ $=\e^{2x}\,\tan x + C$ \points{8}
+\end{solution}
+
+\vfill
+
+\begin{problem}
+求过点$A(1,2,-1), B(2,3,0),C(3,3,2)$ 的三角形$\triangle ABC$ 的面积和它们确定的平面方程.
+\end{problem}
+
+\smallskip
+
+\begin{solution}
+由题设$\overrightarrow{AB}=(1,1,1),\overrightarrow{AC}=(2,1,3)$, \points{2}
+故$\overrightarrow{AB}\times \overrightarrow{AC}=\begin{vmatrix}
+\vec{i}&\vec{j} &\vec{k}\\
+1&1&1\\
+2&1&3\\
+\end{vmatrix}=(2,-1,-1)$, \points{4}
+三角形$\triangle ABC$ 的面积为$S_{\triangle ABC}=\dfrac{1}{2}\big|\overrightarrow{AB}\times
+\overrightarrow{AC}\big|=\dfrac{1}{2}\sqrt{6}.$ \points{6}
+所求平面的方程为$2(x-2)-(y-3)-z=0$, 即$2x-y-z-1=0$ \points{8}
+\end{solution}
+
+\vfill
+
+\newpage
+
+\begin{problem}
+计算四阶行列式 $A = \left|\begin{array}{cccc}
+ 0 & 1 & 2 & 3\\
+ 1 & 2 & 3 & 0\\
+ 2 & 3 & 0 & 1\\
+ 3 & 0 & 1 & 2
+\end{array}\right|$ 的值．
+\end{problem}
+
+\smallskip
+
+\begin{solution}
+$A \? = \left|\begin{array}{cccc}
+ 0 & 1 & 2 & 3\\
+ 1 & 2 & 3 & 0\\
+ 2 & 3 & 0 & 1\\
+ 3 & 0 & 1 & 2
+ \end{array}\right| = \left|\begin{array}{cccc}
+ 0 & 1 & 2 & 3\\
+ 1 & 2 & 3 & 0\\
+ 0 & - 1 & - 6 & 1\\
+ 0 & - 6 & - 8 & 2
+ \end{array}\right| = 1 \cdot (- 1)^{2 + 1} \left|\begin{array}{ccc}
+ 1 & 2 & 3\\
+ - 1 & - 6 & 1\\
+ - 6 & - 8 & 2
+ \end{array}\right|$ \points{4}
+\+ $= -\left|\begin{array}{ccc}
+ 1 & 2 & 3\\
+ 0 & - 4 & 4\\
+ 0 & 4 & 20
+ \end{array}\right| = - \left|\begin{array}{cc}
+ - 4 & 4\\
+ 4 & 20
+ \end{array}\right| = -(-4\cdot20-4\cdot4) = 96$ \points{8}
+\end{solution}
+
+\vfill
+
+\begin{problem}
+利用配方法，将二次型 $f = x_1^2 + 2 x_1 x_2 - 6 x_1 x_3 + 2 x_2^2 - 12
+x_2 x_3 + 9 x^2_3$ 化为标准形 $f = d_1 y^2_1 + d_2 y^2_2 + d_3 y^2_3$ ．
+\end{problem}
+
+\smallskip
+
+\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
+ \+ $= 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
+ \+ $= (x_1 + x_2 - 3 x_3)^2 + x_2^2 - 6 x_2 x_3$ \points{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
+ \+ $= (x_1 + x_2 - 3 x_3)^2 + (x_2 - 3 x_3)^2 - 9 x_3^2$ \points{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．\points{8}
+\end{solution}
+
+\vfill
+
+\newpage
+
+\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
+(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}
+
+\smallskip
+
+\begin{solution}
+$E\xi = n p = 100 \cdot 0.2 = 20, D\xi = n p q = 100 \cdot 0.2 \cdot 0.8 = 16$. \points{2}
+(1) $P (10 < \xi < 30) = P (|\xi - E\xi| < 10) \ge 1 - \frac{D\xi}{10^2}
+ = 1 - \frac{16}{100} = 0.84$. \points{4}
+(2) $P (10 < \xi < 30) \? \approx \Phi_0\left(\frac{30 - 20}{\sqrt{16}}\right)
+ - \Phi_0\left(\frac{10 - 20}{\sqrt{16}}\right)$ \points{6}
+ \+ $= 2 \Phi_0(2.5) - 1 = 2 \cdot 0.9938 - 1 =0.9876$ \points{8}
+\end{solution}
+
+\vfill
+
+\begin{problem}
+从正态总体$N(\mu,\sigma^2)$中抽出样本容量为16的样本，算得其平均数为3160，标准差为100．
+试检验假设$H_0:\mu=3140$是否成立($\alpha = 0.01$)．
+\end{problem}
+
+\smallskip
+
+\begin{solution}
+(1) 待检假设 $H_0 : \mu = 3140$. \points{1}
+(2) 选取统计量 $T = \frac{\widebar{X}-\mu}{S / \sqrt{n}} \sim t(n-1)$. \points{3}
+(3) 查表得到 $t_{\alpha} = t_{\alpha} (n - 1) = t_{0.01} (15) =2.947$. \points{5}
+(4) 计算统计值 $t = \frac{\widebar{x} - \mu_0}{s/\sqrt{n}} =\frac{3160-3140}{100/4} = 0.8$.\points{7}
+(5) 由于 $| t | < t_{\alpha}$, 故接受 $H_0$, 即假设成立. \points{8}
+\end{solution}
+
+\vfill
+
+\newpage
+
+\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}
+
+\smallskip
+
+\begin{solution}
+(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．\points{4}
+(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$2$．\points{8}
+\end{solution}
+
+\vfill
+
+\begin{problem}
+设事件$A$和$B$相互独立，证明$A$和$\widebar{B}$相互独立．
+\end{problem}
+
+\smallskip
+
+\begin{solution}
+\? $P (A \cdot \widebar{B}) = P (A - B) = P (A - A B)$ \points{2}
+\< $= P (A) - P (A B) = P (A) - P (A) P (B)$ \points{4}
+\< $= P (A) (1 - P (B)) = P (A) P (\widebar{B})$ \points{6}
+所以$A$和$\widebar{B}$相互独立．\points{8}
+\end{solution}
+
+\vfill
+
+\makedata{一些可能用到的数据} %附录数据
+
+\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{tabularx}
+
+\end{document}
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@@ -3,4 +3,4 @@
% 重新排版原有的 A4 试卷，不显示答案
\PassOptionsToClass{noanswer}{jnuexam}
-\input{exam-a}
+\input{exam-a-answer}
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@@ -1,359 +0,0 @@
-% -*- coding: utf-8 -*-
-% !TEX program = xelatex
-\documentclass{jnuexam}
-
-%\answerfalse %不显示答案
-
-\begin{document}
-
-\renewcommand{\niandu}{2017--2018}
-\renewcommand{\xueqi}{2}
-\renewcommand{\kecheng}{大学数学}
-\renewcommand{\zhuanye}{理工四学分} % 可以为空白
-\renewcommand{\jiaoshi}{张三，李四，王五} % 教师姓名
-\renewcommand{\shijian}{2018~年~06~月~28~日}
-\renewcommand{\bixiu}{1} % 1 为必修，0 为选修
-\renewcommand{\bijuan}{1} % 1 为闭卷，0 为开卷
-\renewcommand{\shijuan}{A} % A 或 B 或 C 卷
-\renewcommand{\neizhao}{1} % 1 打勾，0 不勾
-\renewcommand{\waizhao}{0} % 1 打勾，0 不勾
-
-\makehead % 生成试卷表头
-
-\makepart{填空题}{共~6~小题，每小题~3~分，共~18~分}
-
-\answertable[3em]{6}{3} % 生成答题栏：行高3em，总共6题，每行3题
-
-\newpageb % B卷分页点
-
-\begin{problem}
-设常数$k>0$，函数$f(x)=\ln x-\dfrac{x}{\e}+k$在$(0,+\infty)$内零点的个数为 \fillout{$2$}．
-\end{problem}
-
-\vfill
-
-\begin{problem}
-设$\va=(2,1,2)$，$\vb=(4,-1,10)$，$\vc=\vb-\lambda\va$，且$\va\bot\vc$，则$\lambda=$ \fillout{$3$}．
-\end{problem}
-
-\vfill
-
-\begin{problem}
-已知二阶行列式 $\left|\begin{array}{cc}
- 1 & 2\\
- - 3 & x
-\end{array}\right|=0$，则 $x=$ \fillout{$-6$}．
-\end{problem}
-
-\vfill
-
-\begin{problem}
-向量组 $\alpha_1=(1,1,0), \alpha_2=(0,1,1), \alpha_3=(1,0,1)$，
-则将向量 $\beta=(4, 5, 3)$ 表示为 $\alpha_1, \alpha_2, \alpha_3$
-的线性组合为 $\beta=$ \fillout{$3\alpha_1+2\alpha_2+\alpha_3$}．
-\end{problem}
-
-\vfill
-
-\begin{problem}
-已知随机变量$\xi$的期望和方差各为$E\xi=3, D\xi=2$, 则$E\xi^2=$ \fillout{$11$}．
-\end{problem}
-
-\vfill
-
-\begin{problem}
-已知$\xi$和$\eta$相互独立且$\xi\sim N(1,4), \eta\sim N(2,5)$，则$\xi-2\eta\sim$ \fillout{$N(-3,24)$}．
-\end{problem}
-
-\vfill
-
-\newpagea % A卷分页点
-
-\makepart{单选题}{共~6~小题，每小题~3~分，共~18~分}
-
-\answertable{6}{6} % 生成答题栏：默认行高，总共8题，每行8题
-
-\newpageb % B卷分页点
-
-\begin{problem}
-在下列等式中，正确的结果是\pickout{C}
-\begin{abcd}
-\item $\int f'(x)\dx=f(x)$
-\item $\int \d f(x)=f(x)$
-\item $\frac{\d}{\dx}\big(\int f(x)\dx\big)=f(x)$
-\item $\d\big(\int f(x)\dx\big)=f(x)$
-\end{abcd}
-\end{problem}
-
-\bigskip
-
-\begin{problem}
-假设$F(x)$是连续函数$f(x)$的一个原函数，则必有\pickout{A}
-\begin{abcd}
-\item $F(x)$是偶函数 $\Leftrightarrow$ $f(x)$是奇函数
-\item $F(x)$是奇函数 $\Leftrightarrow$ $f(x)$是偶函数
-\item $F(x)$是周期函数 $\Leftrightarrow$ $f(x)$是周期函数
-\item $F(x)$是单调函数 $\Leftrightarrow$ $f(x)$是单调函数
-\end{abcd}
-\end{problem}
-
-\bigskip
-
-\begin{problem}
-设矩阵 $A = \left(\begin{array}{ccc}
- 1 & 1 & 0\\
- 1 & x & 0\\
- 0 & 0 & 1
-\end{array}\right)$ 其中两个特征值为 $\lambda_1 = 1$ 和 $\lambda_2
-= 2$，则 $x=$ \pickout{B}
-\begin{abcd}
-\item $2$
-\item $1$
-\item $0$
-\item $-1$
-\end{abcd}
-\end{problem}
-
-\bigskip
-
-\begin{problem}
-二次型 $f = 4 x_1^2 - 2 x_1 x_2 + 6 x_2^2$ 对应的矩阵等于 \pickout{C}
-\begin{abcd}
-\item $\left(\begin{array}{cc}
- 4 & - 2\\
- - 2 & 6
-\end{array}\right)$
-\item $\left(\begin{array}{cc}
- 2 & - 2\\
- - 2 & 3
-\end{array}\right)$
-\item $\left(\begin{array}{cc}
- 4 & - 1\\
- - 1 & 6
-\end{array}\right)$
-\item $\left(\begin{array}{cc}
- 2 & - 1\\
- - 1 & 3
-\end{array}\right)$
-\end{abcd}
-\end{problem}
-
-\bigskip
-
-\begin{problem}
-下列说法\CJKunderline{不正确}的是\pickout{B}
-\begin{abcd}
-\item 大数定律说明了大量相互独立且同分布的随机变量的均值的稳定性
-\item 大数定律说明大量相互独立且同分布的随机变量的均值近似于正态分布
-\item 中心极限定理说明了大量相互独立且同分布的随机变量的和的稳定性
-\item 中心极限定理说明大量相互独立且同分布的随机变量的和近似于正态分布
-\end{abcd}
-\end{problem}
-
-\bigskip
-
-\begin{problem}
-对总体$X$和样本$(X_1,\cdots,X_n)$的说法哪个是\CJKunderline{不正确}的\pickout{D}
-\begin{abcd}
-\item 总体是随机变量
-\item 样本是$n$元随机变量
-\item $X_1, \cdots, X_n$相互独立
-\item $X_1 = X_2 =\cdots = X_n$
-\end{abcd}
-\end{problem}
-
-\bigskip
-
-\newpagea % A卷分页点
-
-\makepart{计算题}{共~6~小题，每小题~8~分，共~48~分}
-
-\newpageb % B卷分页点
-
-\begin{problem}
-求不定积分$\displaystyle\int\e^{2x}\,(\tan x+1)^2\dx$。
-\end{problem}
-
-\bigskip
-
-\begin{solution}
-\everymath{\displaystyle}%
-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
-
-\begin{problem}
-求过点$A(1,2,-1), B(2,3,0),C(3,3,2)$ 的三角形$\triangle ABC$ 的面积和它们确定的平面方程.
