texlive[69115] Master/texmf-dist: litesolution (13dec23)
commits+karl at tug.org
commits+karl at tug.org
Wed Dec 13 22:23:27 CET 2023
Revision: 69115
https://tug.org/svn/texlive?view=revision&revision=69115
Author: karl
Date: 2023-12-13 22:23:27 +0100 (Wed, 13 Dec 2023)
Log Message:
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litesolution (13dec23)
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trunk/Master/texmf-dist/doc/latex/litesolution/README.md
trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-demo.tex
trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-doc.tex
trunk/Master/texmf-dist/doc/latex/litesolution/litesolution.pdf
trunk/Master/texmf-dist/doc/latex/litesolution/litesolution.tex
trunk/Master/texmf-dist/tex/latex/litesolution/litesolution.cls
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trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-gauge.tex
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@@ -1,6 +1,9 @@
# The `LiteSolution` document
+This template provides a light and fresh template, mainly used for typesetting solutions of final, textbook and other exercises.
+The template has been developed based on [ElegantBook](https://ctan.org/pkg/elegantbook) and [VividBooK](https://github.com/Azure1210/VividBooK) and has ported and improved the chapter design module code of the [Legrand Orange Book Template](https://www.overleaf.com/latex/templates/the-legrand-orange-book-template-english/jtctyfmnpppc).
+
Modules of `LiteSolution` provide the following supports:
Modified: trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-demo.tex
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--- trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-demo.tex 2023-12-13 21:23:15 UTC (rev 69114)
+++ trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-demo.tex 2023-12-13 21:23:27 UTC (rev 69115)
@@ -1,85 +1,417 @@
\chapter{A Sample for \pkg{LiteSolution} Template}
-\fancyhead[L]{\color{H6}\kaishu\faIcon{atom}\;2022年\titlelogo{https://sci.hdu.edu.cn}{HDU}「普通物理2」期中试题解析}
+\fancyhead[L]{\color{H6}\kaishu\faIcon{atom}\;2023年\titlelogo{https://sci.hdu.edu.cn}{HDU}「大学物理2」期中模拟}
\fancyhead[R]{\color{H6}\kaishu\rightmark\,}
-\section{选填题（共15分）}
-\begin{choice}{D}[波的能量]
- 一平面简谐波在弹性媒介中传播，在媒质质元从最大位移处回到平衡位置的过程中
+\date{2023年12月3日}{大学物理教学团队}{A0715012}
+{\href{https://qm.qq.com/q/UPbGudx8cK}{\textbf{物理問題作}}}
+{https://ctan.org/pkg/litesolution}{CTAN}
+{https://github.com/xiamyphys/LiteSolution}{GitHub}
+
+\section{选择题（每题3分，共36分）}
+\begin{choice}{D}[弹簧振子]
+ 一劲度系数为$k$的轻弹簧，下端挂一质量为$m$的物体，系统的振动周期为$T_1$. 若将此弹簧截去一半的长度，下端挂一质量为$m/2$的物体，则系统振动周期$T_2$等于
+\begin{tasks}(4)
+ \task $2T_1$
+ \task $T_1$
+ \task $\frac{T_1}{\sqrt2}$
+ \task $\frac{T_1}{2}$
+\end{tasks}
+\end{choice}
+\begin{solution}*
+ 弹簧的劲度系数与长度成反比，所以剪断一半后劲度系数变为$2k$；根据弹簧振子的周期表达式$T=2\pi\sqrt{\frac mk}$可知此时的周期$T_2=2\pi\sqrt{\frac{m/2}{2k}}=\frac{T_1}{2}$.\sokka{D}
+\end{solution}
+
+\begin{choice}{B}[单摆]
+ 把单摆摆球从平衡位置向位移正方向拉开，使摆线与竖直方向成一微小角度$\theta$，然后由静止放手任其振动，从放手时开始计时.若用余弦函数表示其运动方程，则该单摆振动的初相为
+ \begin{tasks}(4)
+ \task $\theta$
+ \task $0$
+ \task $\frac\pi2$