-\end{problem}
-
-\bigskip
-
-\begin{solution}
-由题设$\overrightarrow{AB}=(1,1,1),\overrightarrow{AC}=(2,1,3)$, \score{2}
-故$\overrightarrow{AB}\times \overrightarrow{AC}=\begin{vmatrix}
-\vec{i}&\vec{j} &\vec{k}\\
-1&1&1\\
-2&1&3\\
-\end{vmatrix}=(2,-1,-1)$, \score{4}
-三角形$\triangle ABC$ 的面积为$S_{\triangle ABC}=\dfrac{1}{2}\big|\overrightarrow{AB}\times
-\overrightarrow{AC}\big|=\dfrac{1}{2}\sqrt{6}.$ \score{6}
-所求平面的方程为$2(x-2)-(y-3)-z=0$, 即$2x-y-z-1=0$ \score{8}
-\end{solution}
-
-\vfill
-
-\newpage % A,B卷共同分页点
-
-\begin{problem}
-计算四阶行列式 $A = \left|\begin{array}{cccc}
- 0 & 1 & 2 & 3\\
- 1 & 2 & 3 & 0\\
- 2 & 3 & 0 & 1\\
- 3 & 0 & 1 & 2
-\end{array}\right|$ 的值．
-\end{problem}
-
-\bigskip
-
-\begin{solution}
-$A \? = \left|\begin{array}{cccc}
- 0 & 1 & 2 & 3\\
- 1 & 2 & 3 & 0\\
- 2 & 3 & 0 & 1\\
- 3 & 0 & 1 & 2
- \end{array}\right| = \left|\begin{array}{cccc}
- 0 & 1 & 2 & 3\\
- 1 & 2 & 3 & 0\\
- 0 & - 1 & - 6 & 1\\
- 0 & - 6 & - 8 & 2
- \end{array}\right| = 1 \cdot (- 1)^{2 + 1} \left|\begin{array}{ccc}
- 1 & 2 & 3\\
- - 1 & - 6 & 1\\
- - 6 & - 8 & 2
- \end{array}\right|$ \score{4}
-\+ $= -\left|\begin{array}{ccc}
- 1 & 2 & 3\\
- 0 & - 4 & 4\\
- 0 & 4 & 20
- \end{array}\right| = - \left|\begin{array}{cc}
- - 4 & 4\\
- 4 & 20
- \end{array}\right| = -(-4\cdot20-4\cdot4) = 96$ \score{8}
-\end{solution}
-
-\vfill
-
-\begin{problem}
-用配方法将二次型 $f = x_1^2 + 2 x_1 x_2 - 6 x_1 x_3 + 2 x_2^2 - 12
-x_2 x_3 + 9 x^2_3$ 化为标准形 $f = d_1 y^2_1 + d_2 y^2_2 + d_3 y^2_3$ ．
-\end{problem}
-
-\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
- \+ $= 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}
-
-\vfill
-
-\newpage % A,B卷共同分页点
-
-\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
-(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}
-(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}
-(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
-
-\begin{problem}
-从正态总体$N(\mu,\sigma^2)$中抽出样本容量为16的样本，算得其平均数为3160，标准差为100．
-试检验假设$H_0:\mu=3140$是否成立($\alpha = 0.01$)．
-\end{problem}
-
-\bigskip
-
-\begin{solution}
-(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
-
-\newpagea % A卷分页点
-
-\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}
-
-\bigskip
-
-\begin{solution}
-(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}
-(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}
-
-\vfill
-
-\begin{problem}
-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}
-\? $P (A \cdot \widebar{B}) = P (A - B) = P (A - A B)$ \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{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{tabularx}
-
-\end{document}
Deleted: trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3input.pdf
===================================================================
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Deleted: trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3input.tex
===================================================================
--- trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3input.tex 2023-12-11 00:43:20 UTC (rev 69088)
+++ trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3input.tex 2023-12-11 21:08:45 UTC (rev 69089)
@@ -1,9 +0,0 @@
-% -*- coding: utf-8 -*-
-% !TEX program = xelatex
-
-% 直接包含 A4 试卷的 PDF 文件，生成双栏的 A3 试卷
-\documentclass[a3input]{jnuexam}
-\begin{document}
-\includepdf[pages=-,nup=2x1,offset=0 0,delta=0 0]{exam-b-empty}
-%\includepdf[pages=-,nup=2x1,offset=0 0,delta=0 0,frame]{exam-b}
-\end{document}
Deleted: trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3split.pdf
===================================================================
(Binary files differ)
Deleted: trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3split.tex
===================================================================
--- trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3split.tex 2023-12-11 00:43:20 UTC (rev 69088)
+++ trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3split.tex 2023-12-11 21:08:45 UTC (rev 69089)
@@ -1,6 +0,0 @@
-% -*- coding: utf-8 -*-
-% !TEX program = xelatex
-
-% 将原有的 A4 试卷改为 A3 试卷双栏排版
-\PassOptionsToClass{a3paper,noanswer}{jnuexam}
-\input{exam-b}
Added: trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-answer.pdf
===================================================================
(Binary files differ)
Index: trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-answer.pdf
===================================================================
--- trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-answer.pdf 2023-12-11 00:43:20 UTC (rev 69088)
+++ trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-answer.pdf 2023-12-11 21:08:45 UTC (rev 69089)
Property changes on: trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-answer.pdf
___________________________________________________________________
Added: svn:mime-type
## -0,0 +1 ##
+application/pdf
\ No newline at end of property
Added: trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-answer.tex
===================================================================
--- trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-answer.tex (rev 0)
+++ trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-answer.tex 2023-12-11 21:08:45 UTC (rev 69089)
@@ -0,0 +1,6 @@
+% -*- coding: utf-8 -*-
+% !TEX program = xelatex
+
+% 从 A 卷自动生成 B 卷，将题目乱序排列
+\PassOptionsToClass{random}{jnuexam}
+\input{exam-a-answer}
Property changes on: trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-answer.tex
___________________________________________________________________
Added: svn:eol-style
## -0,0 +1 ##
+native
\ No newline at end of property
Modified: trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-empty.pdf
===================================================================
(Binary files differ)
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===================================================================
--- trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-empty.tex 2023-12-11 00:43:20 UTC (rev 69088)
+++ trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b-empty.tex 2023-12-11 21:08:45 UTC (rev 69089)
@@ -3,4 +3,4 @@
% 重新排版原有的 A4 试卷，不显示答案
\PassOptionsToClass{noanswer}{jnuexam}
-\input{exam-b}
+\input{exam-b-answer}
Deleted: trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b.pdf
===================================================================
(Binary files differ)
Deleted: trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b.tex
===================================================================
--- trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b.tex 2023-12-11 00:43:20 UTC (rev 69088)
+++ trunk/Master/texmf-dist/doc/latex/jnuexam/exam-b.tex 2023-12-11 21:08:45 UTC (rev 69089)
@@ -1,6 +0,0 @@
-% -*- coding: utf-8 -*-
-% !TEX program = xelatex
-
-% 从 A 卷自动生成 B 卷，将题目逆序排列
-\PassOptionsToClass{reverse}{jnuexam}
-\input{exam-a}
Modified: trunk/Master/texmf-dist/doc/latex/jnuexam/jnuexam.pdf
===================================================================
(Binary files differ)
Modified: trunk/Master/texmf-dist/doc/latex/jnuexam/jnuexam.tex
===================================================================
--- trunk/Master/texmf-dist/doc/latex/jnuexam/jnuexam.tex 2023-12-11 00:43:20 UTC (rev 69088)
+++ trunk/Master/texmf-dist/doc/latex/jnuexam/jnuexam.tex 2023-12-11 21:08:45 UTC (rev 69089)
@@ -77,6 +77,8 @@
\titlepage
\end{frame}
+\section{模板介绍}
+
\begin{framex}
\frametitle{简单介绍}
本文档介绍 \verb!jnuexam! 文档类。这个文档类提供暨南大学考试试卷的 LaTeX 模板。
@@ -84,7 +86,7 @@
这个模板将格式和内容分开，而且可以从一份 \verb!tex! 文件编译出四份试卷(A卷 / B卷 / A卷答案 / B卷答案)，使用方便。
\par
这个模板的最新版本可以在下面地址下载：\newline
- \href{https://lvjr.bitbucket.io/jnuexam.html?\the\year}{https://lvjr.bitbucket.io/jnuexam.html}
+ \color{blue}{\href{https://lvjr.bitbucket.io/jnuexam.html?\the\year}{\ttfamily https://lvjr.bitbucket.io/jnuexam.html}}
\end{framex}
\begin{framex}
@@ -98,11 +100,42 @@
推荐用较先进的 \verb!XeLaTeX! 编译。
\end{framex}
+\section{试卷结构}
+
\begin{framex}
\frametitle{试卷结构}
+下面是 \verb!jnuexam! 试卷文档的基本结构：
\begin{code}
\documentclass{jnuexam}
+% 导言区
\begin{document}
+% 正文区
+\end{document}
+\end{code}
+导言区用于设定装订线和草稿纸等等选项。\par
+正文区用于填写试卷表头和输入试卷内容。
+\end{framex}
+
+\begin{framex}
+\frametitle{装订草稿}
+在文档的导言区可以设定装订线和草稿纸。比如：
+\begin{code}
+\setexam{
+ binding = 2, % 装订线
+ scratch = 1, % 草稿纸
+}
+\end{code}
+其中 \verb!binding! 取 0 表示没有装订线，
+取 1 表示仅空白试卷有，取 2 表示空白试卷和试卷答案都有。\par
+而 \verb!scratch! 的取值表示草稿纸数量，以 A3 大小双面印刷计算。
+草稿纸仅在空白试卷中出现，试卷答案里不会带草稿纸。
+\end{framex}
+
+\begin{framex}
+\frametitle{试卷正文}
+\begin{code}
+\documentclass{jnuexam}
+\begin{document}
......
\makehead %生成试卷表头
......
@@ -145,14 +178,19 @@
\makepart{判断题}{题数分值}
\begin{problem}
-第一道判断题描述。\true
+第一道判断题描述。\tickout{t}
\end{problem}
\begin{problem}
-第二道判断题描述。\false
+第二道判断题描述。\tickout{f}
\end{problem}
\end{code}
-其中 \verb!\true! 和 \verb!\false! 命令分别表示正确和错误。
+其中 \verb!\tickout{t}! 和 \verb!\tickout{f}!
+分别表示打勾（\textcolor{blue}{$\checkmark$}）和打叉（\textcolor{blue}{\large$\times$}）。
+还可用大写的 \verb!\tickout{T}! 和 \verb!\tickout{F}!，
+分别表示输出 \textcolor{blue}{\textsf{T}} 和 \textcolor{blue}{\textsf{F}}。
+\par
+答案必须放在 \verb!\tickout! 命令里；这样才能在生成空白试卷时隐藏它。
\end{framex}
\begin{framex}
@@ -274,14 +312,14 @@
\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，
-可以用 \verb!\score! 命令给出各步骤得分。比如：
+可以用 \verb!\points! 命令给出各步骤得分。比如：
\begin{code}
\begin{solution}
-$1+1=2$ \score{4}
-$2+2=4$ \score{8}
+$1+1=2$ \points{4}
+$2+2=4$ \points{8}
\end{solution}
\end{code}
-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。
+评分命令 \verb!\points! 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}
@@ -289,11 +327,11 @@
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
+\leavevmode\phantom{我们有$(a+b)^2$}${}= a^2 + 2ab + b^2$ \hfill$\cdots\cdots$ 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}
+ \+$= a^2+2ab+b^2$ \points{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。
@@ -306,11 +344,11 @@
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
+\leavevmode\phantom{我们\,}${}= a^2 + 2ab + b^2$ \hfill$\cdots\cdots$ 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}
+ \< $= a^2+2ab+b^2$ \points{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!\<! 表示跳到之前保存位置的左侧（左移一个等号的宽度）。
@@ -345,26 +383,28 @@
附录数据必须放在 \verb!\makedata! 命令后面；否则在从A卷生成B卷时会出问题。
\end{framex}
+\section{模板选项}
+
\begin{framex}
\frametitle{空白试卷}
-假设 \verb!exam-a.tex! 是含答案的试卷。新建一个包含以下内容的 \verb!exam-a-empty.tex! 文档，
+假设 \verb!exam-a-answer.tex! 是含答案的试卷。新建一个包含以下内容的 \verb!exam-a-empty.tex! 文档，
编译后将得到不含答案的空白试卷。
\begin{code}
\PassOptionsToClass{noanswer}{jnuexam}
-\input{exam-a}
+\input{exam-a-answer}
\end{code}
也就是说，给 \verb!jnuexam! 文档类加上 \verb!noanswer! 选项后，编译时将会自动隐藏试卷答案。
\end{framex}
\begin{framex}
-\frametitle{逆序出题}
-假设 \verb!exam-a.tex! 是含答案的A卷。新建一个包含以下内容的 \verb!exam-b.tex! 文档，
-编译后将得到逆序出题的B卷。
+\frametitle{乱序出题}
+假设 \verb!exam-a-answer.tex! 是含答案的A卷。新建一个包含以下内容的 \verb!exam-b-answer.tex! 文档，
+编译后将得到乱序出题的B卷。
\begin{code}
-\PassOptionsToClass{reverse}{jnuexam}
-\input{exam-a}
+\PassOptionsToClass{random}{jnuexam}
+\input{exam-a-answer}
\end{code}
-也就是说，给 \verb!jnuexam! 文档类加上 \verb!reverse! 选项后，编译时将会逆序排列各题型的小题。
+也就是说，给 \verb!jnuexam! 文档类加上 \verb!random! 选项后，编译时将会乱序排列各题型的小题。
\end{framex}
\begin{framex}
@@ -383,48 +423,35 @@
\hline
\end{tabularx}\par
当然，竖直空白命令可以连续使用多个，以得到所需的空白。
+\vfill
+在试卷中可以使用分页命令 \verb!\newpage!，
+\alert{不要}使用其他分页命令，比如 \verb!\clearpage! 等，
+以免导致 B 卷格式错乱。
\end{framex}
\begin{framex}
-\frametitle{分页命令}
-分页命令 \verb!\newpage! 同样可以使用。由于A卷和B卷的小题顺序相反，
-其中的分页位置通常也不同。因此这里另外提供 \verb!\newpagea! 和 \verb!\newpageb! 命令，
-分别只对 A 卷和 B 卷有效。
-\par
-\renewcommand{\arraystretch}{1.3}%
-\begin{tabularx}{\linewidth}{l<{\qquad}X}
- \hline
- \texttt{\string\newpage} & 分页，对A卷和B卷均有效 \\
- \hline
- \texttt{\string\newpagea} & 分页，仅对A卷有效 \\
- \hline
- \texttt{\string\newpageb} & 分页，仅对B卷有效 \\
- \hline
-\end{tabularx}\par
-在试卷中\alert{不要}使用其他分页命令，比如 \verb!\clearpage! 等。
+\frametitle{双栏试卷}
+假设 \verb!exam-a-empty.tex! 是原来试卷的 TeX 文件。新建一个包含以下内容的文档，
+编译后将得到的 A3 纸张的试卷。
+\begin{code}
+\PassOptionsToClass{a3paper}{jnuexam}
+\input{exam-a-empty}
+\end{code}
+也就是说，给 \verb!jnuexam! 文档类加上 \verb!a3paper! 选项后，编译时将会按照 A3 纸张排版出双栏试卷。
\end{framex}
\begin{framex}
-\frametitle{分页例子}
-关于分页命令的使用，可以看下面的典型例子：
+\frametitle{双栏试卷}
+假设 \verb!exam-a-empty.pdf! 是原来试卷的 PDF 文件。新建一个包含以下内容的文档，
+编译后将得到的 A3 纸张的试卷。
\begin{code}
-\makepart{某题型}{题型分值}
-\newpageb
-\begin{problem}第一题\end{problem}\vfill
-\begin{problem}第二题\end{problem}\vfill
-\newpage
-\begin{problem}第三题\end{problem}\vfill
-\begin{problem}第四题\end{problem}\vfill
-\newpagea
+\documentclass[a3input]{jnuexam}
+\begin{document}
+\includepdf[pages=-,nup=2x1]{exam-a-empty}
+\end{document}
\end{code}
-这样编译得到的A卷就是这样的顺序：
-\begin{code}
-第一题 第二题 分页 第三题 第四题 分页
-\end{code}
-而编译得到的B卷就是这样的顺序：
-\begin{code}
-第四题 第三题 分页 第二题 第一题 分页
-\end{code}
+%\includepdf[pages=-,nup=2x1,offset=0 0,delta=0 0]{exam-a-empty}
+这种用法直接读入 A4 试卷的 PDF 文件，生成双栏的 A3 试卷，适合没有 TeX 文件时使用。
\end{framex}
\end{document}
Modified: trunk/Master/texmf-dist/tex/latex/jnuexam/jnuexam.cls
===================================================================
--- trunk/Master/texmf-dist/tex/latex/jnuexam/jnuexam.cls 2023-12-11 00:43:20 UTC (rev 69088)
+++ trunk/Master/texmf-dist/tex/latex/jnuexam/jnuexam.cls 2023-12-11 21:08:45 UTC (rev 69089)
@@ -6,25 +6,39 @@
% ----------------------------------------------------------------------------
\NeedsTeXFormat{LaTeX2e}
-\ProvidesClass{jnuexam}[2020/11/06 v1.0 An exam class for Jinan University]
+\ProvidesClass{jnuexam}[2023/12/11 v2.0 An exam class for Jinan University]
-\newif\ifmifengxian \mifengxiantrue % 是否加密封线
-\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
+%% 旧版本的 LaTeX 不能识别 2022-11-01 这种日期格式
+%\@ifl at t@r\fmtversion{2022-11-01}{}{
+\@ifl at t@r\fmtversion{2022/11/01}{}{
+ \ClassError{jnuexam}{%
+ Your current TeX distribution is quite old.\MessageBreak
+ We need CTeX 3.0+ or MiKTeX 2023+ or TeXLive 2023+%
+ }{Please update your TeX distribution first.}
+}
-\DeclareOption{a3paper}{\sidebysidetrue}
-\DeclareOption{a3input}{\sidebysidetrue\mifengxianfalse}
-\DeclareOption{reverse}{\reversetrue}
+\newif\ifplain \plainfalse % 是否添加装订线和草稿纸
+\newif\iftwoinone \twoinonefalse % 是否使用 A3 纸张
+\newif\ifoneside \onesidefalse % 是否单面印刷试卷
+\newif\ifresetnumber \resetnumbertrue % 是否对各题型小题分别编号
+\newif\ifrandom \randomfalse % 是否乱序出题
+\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\ifsolidot \solidotfalse % 是否替换空心句号为实心句号
+\newif\ifcellspace \cellspacefalse % 增加表格列间距
+\newif\ifmedmath \medmathfalse % 切换公式尺寸
+
+\DeclareOption{plain}{\plaintrue}
+\DeclareOption{a3paper}{\twoinonetrue}
+\DeclareOption{a3input}{\twoinonetrue\plaintrue}
+\DeclareOption{oneside}{\onesidetrue}
+\DeclareOption{random}{\randomtrue}
\DeclareOption{noanswer}{\answerfalse}
\DeclareOption{amsfonts}{\amsfontstrue}
\DeclareOption{sourcehan}{\sourcehantrue}
-\DeclareOption{display}{\displaytrue}
-\DeclareOption{collection}{\collectiontrue\mifengxianfalse}
+\DeclareOption{solidot}{\solidottrue}
+\DeclareOption{medmath}{\cellspacetrue\medmathtrue}
\DeclareOption*{\PassOptionsToClass{\CurrentOption}{ctexart}} %其它选项
@@ -32,33 +46,62 @@
% 四号 小四号 五号 小五号
% 14bp 12bp 10.5bp 9bp
+% 实际上，在旧版本 ctex 中只能用 cs4size 和 c5size 选项
+% 而新版本 ctex 中，可以利用 zihao 选项指定各种中文字号
+\PassOptionsToPackage{CJKnumber}{xeCJK}
\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\ }