+ \task $-\pi$
+ \end{tasks}
+\end{choice}
+
+\vskip-2ex
+\begin{paracol}{2}
+\begin{choice}{A}[平面简谐波的波函数]
+ 一平面简谐波，波速$u=5\mathrm{m/s}$，$t=3\mathrm{s}$时波形曲线如图，则$x=0$处质点的振动方程可能为
+\begin{tasks}(2)
+ \task $y=2\ee{-2}\cos\ab(\frac12\pi t-\frac12\pi)$
+ \task $y=2\ee{-2}\cos\ab(\frac12\pi t+\frac12\pi)$
+ \task $y=2\ee{-2}\cos(\pi t+\pi)$
+ \task $y=2\ee{-2}\cos\ab(\pi t-\frac32\pi)$
+\end{tasks}
+\end{choice}
+\switchcolumn\centering
+\vfill
+\begin{tikzpicture}
+ \draw [->,thick] (-1.2,0) -- (4.8,0) node [anchor=south] {$x(\mathrm{m})$};
+ \draw [->,thick] (0,-1.8) -- (0,1.8) node [anchor=west] {$y(\mathrm{m})$};
+ \draw [very thick] (-.8,0) sin (0,-1) node [anchor=north west] {$-2\ee{-2}$} cos (.8,0) sin (1.6,1) cos (2.4,0) sin (3.2,-1) cos (4,0) sin (4.8,1);
+ \draw [thick,->] (1.6,1.2) -- (2.4,1.2) node [midway,anchor=south] {$u$};
+ \foreach \a in {5,10,...,25}
+ \node [anchor=north,xshift=0.16*\a cm] at (0,0) {\a};
+\end{tikzpicture}
+\vfill
+\end{paracol}
+\vspace{-.75em}
+\begin{solution}*
+ 由图可知$A=0.02\mathrm{m},\ \omega=\frac{2\pi u}{\lambda}=\frac12\pi$.由于原点处$v>0$，所以初相$\varphi=-\frac\pi2$.\sokka{A}
+\end{solution}
+
+\begin{choice}{A}[增透膜]
+ 一艘油船行经我国台湾岛东部海域时发生石油泄漏，在海面上形成大片油膜，太阳光在头顶正射时，救授人员乘直升飞机从上往下看，发现油膜对$552\mathrm{nm}$波长的可见光反射形成干涉相长而最亮，则可以推测该区域油膜厚度可能为多少？（设石油折射率$n=1.2$，海水折射率$n=1.3$）
+\begin{tasks}(4)
+ \task $460\mathrm{nm}$
+ \task $552\mathrm{nm}$
+ \task $345\mathrm{nm}$
+ \task $425\mathrm{nm}$
+\end{tasks}
+\end{choice}
+\begin{solution}
+ \begin{itemize}
+ \item 由于$n_{\text{空}}>n_{\text{海}}>n_{\text{油}}$，所以石油两个表面反射光光程差为$\delta=2ne$.
+ \item 使反射光干涉相长，即$2ne=k\lambda$. A选项刚好满足$k=2$时，$e_{\min}=2\cdot\frac{\lambda}{2n}=460\mathrm{nm}$.
+ \end{itemize}
+\end{solution}
+
+\begin{choice}{C}[光程和光程差]
+ 在相同的时间内，一束波长为$\lambda$的单色光在空气中和在玻璃中
\begin{tasks}(2)
- \task 它的势能转换成动能
- \task 它的动能转换成势能
- \task 它从相邻的一段质元获得能量，其能量逐渐增加
- \task 它把能量传给相邻的一段质元，其能量逐渐减小
+ \task 传播的路程相等，走过的光程相等
+ \task 传播的路程相等，走过的光程不相等
+ \task 传播的路程不相等，走过的光程相等
+ \task 传播的路程不相等，走过的光程不相等
\end{tasks}
\end{choice}
\begin{solution}*
- 波在传播过程中介质质元振动的动能和势能同时变化.\sokka{C}
+ 光程的定义：在相同时间内光线在真空中传播的距离.题目中光传播时间相同，故光程相等；又因为光在两种介质中的传播速度不同，所以在相同的时间内传播的路程不相等.\sokka{C}
\end{solution}
-\begin{choice}{D}[双缝干涉]
- 在双缝干涉实验中，两缝间距离为$d$，双缝与屏幕之间的距离为$D$（$D\gg d$）.波长为$\lambda$的平行单色光垂直照射到双缝上，屏幕上干涉条纹中相邻暗纹之间的距离是
- \begin{tasks}(4)
- \task $\frac{2\lambda D}{d}$
- \task $\frac{\lambda d}{D}$
- \task $\frac{dD}{\lambda}$
- \task $\frac{\lambda D}{d}$
+\begin{choice}{C}[多普勒效应]
+ 一观察者站在铁路旁，一火车以$30\mathrm{m/s}$的速度向他驶来并发出频率为$440\mathrm{Hz}$的汽笛声. 已知空气中声速为$330\mathrm{m/s}$，问观察者听到的火车频率为
+\begin{tasks}(4)
+ \task $403\mathrm{Hz}$
+ \task $480\mathrm{Hz}$
+ \task $484\mathrm{Hz}$
+ \task $528\mathrm{Hz}$
+\end{tasks}
+\end{choice}
+\begin{solution}
+ 已知多普勒效应观察者（Observer）和发射源（Source）的的频率关系为
+ \[\nu=\frac{u\pm v_o}{u\mp v_s}\nu_0\]
+
+ $v_o$为观察者速度，接近为$+$，远离为$-$；$v_s$为发射源速度，接近为$-$，远离为$+$.观察者静止，其所听频率为
+ \[\nu=\frac{330}{330-30}\times 440\mathrm{Hz}=484\mathrm{Hz}\]
+
+ \sokka{C}
+\end{solution}
+
+\vskip-2ex
+\begin{paracol}{2}
+\begin{choice}{C}[弗琅禾费衍射]
+ 在如图所示的单缝弗琅禾费衍射实验中，若将单缝沿透镜光轴方向向透镜平移，则屏幕上的衍射条纹
+\begin{tasks}(5)
+ \task 间距变大\!