+% 在旧版本 xeCJK 中，必须用 CJKnumber 选项载入 CJKnumb 包，后面才载入会报错
+% 但在新版本 xeCJK 中 CJKnumber 选项已经被废弃，需要在后面自行载入它
+% 在 xeCJK 中已经禁止载入 CJK，但是在新版本 ctex 宏包中却失效了
+% 我们假装 CJK 已经载入，再载入 CJKnumb，避免出现 \CJKglue 重复定义的问题
+% 注意用 PDFLaTeX 编译时需要用到 CJK，所以只在未定义时才作修改
+\ifdefined\else
+ \@namedef{ver at CJK.sty}{}
+ %\@namedef{opt at CJK.sty}{}
\fi
+\RequirePackage{CJKnumb}
-\ifsidebyside
+% 新版本 xeCJK 已经废弃并禁用 CJKfntef，改用 xeCJKfntef 取代，我们需要载入后者
+% 注意要保证能在较旧的 TeX 系统中编译，我们只能用 \ifXeTeX 而不能用 \ifxetex
+% 因为旧版本 iftex 宏包只有 \ifXeTeX 命令，而 ifxetex 宏包才有 \ifxetex 命令
+% 在 2019 年 10 月，LaTeX 开发团队接管了 iftex 宏包，新版本同时提供这两个命令
+\RequirePackage{CJKfntef}
+\RequirePackage{iftex}
+\ifXeTeX\@ifpackagelater{xeCJK}{2020/02/10}{\RequirePackage{xeCJKfntef}}{}\fi
+
+\iftwoinone
\RequirePackage[a3paper,landscape,twocolumn,columnsep=60mm,left=30mm,right=30mm,top=25mm,bottom=25mm]{geometry}
\else
\RequirePackage[a4paper,left=30mm,right=30mm,top=25mm,bottom=25mm]{geometry}
\fi
-\RequirePackage{tabularx}
-\RequirePackage{lastpage}
-\RequirePackage{fancyhdr}
-\RequirePackage{xcolor}
+\RequirePackage{amsmath}
+\RequirePackage{array}
+\RequirePackage{calc}
\RequirePackage{comment}
+\RequirePackage[inline]{enumitem}
\RequirePackage{environ}
\RequirePackage{etoolbox}
-\RequirePackage{calc}
-\RequirePackage{iftex}
+\RequirePackage{fancyhdr}
+\RequirePackage{zref-user,zref-lastpage}
+\RequirePackage{tabularx}
+\RequirePackage{xcolor}
+\RequirePackage{xkeyval}
+\ifplain
+ \allowdisplaybreaks[4]
+\fi
+
+\ifamsfonts
+ \RequirePackage{amssymb}
+\else
+ \RequirePackage[utopia]{mathdesign} % charter, utopia
+ \renewcommand\bfdefault{bx}
+ \let\oldoiint\oiint\renewcommand{\oiint}{\oldoiint\nolimits}
+ \DeclareTextCommandDefault{\nobreakspace}{\leavevmode\nobreak\ }
+\fi
+
\newcolumntype{Y}{>{\centering\arraybackslash}X}
\newcolumntype{n}[1]{>{\centering\arraybackslash}m{#1}}
@@ -67,55 +110,151 @@
\setlength{\lineskip}{4pt}
%% ---------------------------------------------------------------------------
-%% 密封线命令 \mifengxian
-%% 草稿纸命令 \caogaozhi
-%% 这两个命令仅在 A3 纸张中用到，且需要编译两次才能得到正确结果
+%% 装订线命令 \addbindingline
+%% 草稿纸命令 \addscratchpaper
+%% 这两个命令需要编译两次才能得到正确结果
%% ---------------------------------------------------------------------------
-\newcommand{\mifengxianleft}{
+\newcommand{\setexam}[1]{\setkeys{jnuexam at setup}{#1}}
+
+\ifplain\else
+ \RequirePackage{tikz}
+ \RequirePackage{everypage}
+\fi
+
+\newcommand{\bindinglineleft}{
\path (current page.north west) +(25mm,-25mm) coordinate (a1);
\path (current page.south west) +(25mm,25mm) coordinate (a2);
\draw[dashed] (a1) -- node[left=1mm,text width=1em,inner sep=0pt]{\1{线}\0\1{订}\0\1{装}} (a2);
}
-\newcommand{\mifengxianright}{
+\newcommand{\bindinglineright}{
\path (current page.north east) +(-25mm,-25mm) coordinate (b1);
\path (current page.south east) +(-25mm,25mm) coordinate (b2);
\draw[dashed] (b1) -- node[right=1mm,text width=1em,inner sep=0pt]{\2{装}\0\2{订}\0\2{线}} (b2);
}
-\newcommand{\mifengxianone}{%
+\newcommand{\bindinglineone}{%
\def\0{\\[70mm]}\def\1{\rotatebox{90}}\def\2{\rotatebox{-90}}%
\begin{tikzpicture}[remember picture,overlay,very thick]
- \ifnumodd{\value{page}}{\mifengxianleft}{\mifengxianright}
+ \ifnumodd{\value{page}}{\bindinglineleft}{\bindinglineright}
\end{tikzpicture}}
-\newcommand{\mifengxiantwo}{%
+\newcommand{\bindinglinetwo}{%
\def\0{\\[70mm]}\def\1{\rotatebox{90}}\def\2{\rotatebox{-90}}%
\begin{tikzpicture}[remember picture,overlay,very thick]
- \mifengxianleft\mifengxianright
+ \bindinglineleft\bindinglineright
\end{tikzpicture}}
-\newcommand{\caogaozhi}{%
+\iftwoinone
+ \let\bindingline=\bindinglinetwo
+\else
+ \let\bindingline=\bindinglineone
+\fi
+
+\def\zhuangdingxian{1}
+\define at key{jnuexam at setup}{binding}[2]{\def\zhuangdingxian{#1}}
+
+\newcommand{\addbindingline}{%
+ \ifcase\zhuangdingxian % 0
+ \or % 1
+ \ifbool{answer}{}{\bindingline}%
+ \or % 2
+ \bindingline
+ \fi
+}
+
+\ifplain\else
+ \AddEverypageHook{\addbindingline}
+\fi
+
+\newcommand{\scratchpaperone}{%
+\begin{tikzpicture}[remember picture,overlay,font=\sffamily\fontsize{120pt}{120pt}\selectfont]
+ \node[text=lightgray!40,text width=120pt] at (current page.center) {草\\ 稿\\ 纸};
+\end{tikzpicture}}
+
+\newcommand{\scratchpapertwo}{%
\begin{tikzpicture}[remember picture,overlay,font=\sffamily\fontsize{180pt}{180pt}\selectfont]
\node[text=lightgray!40] at (current page.center) {草\quad 稿\quad 纸};
\end{tikzpicture}}
-\ifmifengxian
- \RequirePackage{tikz}
- \RequirePackage{everypage}
- \ifsidebyside
- \AddEverypageHook{\mifengxiantwo}
- % 在 twocolumn 文档中，\newpage 可能是到下一栏，\clearpage 才能保证到下一页
- \preto{\@enddocumenthook}{\clearpage\pagestyle{empty}\caogaozhi\clearpage\caogaozhi\addtocounter{page}{-2}}
- % 在旧版本 ctex 宏包中不能用 \AtEndDocument 添加含中文的内容，即不能用 \appto 而要用 \preto
- %\AtEndDocument{\clearpage\pagestyle{empty}\caogaozhi\clearpage\caogaozhi\addtocounter{page}{-2}}
- \else
- \AddEverypageHook{\mifengxianone}
- \fi
+\def\caogaozhi{0}
+\define at key{jnuexam at setup}{scratch}[1]{\def\caogaozhi{#1}}
+
+\newcounter{my at empty@page} % 空白页数
+\newcounter{my at scratch@page} % 草稿页数
+\newcounter{my at extra@page} % 空白页数+草稿页数
+
+% 添加空白页，使得草稿纸前页数在单面印刷时为偶数，双面印刷时为4的倍数
+% 添加草稿纸，页数在单面印刷时等于设置值的两倍，双面印刷时等于设置值的4倍
+\newcommand{\doscratchpaperone}{%
+ \clearpage\pagestyle{empty}\let\addbindingline=\relax
+ \ifnumgreater{\caogaozhi}{0}{%
+ \ifbool{oneside}{%
+ \ifnumodd{\value{page}}%
+ {\setcounter{my at empty@page}{0}}%
+ {\setcounter{my at empty@page}{1}}%
+ \setcounter{my at scratch@page}{\caogaozhi*2}%
+ }{%
+ \setcounter{my at empty@page}{(\value{page}+2)/4*4+1-\value{page}}%
+ \setcounter{my at scratch@page}{\caogaozhi*4}%
+ }%
+ \my at add@extra at page
+ }{}%
+}
+
+% 注意在 twocolumn 文档中，\newpage 到下一栏，\clearpage 到下一页
+% 另外注意第一行的 \clearpage 等不能移动到 \my at add@extra at page 命令里
+% 因为 \clearpage 后本页已经结束，page 表示的是下一页的页码
+\newcommand{\doscratchpapertwo}{%
+ \clearpage\pagestyle{empty}\let\addbindingline=\relax
+ \ifnumgreater{\caogaozhi}{0}{%
+ \ifbool{oneside}{%
+ \setcounter{my at empty@page}{0}%
+ \setcounter{my at scratch@page}{\caogaozhi}%
+ }{%
+ \setcounter{my at empty@page}{(\value{page}/2)-(\value{page}/2)/2*2}%
+ \setcounter{my at scratch@page}{\caogaozhi*2}%
+ }%
+ \my at add@extra at page
+ }{}%
+}
+
+\newcommand{\my at add@extra at page}{%
+ \setcounter{my at extra@page}{\value{my at empty@page}+\value{my at scratch@page}}%
+ \whileboolexpr{
+ test{\ifnumgreater{\value{my at empty@page}}{0}}
+ }{%
+ \addtocounter{my at empty@page}{-1}%
+ \leavevmode\clearpage
+ }%
+ \whileboolexpr{
+ test{\ifnumgreater{\value{my at scratch@page}}{0}}
+ }{%
+ \addtocounter{my at scratch@page}{-1}%
+ \leavevmode\scratchpaper\clearpage
+ }%
+ \addtocounter{page}{-\value{my at extra@page}}%
+}
+
+\iftwoinone
+ \let\scratchpaper=\scratchpapertwo
+ \let\doscratchpaper=\doscratchpapertwo
+\else
+ \let\scratchpaper=\scratchpaperone
+ \let\doscratchpaper=\doscratchpaperone
\fi
-\ifsidebyside
+% 在旧版本 ctex 宏包中不能用 \AtEndDocument 添加含中文的内容，即不能用 \appto 而要用 \preto
+\newcommand{\addscratchpaper}{%
+ \preto{\@enddocumenthook}{\doscratchpaper}%
+}
+
+\ifplain\else\ifanswer\else
+ \addscratchpaper
+\fi\fi
+
+\iftwoinone
\RequirePackage{pdfpages}
% 新版本 pdfpages 将 \includepdf 放在文档开头时会有命令未定义的错误
% 见 https://tex.stackexchange.com/questions/352007/ieeetran-and-pdfpages
@@ -167,7 +306,7 @@
\hline
\textbf{考试方式} \\ 开卷~[\isquad{\bijuan}]\quad 闭卷~[\ischeck{\bijuan}] \\
\hline
- \textbf{试卷类别}~(\,A, B, C\,)\\\relax [\,\shijuan\,]\hfill 共~\pageref{LastPage}~页
+ \textbf{试卷类别}~(\,A, B, C\,)\\\relax [\,\shijuan\,]\hfill 共~\zpageref{LastPage}~页
\end{tabular}
}
@@ -181,9 +320,12 @@
\newcommand{\makehead}{
\thispagestyle{plain}
- \centerline{\LARGE\bfseries 暨\quad 南\quad 大\quad 学\quad 考\quad 试\quad 试\quad 卷}
+ \centerline{%
+ \LARGE\bfseries\ifanswer\color{red!80!black}\fi
+ 暨\quad 南\quad 大\quad 学\quad 考\quad 试\quad 试\quad 卷\ifanswer\quad 答\quad 案\fi
+ }%
\vspace{2em}%
- \ifreverse
+ \ifrandom
\ifx\shijuan\my at temp@a\renewcommand{\shijuan}{B}\fi
\ifx\shijuan\my at temp@c\renewcommand{\shijuan}{D}\fi
\fi
@@ -204,9 +346,9 @@
\hline
\textbf{题\quad 号} & 一 & 二 & 三 & 四 & 五 & 六 & 总分\\
\hline
- \textbf{得\quad 分} & & & & & & & \\
+ \textbf{得\quad 分}\rule[-0.75em]{0pt}{2.5em} & & & & & & & \\
\hline
- \textbf{评阅人} & & & & & & & \\
+ \textbf{评阅人}\rule[-0.75em]{0pt}{2.5em} & & & & & & & \\
\hline
\end{tabularx}
}
@@ -219,12 +361,27 @@
\newcommand{\my at headleft}{\hspace{-0.3em}《\kecheng》\kern-0.3em 试卷\,\shijuan}
\newcommand{\my at headright}{姓名\hspace{6em}学号\hspace{6em}}
\newcommand{\my at headtext}{\my at headleft\hfill \my at headright}
-\newcommand{\my at foottext}{第~\thepage~页\quad 共~\pageref{LastPage}~页}
+\newcommand{\my at foottext}{第~\thepage~页\quad 共~\zpageref{LastPage}~页}
+% fancy page style
+\fancyhf{} % 清空页眉页脚
+\iftwoinone
+ \renewcommand{\headrulewidth}{0pt}%
+ \lhead{\small\underline{\my at columnbox{\my at headtext}\strut}}
+ \rhead{\small\underline{\my at columnbox{\my at headtext}\strut}}
+ \lfoot{\small\my at columnbox{\my at foottext}}
+ \rfoot{\small\my at columnbox{\stepcounter{page}\my at foottext}}
+\else
+ \lhead{\small\my at headleft}
+ \rhead{\small\my at headright}
+ \cfoot{\small\my at foottext}
+\fi
+
+% plain page style
\fancypagestyle{plain}{
\renewcommand{\headrulewidth}{0pt}%
\fancyhf{}
- \ifsidebyside
+ \iftwoinone
\rhead{\small\underline{\my at columnbox{\my at headtext\strut}}}
\lfoot{\small\my at columnbox{\my at foottext}}
\rfoot{\small\my at columnbox{\stepcounter{page}\my at foottext}}
@@ -232,18 +389,11 @@
\cfoot{\small\my at foottext}
\fi
}
-\fancyhf{}
-\pagestyle{fancy}
-\ifsidebyside
- \renewcommand{\headrulewidth}{0pt}%
- \lhead{\small\underline{\my at columnbox{\my at headtext}\strut}}
- \rhead{\small\underline{\my at columnbox{\my at headtext}\strut}}
- \lfoot{\small\my at columnbox{\my at foottext}}
- \rfoot{\small\my at columnbox{\stepcounter{page}\my at foottext}}
+
+\ifplain
+ \pagestyle{plain}
\else
- \lhead{\small\my at headleft}
- \rhead{\small\my at headright}
- \cfoot{\small\my at foottext}
+ \pagestyle{fancy}
\fi
%% ---------------------------------------------------------------------------
@@ -251,44 +401,174 @@
%% 附录命令 \makedata
%% 题目环境 problem
%% 解答环境 solution
-%% 逆序选项 reverse
+%% 乱序选项 random
%% ---------------------------------------------------------------------------
+\newif\ifonlyoneproblem \onlyoneproblemfalse % 此部分仅有一道题时不显示题目编号
\xdef\allproblems{}
\xdef\lastproblem{}
-\newcounter{problem}
+\newcounter{problem} % 当前题型的小题编号
+\newcounter{problemreal} % 实际显示的小题编号，在各题型小题统一编号时使用
+\newcounter{totalproblems} % 之前各题型小题总数，在各题型小题统一编号时使用
\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
+\newcommand{\hangtext}{}
+\newlength{\hanglength}
+\colorlet{part number}{black}
+\colorlet{problem number}{blue!80!black}
+\colorlet{solution name}{blue!80!black}
-\newcommand{\printproblems}{\ifreverse\lastproblem\allproblems\fi\xdef\allproblems{}\xdef\lastproblem{}}
+\ifrandom
+ \RequirePackage{pgf}
+ \RequirePackage{pgffor}
+\fi
+\newcounter{my at shuffle@temp at cnt}
+\newcounter{my at list@temp at cnt}
+
+\newcommand\my at list@print[1]{%
+ \par\renewcommand*{\do}[1]{(##1)}%
+ \dolistloop#1%
+}
+
+\newcommand\my at list@remove[2]{%
+ %\my at list@print\my at shuffle@list
+ \setcounter{my at list@temp at cnt}{0}%
+ \global\let\my at tmpa@list=#1%
+ \gdef#1{}%
+ \par\renewcommand*{\do}[1]{%
+ \stepcounter{my at list@temp at cnt}%
+ \ifnumequal{\value{my at list@temp at cnt}}{#2}{%
+ \def\my at list@item{##1}%
+ %[##1]%
+ }{
+ \listxadd#1{##1}%
+ }%
+ }%
+ \dolistloop\my at tmpa@list
+}
+
+%% 随机数种子不能超过 2147483647 = "7FFFFFFF
+\def\my at random@seed{19061116}
+\define at key{jnuexam at setup}{seed}[19061116]{\def\my at random@seed{#1}}
+
+\newcommand\my at shuffle@problems{%
+ %% 当\pgfmathrandom的参数为3的倍数时，对相邻种子生成的多个随机数分布不均匀
+ %\pgfmathsetseed{\numexpr\my at random@seed+\value{section}-1\relax}%
+ %% 因此我们改用下面的方法，用随机数种子生成下一个随机数种子
+ \pgfmathsetseed{\my at random@seed}%
+ \pgfmathrandominteger\my at random@seed{1}{2147483647}%
+ \ifnumgreater{\value{problem}}{2}{%
+ \gdef\my at shuffle@list{}%
+ \foreach \i in {1,...,\value{problem}} {\listxadd\my at shuffle@list{\i}}%
+ %% 首尾两个小题的位置总要改变
+ \pgfmathrandom{2,\value{problem}}%
+ \my at list@remove\my at shuffle@list{\pgfmathresult}%
+ \global\csletcs{my at problem@b at 1}{my at problem@a@\my at list@item}%
+ \ifnumequal{\my at list@item}{\value{problem}}{
+ \pgfmathrandom{\numexpr\value{problem}-1\relax}%
+ \my at list@remove\my at shuffle@list{\pgfmathresult}%
+ \global\csletcs{my at problem@b@\the\value{problem}}{my at problem@a@\my at list@item}%
+ }{%
+ \pgfmathrandom{\numexpr\value{problem}-2\relax}%
+ \my at list@remove\my at shuffle@list{\pgfmathresult}%
+ \global\csletcs{my at problem@b@\the\value{problem}}{my at problem@a@\my at list@item}%
+ }%
+ %% 其他小题的位置没有任何限制
+ \setcounter{my at shuffle@temp at cnt}{1}%
+ \whileboolexpr{%
+ test{\ifnumless{\value{my at shuffle@temp at cnt}}{\numexpr\value{problem}-1\relax}}%
+ }{%
+ \stepcounter{my at shuffle@temp at cnt}%
+ \pgfmathrandom{\numexpr\value{problem}-\value{my at shuffle@temp at cnt}\relax}%
+ \my at list@remove\my at shuffle@list{\pgfmathresult}%
+ \global\csletcs{my at problem@b@\the\value{my at shuffle@temp at cnt}}{%
+ my at problem@a@\my at list@item
+ }%
+ }%
+ }{%
+ \ifnumequal{\value{problem}}{2}{%
+ \global\csletcs{my at problem@b at 1}{my at problem@a at 2}%
+ \global\csletcs{my at problem@b at 2}{my at problem@a at 1}%
+ }{}%
+ }%
+}
+
+\newcommand{\printproblems}{%
+ \ifrandom
+ \my at appto@problems
+ \my at shuffle@problems
+ \setcounter{problem}{0}%
+ \allproblems
+ \fi
+ \xdef\allproblems{}%
+ \xdef\lastproblem{}%
+}
+
\newcommand{\makepart}[2]{%
\printproblems
+ \setcounter{totalproblems}{\value{totalproblems}+\value{problem}}%
\setcounter{problem}{0}%
\stepcounter{section}%
\vspace{1em}%
- \noindent\textbf{\Chinese{section}、#1}（#2）%
- \par\vspace{1em}%
+ \noindent\textbf{\textcolor{part number}{\Chinese{section}}、#1}（#2）%
+ \par\nopagebreak
+ \if\relax\detokenize{#1}\relax % #1 is empty
+ \onlyoneproblemtrue
+ \else
+ \onlyoneproblemfalse
+ \vspace{1em}%
+ \fi
+ %其中设定了\@nobreaktrue，保证在列表前也不分页，详情见 source2e
+ \@afterheading
}
\newcommand{\makedata}[1]{%
- \printproblems\my at stop@reverse
+ \printproblems\my at stop@random
\centerline{\textbf{附录}\quad #1}\smallskip
}
-\preto{\@enddocumenthook}{\printproblems\my at stop@reverse}
+\preto{\@enddocumenthook}{\printproblems\my at stop@random}
\newcommand\ignorepars{\@ifnextchar\par{\expandafter\ignorepars\@gobble}{}}
-\newenvironment{problemreal}{%
- \stepcounter{problem}\setcounter{choice}{0}\setcounter{step}{0}%
- \textbf{\textsf{{\color{blue}\arabic{problem}}.}}\;\,\ignorespaces
-}{\par}
+% 局部定义，仅在当前题目内有效
+\define at key{jnuexam at problem}{points}[-1]{\def\my at problem@points{#1}}
+\define at key{jnuexam at problem}{level}[]{\def\my at problem@level{#1}}
+\define at key{jnuexam at problem}{year}[]{\def\my at problem@year{#1}}
+
+\newcommand{\problempointstext}[1]{（#1 分）}
+
+\newcommand\my at hook@exec at other@keys{}
+
+\newcommand{\execute at problem@keys}[1]{%
+ \setkeys{jnuexam at problem}{#1}%
+ \my at hook@exec at other@keys
+ \ifdefvoid{\my at problem@points}{}{\problempointstext{\my at problem@points}}%
+}
+
+\newenvironment{problemreal}[1][]{%
+ \stepcounter{problem}\setcounter{choice}{0}%
+ \ifresetnumber
+ \ifonlyoneproblem
+ \renewcommand{\hangtext}{\qquad}%
+ \else
+ \renewcommand{\hangtext}{\textbf{\textsf{\textcolor{problem number}{\arabic{problem}}.}}\;\,}%
+ \fi
+ \else
+ \setcounter{problemreal}{\value{totalproblems}+\value{problem}}%
+ \renewcommand{\hangtext}{\textbf{\textsf{\textcolor{problem number}{\arabic{problemreal}}.}}\;\,}%
+ \fi
+ \settowidth{\hanglength}{\hangtext}%
+ \description[leftmargin=\hanglength,labelwidth=0pt,labelsep=0pt,topsep=0pt,parsep=0pt]
+ \item[\hangtext]\execute at problem@keys{#1}%
+}{\enddescription}
\newenvironment{solutionreal}{%
- \setcounter{step}{0}%
- \textbf{\textsf{{\color{blue}\solutionname}.}}\;\,\ignorepars
-}{\par}
+ \renewcommand{\hangtext}{\textbf{\textsf{\textcolor{solution name}{\solutionname}.}}\;\,}%
+ \settowidth{\hanglength}{\hangtext}%
+ \description[leftmargin=\hanglength,labelwidth=0pt,labelsep=0pt,topsep=0pt,parsep=0pt]
+ \item[\hangtext]
+}{\enddescription}
\let \oldnewpage = \newpage
\let \oldvfill = \vfill
@@ -296,42 +576,45 @@
\let \oldmedskip = \medskip
\let \oldbigskip = \bigskip
-\ifreverse
+\ifrandom
+ \newcommand\my at appto@problems{%
+ \xappto\allproblems{\expandonce\lastproblem}%
+ }%
\NewEnviron{problem}{%
- \xdef\allproblems{%
- \unexpanded\expandafter{\lastproblem}%
- \unexpanded\expandafter{\allproblems}%
- }%
- \xdef\lastproblem{%
+ \stepcounter{problem}%
+ \my at appto@problems
+ \csxdef{my at problem@a@\the\value{problem}}{%
\unexpanded{\begin{problemreal}}%
\unexpanded\expandafter{\BODY}%
\unexpanded{\end{problemreal}}%
}%
+ \csxdef{my at problem@b@\the\value{problem}}{%
+ \expandonce{\csname my at problem@a@\the\value{problem}\endcsname}%
+ }%
+ \xdef\lastproblem{%
+ \expandonce{\csname my at problem@b@\the\value{problem}\endcsname}%
+ }%
}
\NewEnviron{solution}{%
- \xdef\lastproblem{%
- \unexpanded\expandafter{\lastproblem}%
+ \csxappto{my at problem@a@\the\value{problem}}{%
\unexpanded{\begin{solutionreal}}%
- \unexpanded\expandafter{\BODY}%
+ \expandonce{\BODY}%
\unexpanded{\end{solutionreal}}%
}%
}
- \renewcommand{\newpage}{\xdef\lastproblem{\noexpand\oldnewpage\unexpanded\expandafter{\lastproblem}}}
- \renewcommand{\vfill}{\xdef\lastproblem{\unexpanded\expandafter{\lastproblem\oldvfill}}}
- \renewcommand{\smallskip}{\xdef\lastproblem{\unexpanded\expandafter{\lastproblem\oldsmallskip}}}
- \renewcommand{\medskip}{\xdef\lastproblem{\unexpanded\expandafter{\lastproblem\oldmedskip}}}
- \renewcommand{\bigskip}{\xdef\lastproblem{\unexpanded\expandafter{\lastproblem\oldbigskip}}}
- \let \newpagea = \relax
- \let \newpageb = \newpage
+ \renewcommand{\newpage}{\gappto\lastproblem{\oldnewpage}}
+ \renewcommand{\vfill}{\csgappto{my at problem@a@\the\value{problem}}{\oldvfill}}
+ \renewcommand{\smallskip}{\csgappto{my at problem@a@\the\value{problem}}{\oldsmallskip}}
+ \renewcommand{\medskip}{\csgappto{my at problem@a@\the\value{problem}}{\oldmedskip}}
+ \renewcommand{\bigskip}{\csgappto{my at problem@a@\the\value{problem}}{\oldbigskip}}
\else
- \newenvironment{problem}{\problemreal}{\endproblemreal}
- \newenvironment{solution}{\solutionreal}{\endsolutionreal}
- \let \newpagea = \newpage
- \let \newpageb = \relax
+ \newenvironment{problem}[1][]{\problemreal[#1]}{\endproblemreal}
+ %\newenvironment{solution}{\solutionreal}{\endsolutionreal}
+ \NewEnviron{solution}{\begin{solutionreal}\BODY\end{solutionreal}}
\fi
-\newcommand{\my at stop@reverse}{%
- \ifreverse
+\newcommand{\my at stop@random}{%
+ \ifrandom
\renewenvironment{problem}{\problemreal}{\endproblemreal}%
\renewenvironment{solution}{\solutionreal}{\endsolutionreal}%
\let \newpage = \oldnewpage
@@ -416,7 +699,7 @@
%% ---------------------------------------------------------------------------
%% 答案切换命令 \answer
-%% 判断命令 \true 和 \false
+%% 判断命令 \tickin 和 \tickout
%% 填空命令 \fillin 和 \fillout
%% 选择命令 \pickin 和 \pickout
%% ---------------------------------------------------------------------------
@@ -423,17 +706,28 @@
\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*{\cdotfill}{\leavevmode\xleaders\hbox to 0.5em{\hss$\cdot$\hss}\hfill\kern0pt\relax}
-\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*{\tick at box}[1]{[\makebox[1.5em]{\color{blue}\answer{#1}}]}
+\newcommand*{\tick at text@t}{$\checkmark$}
+\newcommand*{\tick at text@f}{{\large$\times$}}
+\newcommand*{\tick at text@T}{\sffamily T}
+\newcommand*{\tick at text@F}{\sffamily F}
+\newcommand*{\tickin}[1]{\tick at box{\csname tick at text@#1\endcsname}}
+\newcommand*{\tickout}[1]{\unskip\nobreak\cdotfill\tick at box{\csname tick at text@#1\endcsname}}
+% 过时命令，不要再使用
+%\newcommand*{\true}{\tickout{t}}
+%\newcommand*{\false}{\tickout{f}}
-\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}
+\newcommand*{\ulinefill}[1]{\xleaders\hbox{\underline{\vphantom{#1}\kern1pt}}\hfill\kern0pt}
+\newcommand*{\minwidthbox}[2]{\makebox[{\ifdim#1<\width\width\else#1\fi}]{#2}}
+\newcommand*{\fillout}[1]{\allowbreak\hbox{}\nobreak\ulinefill{#1}\underline{\color{blue}\answer{#1}}\ulinefill{#1}}
+\newcommand*{\fillin}[1]{\underline{\hspace{1em}\color{blue}\minwidthbox{2em}{\answer{#1}}\hspace{1em}}}
+
+\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}
+
%% ---------------------------------------------------------------------------
%% 选择题四个选项排版环境，根据四个选项的长度自动排成一行、两行或四行
%% 其中 abcd 环境各列平分整行宽度，而 abcd* 环境各列平分剩余空白
@@ -440,6 +734,7 @@
%% ---------------------------------------------------------------------------
\newlength{\my at item@len}
+\newlength{\my at label@len}
\newcommand\my at item@temp{%
\unskip\cr\stepcounter{choice}(\Alph{choice})\ %
@@ -449,7 +744,13 @@
\stepcounter{choice}(\Alph{choice})\ \ignorespaces
}
\newcommand\my at item@par{%
- \par\stepcounter{choice}(\Alph{choice})\ \ignorespaces
+ \stepcounter{choice}%
+ \def\my at label@text{(\Alph{choice})\ }%
+ \settowidth{\my at label@len}{\my at label@text}%
+ \par \parshape 2 \hanglength \linewidth
+ \dimexpr\hanglength + \my at label@len\relax
+ \dimexpr\linewidth - \my at label@len\relax
+ \my at label@text\ignorespaces
}
\NewEnviron{abcd}{
@@ -514,13 +815,14 @@
}
%% ---------------------------------------------------------------------------
-%% 解答题步骤命令 \step
+%% 设定有序列表使用悬挂缩进，并指定前两级有序列表的标签格式
+%% 标签宽度按最宽者自动设定，左边距自动计算，竖直空白全部去掉
+%% 最后的 itemjoin 设定行内有序列表 enumerate* 两项之间的空白
%% ---------------------------------------------------------------------------
-\newcommand{\step}{%
- \stepcounter{step}%
- \textsf{(\arabic{step})}\;\,%
-}
+\setlist[enumerate]{labelindent=0pt,labelsep=0.2em,itemindent=0pt,leftmargin=*,nosep,itemjoin=\quad}
+\setlist[enumerate,1]{label=(\arabic*)}
+\setlist[enumerate,2]{label=(\alph*),widest*=1}
%% ---------------------------------------------------------------------------
%% 自由对齐命令 \tabpoint, \tabto, \tableft
@@ -593,48 +895,31 @@
\let \< = \tableft
%% ---------------------------------------------------------------------------
-%% 评分命令 \score
+%% 评分命令 \points
%% ---------------------------------------------------------------------------
\PassOptionsToPackage{tbtags}{amsmath}
\RequirePackage{amsmath}
-\newcommand{\myscore}[1]{\textcolor{red}{#1\kern0.15em 分}}
+\newcommand{\mypoints}[1]{\textcolor{red}{#1\kern0.15em 分}}
-\newcommand{\scoretext}[1]{\mbox{}\nobreak\hfill$\cdots\cdots$\myscore{#1}\par\noindent\ignorespaces}
-\newcommand{\scoreeqno}[1]{\eqno{\cdots\cdots\text{\myscore{#1}}}}
-\newcommand{\scoretag}[1]{\tag*{$\cdots\cdots$\myscore{#1}}}
+\newcommand{\pointstext}[1]{\mbox{}\nobreak\hfill$\cdots\cdots$\mypoints{#1}\par\noindent\ignorespaces}
+\newcommand{\pointseqno}[1]{\eqno{\cdots\cdots\text{\mypoints{#1}}}}
+\newcommand{\pointstag}[1]{\tag*{$\cdots\cdots$\mypoints{#1}}}
-\newrobustcmd{\score}[1]{%
+\newrobustcmd{\points}[1]{%
\ifbool{mmode}{%
- \ifdefstrequal{\tag}{\dft at tag}{\scoreeqno{#1}}{\scoretag{#1}}%
+ \ifdefstrequal{\tag}{\dft at tag}{\pointseqno{#1}}{\pointstag{#1}}%
}{%
- \scoretext{#1}%
+ \pointstext{#1}%
}%
}
+%\let\score=\points % \score 命令已经废弃，不要再使用
%% ---------------------------------------------------------------------------
-%% 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}{
- \RequirePackage[math]{cellspace}
- \setlength\cellspacetoplimit{2pt}
- \setlength\cellspacebottomlimit{2pt}
- \addparagraphcolumntypes{X}
- % Fix cellspace bug before version 1.7
- % See https://tex.stackexchange.com/a/385581
- \@ifpackagelater{cellspace}{2017/08/12}{}{
- \patchcmd{\@endpbox}{\color at endgroup}{\expandafter\color at endgroup}{}{}
- }
- \display
-}{}
-
-%% ---------------------------------------------------------------------------
%% 载入个人定制文件 jnuexam.cfg
%% 中文字体切换选项 sourcehan
+%% 实心句号替换选项 solidot
%% ---------------------------------------------------------------------------
\InputIfFileExists{jnuexam.cfg}{}{}
@@ -647,51 +932,150 @@
%\setCJKsansfont{思源黑体}
}
-\ifbool{sourcehan}{
- % https://sourceforge.net/p/xetex/code/ci/master/tree/source/texk/web2c/xetexdir/NEWS
- \ifbool{XeTeX}{ % TeXLive 2015
- \ifdimless{\the\XeTeXversion\XeTeXrevision pt}{0.99992pt}{}{\my at set@sourcehan}
- }{}
- \ifbool{LuaTeX}{\my at set@sourcehan}{}
-}{}
+\AtBeginDocument{%
+ \ifbool{sourcehan}{%
+ % https://sourceforge.net/p/xetex/code/ci/master/tree/source/texk/web2c/xetexdir/NEWS
+ \ifbool{XeTeX}{% TeXLive 2015
+ \ifdimless{\the\XeTeXversion\XeTeXrevision pt}{0.99992pt}{}{\my at set@sourcehan}%
+ }{}%
+ \ifbool{LuaTeX}{\my at set@sourcehan}{}%
+ }{}%
+}
+%% 这里不能用 \ifbool，因为涉及到 catcode 的改变
+\ifsolidot
+ \ifXeTeX
+ \catcode`。=\active\def。{．}%
+ \else\ifLuaTeX
+ \catcode`。=\active\def。{．}%
+ \fi
+ \fi
+\fi
+
%% ---------------------------------------------------------------------------
-%% 试卷题库选项 collection
+%% 在 tabular 和 array 等表格环境中添加列间距
+%% 避免单元格里出现的分式太过接近上面和下面行
%% ---------------------------------------------------------------------------
-\ifcollection
- \RequirePackage{hyperref}
- \hypersetup{
- pdfstartview={FitH},
- bookmarksnumbered=true,
- unicode=true,
- hidelinks=true
- %colorlinks=true,
- %linkcolor=black
+\newcommand{\my at do@cellspace}{%
+ \RequirePackage[math]{cellspace}%
+ \setlength\cellspacetoplimit{2pt}%
+ \setlength\cellspacebottomlimit{2pt}%
+ \addparagraphcolumntypes{X}%
+ \newcolumntype{0}[1]{>{\bcolumn ##1\@nil}##1<{\ecolumn}}%
+ \newcolumntype{5}[1]{>{$}0{##1}<{$}}%
+ % Fix cellspace bug before version 1.7
+ % See https://tex.stackexchange.com/a/385581
+ \@ifpackagelater{cellspace}{2017/08/12}{}{%
+ \patchcmd{\@endpbox}{\color at endgroup}{\expandafter\color at endgroup}{}{}%
+ }%
+}
+
+\AtBeginDocument{%
+ \ifcellspace \my at do@cellspace \fi
+}
+
+%% ---------------------------------------------------------------------------
+%% 统一行间公式和行内公式的巨算符和分式的尺寸
+%% 利用开头的 medmath 选项可以启用此部分设定
+%% ---------------------------------------------------------------------------
+
+\newcommand{\my at do@medmath}{%
+ \RequirePackage[mediummath]{nccmath}%
+ %% 补充 \oiint 命令的调整
+ \ifdef{\oiint}{%
+ \let\NCC at op@oiint=\oiint
+ \DeclareRobustCommand{\oiintop}{\mathop{\medmath{\NCC at op@oiint}}}%
+ \def\oiint{\DOTSI\NCC at op@prepare{\oiintop}}%
+ }{}%
+ %% 设定 nccmath 的积分号校正尺寸
+ %% 后面已经改用相对尺寸，不再需要
+ %\ifamsfonts
+ % \medintcorr{0.5em}
+ %\else
+ % \medintcorr{0.3em}
+ %\fi
+ %% 使用 nccmath 宏包后，cases 环境包含定积分时将无法编译，这里重新定义此环境
+ %% 注意相比 amsmath 的原始定义，我们这里将 \quad 从两列之间移动到第二列最前面
+ %% 这是因为，我们常将 cases 用于只有一列的方程组，这样处理末尾不会有多余空白
+ \ifbool{cellspace}{%
+ \renewenvironment{cases}{%
+ \left\{\linespread{1.0}\selectfont\def\arraystretch{1.2}%
+ \begin{array}{@{}5l@{}>{\quad}5l@{}}%
+ }{%
+ \end{array}\right.%
+ }%
+ }{%
+ \renewenvironment{cases}{%
+ \left\{\linespread{1.0}\selectfont\def\arraystretch{1.2}%
+ \begin{array}{@{}l@{}>{\quad}l@{}}%
+ }{%
+ \end{array}\right.%
+ }%
}
- \appto{\endproblem}{\medskip}
- \appto{\endsolution}{\medskip}
- \preto{\problem}{\ifnum\value{problem}=9 \setcounter{problem}{-1}\fi}
- \pagestyle{plain}
- \let\chapter=\part
- \@addtoreset{section}{part}
- \RequirePackage{xkeyval}
- \define at key{jnuexam}{level}[]{\def\my at collection@level{#1}}
- \define at key{jnuexam}{year}[]{\def\my at collection@year{#1}}
- \define at key{jnuexam}{name}[]{\def\my at collection@name{#1}}
- \newcommand{\info}[1]{%
- \setkeys{jnuexam}{level,year,name,#1}% 重置之前的数据
- \my at collection@info at do
+ %% 当 minipage 或 \parbox 仅包含行间公式时，盒子的右边距丢失，这里修正它
+ %% 其他类似问题的描述见 http://tex.stackexchange.com/q/22170
+ \let\start at gather=\NCC at startgather
+ \let\start at align=\NCC at startalign
+ \let\start at multline=\NCC at startmultline
+ \let\mathdisplay=\NCC at startdisplay
+}
+
+\newcommand{\my at do@medmath at fix}{%
+ %% 修正在角标处的非积分巨算符尺寸
+ %% 参考了 scalerel 宏包对数学样式的保存方法
+ % 非角标巨算符保持为 \displaystyle 巨算符的 80% 大小
+ % 一级角标巨算符修改为 \textstyle 巨算符的 80% 大小
+ % 二级角标巨算符修改为 \scriptstyle 巨算符的 80% 大小
+ \def\@my at style@D{\displaystyle}%
+ \def\@my at style@T{\displaystyle}%
+ \def\@my at style@S{\textstyle}%
+ \def\@my at style@s{\scriptstyle}%
+ \def\my at style@saved{\csname @my at style@\@my at style@switch\endcsname}%
+ \newcommand{\my at style@this}[1]{%
+ \mathchoice{\def\@my at style@switch{D}##1}{\def\@my at style@switch{T}##1}
+ {\def\@my at style@switch{S}##1}{\def\@my at style@switch{s}##1}%
+ }%
+ \DeclareRobustCommand*\medmath[1]{\NCC at select@msize
+ \mathord{\my at style@this{\raise\@tempdima\hbox{\NCC at prepare@msize$\my at style@saved ##1$}}}%
}
- \newcommand{\my at collection@info at do}{%
- \bigskip\hrule\nopagebreak\vspace{0.4em}\nopagebreak
- \ifdefempty{\my at collection@level}{}{Level = \my at collection@level\quad}%
- \ifdefempty{\my at collection@year}{}{Year = \my at collection@year\quad}%
- \ifdefempty{\my at collection@name}{}{Name = \my at collection@name\quad}%
- \nopagebreak\vspace{0.4em}\nopagebreak\hrule\medskip
- }
-\fi
+ %% 修正在角标处的积分算符尺寸和上下限位置
+ \newlength{\@my at em}%
+ \setlength{\@my at em}{1em}%
+ \ifamsfonts
+ \medintcorr{0.5\@my at em}%
+ \else
+ \medintcorr{0.3\@my at em}%
+ \fi
+ \newcommand{\my at style@unit}[1]{%
+ \mathchoice{\setlength{\@my at em}{1em}##1}{\setlength{\@my at em}{1em}##1}
+ {\setlength{\@my at em}{0.5em}##1}{\setlength{\@my at em}{0.3em}##1}%
+ }%
+ \let\my at saved@op at printm=\NCC at op@printm
+ \def\NCC at op@printm{\my at style@unit{\my at saved@op at printm}}%
+ %% 修正在角标位置的分式和嵌套分式
+ %% 总是使用当前样式尺寸来排版分式的分子和分母
+ \newcommand{\my at larger@frac}[2]{%
+ \mathchoice{\genfrac{}{}{}{0}{##1}{##2}}{\genfrac{}{}{}{0}{##1}{##2}}%
+ {\genfrac{}{}{}{1}{##1}{##2}}{\genfrac{}{}{}{2}{##1}{##2}}%
+ }%
+ \DeclareRobustCommand{\frac}[2]{%
+ \mathchoice{\mfrac{##1}{##2}}{\mfrac{##1}{##2}}%
+ {\my at larger@frac{##1}{##2}}{\my at larger@frac{##1}{##2}}%
+ }%
+ \patchcmd{\NCC at prepare@msize}{%
+ \def\frac{\protect\NCC at innerfrac{}}%
+ }{%
+ \let\frac=\my at larger@frac
+ }{}{}%
+}
+\AtBeginDocument{%
+ \ifmedmath
+ \my at do@medmath \my at do@medmath at fix
+ \fi
+}
+
%% ---------------------------------------------------------------------------
%% 载入常用宏包，定义常用命令
%% ---------------------------------------------------------------------------
@@ -713,26 +1097,55 @@
\xeCJKsetcharclass{"2460}{"24FF}{1} % 带圈数字字母，括号数字字母，带点数字等
}{}
-\RequirePackage{CJKfntef}
\RequirePackage{multirow}
-\RequirePackage{diagbox}
\RequirePackage{tabu}
+\RequirePackage{diagbox}
+%% 修正 \diagbox 在 array 环境中使用的问题
+\newrobustcmd{\diagboxtwo}[3][]{%
+ \ifbool{mmode}{%
+ \hbox{\let\tabcolsep=\arraycolsep\diagbox[#1]{$#2$}{$#3$}}%
+ }{%
+ \diagbox[#1]{#2}{#3}%
+ }
+}
+\newrobustcmd{\diagboxthree}[4][]{%
+ \ifbool{mmode}{%
+ \hbox{\let\tabcolsep=\arraycolsep\diagbox[#1]{$#2$}{$#3$}{$#4$}}%
+ }{%
+ \diagbox[#1]{#2}{#3}{#4}%
+ }
+}
+
+\RequirePackage{mathtools} % \mathllap 命令，pmatrix* 环境等
\RequirePackage{extarrows}
\RequirePackage{relsize}
\newcommand{\Int}{\mathop{\mathlarger{\int}}}
+\AtBeginDocument{%
+ \let\my at saved@lim=\lim \def\lim{\my at saved@lim\limits}%
+ \let\my at saved@sum=\sum \def\sum{\my at saved@sum\limits}%
+ \let\my at saved@prod=\prod \def\prod{\my at saved@prod\limits}%
+}
+%\newcommand{\limit}{\lim\limits} % 废弃命令，不要再使用
+
\newcommand{\e}{\mathrm{e}}
-\newcommand{\limit}{\lim\limits}
\newcommand{\R}{\mathbb{R}}
+\DeclareMathOperator{\arccot}{arccot}
\DeclareMathOperator{\Corr}{\rho}
\DeclareMathOperator{\Cov}{Cov}
+\DeclareMathOperator{\diag}{diag}
\DeclareMathOperator{\grad}{grad}
\DeclareMathOperator{\Prj}{Prj}
+\DeclareMathOperator{\tr}{tr}
\DeclareMathOperator{\Var}{Var}
+\DeclareMathOperator{\diver}{div}
+\let\division=\div
+\let\div=\diver
+
\newcommand{\diff}{\mathop{}\!\mathrm{d}}
\newcommand{\dx}{\diff x}
\newcommand{\dy}{\diff y}
@@ -750,6 +1163,23 @@
\renewrobustcmd{\d}{\ifbool{mmode}{\diff}{\oldd}}%
}
+\let\pd=\partial
+\newcommand{\pdf}{\pd f}
+\newcommand{\pdg}{\pd g}
+\newcommand{\pdh}{\pd h}
+\newcommand{\pdl}{\pd l}
+\newcommand{\pdn}{\pd n}
+\newcommand{\pdu}{\pd u}
+\newcommand{\pdv}{\pd v}
+\newcommand{\pdx}{\pd x}
+\newcommand{\pdy}{\pd y}
+\newcommand{\pdz}{\pd z}
+\newcommand{\pdF}{\pd F}
+\newcommand{\pdL}{\pd L}
+\newcommand{\pdP}{\pd P}
+\newcommand{\pdQ}{\pd Q}
+\newcommand{\pdR}{\pd R}
+
% from mathabx package
\DeclareFontFamily{U}{mathx}{\hyphenchar\font45}
\DeclareFontShape{U}{mathx}{m}{n}{<-> mathx10}{}
@@ -778,7 +1208,44 @@
\def\T{\mathrm{T}\kern-.5pt}
-\newrobustcmd{\wfrac}[3][2pt]{%
- {\begingroup\hspace{#1}#2\hspace{#1}\endgroup\over\hspace{#1}#3\hspace{#1}}%
+% 分数线长一点的分数，\wfrac[2pt]{x}{y} 表示左右加 2pt
+% 和前面的 medmath 一样，将代码放在 \AtBeginDocument 里
+\AtBeginDocument{%
+ \newrobustcmd{\wfrac}[3][2pt]{%
+ \frac{\hspace{#1}#2\hspace{#1}}{\hspace{#1}#3\hspace{#1}}%
+ }%
+ \newrobustcmd{\wdfrac}[3][2pt]{%
+ \dfrac{\hspace{#1}#2\hspace{#1}}{\hspace{#1}#3\hspace{#1}}%
+ }%
+ \newrobustcmd{\wtfrac}[3][2pt]{%
+ \tfrac{\hspace{#1}#2\hspace{#1}}{\hspace{#1}#3\hspace{#1}}%
+ }%
}
+% 使用 stix font 中的 white arrows
+\ifxetex
+ %\IfFileExists{STIX-Regular.otf}{% 在 TeXLive 中无效
+ \IfFileExists{stix.sty}{%
+ \newfontfamily{\mystix}{STIX} % stix v1.1
+ }{%
+ \newfontfamily{\mystix}{STIXGeneral} % stix v1.0
+ }
+ \newrobustcmd\leftwhitearrow{%
+ \mathrel{\text{\normalfont\mystix\symbol{"21E6}}}%
+ }
+ \newrobustcmd\upwhitearrow{%
+ \mathrel{\text{\normalfont\mystix\symbol{"21E7}}}%
+ }
+ \newrobustcmd\rightwhitearrow{%
+ \mathrel{\text{\normalfont\mystix\symbol{"21E8}}}%
+ }
+ \newrobustcmd\downwhitearrow{%
+ \mathrel{\text{\normalfont\mystix\symbol{"21E9}}}%
+ }
+\else
+ \let \leftwhitearrow = \Leftarrow
+ \let \rightwhitearrow = \Rightarrow
+ \let \upwhitearrow = \Uparrow
+ \let \downwhitearrow = \Downarrow
+\fi
+
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