+ \task 间距变小\!
+ \task 不变化
+ \task* 间距不变，明暗纹交替
+\end{tasks}
+\end{choice}
+\switchcolumn\centering
+\vfill
+\begin{tikzpicture}
+ \filldraw [fill=gray] (0,1.5) rectangle (0.2,0.5);
+ \filldraw [fill=gray] (0,-1.5) rectangle (0.2,-0.5);
+ \draw [thick,densely dashed] (-1.2,0)--(3,0);
+ \draw [thick] (3,1.5)--(3,-1.5);
+ \filldraw [fill=white] (1.5,0) ellipse (0.1 and 1.2);
+ \foreach \y in {0.5,-0.5}
+ \draw [thick,->,yshift=\y cm] (-1,0)--(0,0);
+ \length{(1.5,-1.4)}{(3,-1.4)}{$f$}{(0.25,0)}
+ \node [anchor=south] at (-1.2,0.5) {$\lambda$};
+ \node [anchor=south] at (1.5,1.2) {$L$};
+ \node [anchor=west] at (3,1.5) {Screen};
+\end{tikzpicture}
+\vfill
+\end{paracol}
+\vspace{-.75em}
+\begin{solution}*
+ 条纹间距只与波长、焦距、缝宽有关，入射光方向不变，所以条纹间距、位置不变. \sokka{C}
+\end{solution}
+
+\begin{choice}{A}[牛顿环]
+ 牛顿环干涉装置上平凸透镜在垂直于平板玻璃的方向上，逐渐向下平移（靠近玻璃板）时，反射光形成的干涉条纹的变化情况是
+ \begin{tasks}(2)
+ \task 环纹向边缘扩散，环数不变
+ \task 环纹向边缘扩散，环数增加
+ \task 环纹向中心靠拢，环数不变
+ \task 环纹向中心靠拢，环数增加
\end{tasks}
\end{choice}
\begin{solution}*
- 由明纹公式$x=k\lambda D/d$得暗纹间距$\Delta x=\lambda D/d$.\sokka{D}
+ 对于某条环，其光程差是确定的，所以环数不变；向边缘扩散光程差增大，可抵消透镜下移时导致的光程差减小.\sokka{A}
\end{solution}
-\begin{problem}[简谐振动的合成][3]
- 一个质点同时参与两个在同一直线上的简谐振动：$x_1=4\cos{\ab(2t+\frac{\pi}{6})},\ x_2=2\cos{\ab(2t-\frac{\pi}{6})}$.该质点合振动的振幅大小为\ans{$2\sqrt{7}$}.
+\begin{choice}{B}[最大分辨力]
+ 假设用FAST装置探测波长为$20\mathrm{cm}$的宇宙射电信号，FAST望远镜的镜面直径为$500\mathrm{m}$，则装置的最小分辨角为
+\begin{tasks}(4)
+ \task $9.76\ee{-4}$
+ \task $4.88\ee{-4}$
+ \task $2.44\ee{-4}$
+ \task $4.00\ee{-4}$
+\end{tasks}
+\end{choice}
+\begin{solution}*
+ $\theta=\frac{1.22\lambda}{D}=4.88\ee{-4}\mathrm{rad}$.\sokka{B}
+\end{solution}
+
+\vskip-2ex
+\begin{paracol}{2}
+ \begin{choice}{B}[双缝干涉]
+ 在双缝干涉实验中，屏幕$E$上的$P$点是明纹.若将缝$S_2$盖住，并在$S_1S_2$连线的垂直平分面处放一高折射率反射面$M$，如图所示.则此时$P$点
+ \begin{tasks}(2)
+ \task $P$点仍为明条纹
+ \task $P$点为暗条纹
+ \task 不能确定$P$点是明纹还是暗纹
+ \task 无干涉条纹
+ \end{tasks}
+ \end{choice}
+ \switchcolumn\centering
+ \vfill
+ \begin{tikzpicture}[decoration={markings,mark=between positions .2 and .8 step 18mm with {\arrow{stealth}}}]
+ \draw [very thick] (0,1.5)--(0,-1.5);
+ \draw [thick,postaction=decorate] (-1,0)--(0,0.5)--(2,0.8);
+ \draw [thick] (0,0.5)--(0.8,0);
+ \draw [thick,postaction=decorate] (-1,0)--(0,-0.5)--(2,0.8);
+ \draw [very thick] (0,0)--(1.8,0);
+ \draw [very thick] (2,1.8)--(2,-1.8);
+ \fill [pattern=north west lines] (0,0) rectangle (1.8,-0.2);
+ \fill [pattern=north east lines] (2,1.8) rectangle (2.2,-1.8);
+ \node [anchor=east] at (-1,0) {$S$} node [anchor=south east] at (0,0.5) {$S_1$} node [anchor=north east] at (0,-0.5) {$S_2$};
+ \node [anchor=north] at (0.8,-0.2) {$M$} node [anchor=west] at (2.2,0.8) {$P$} node [anchor=north] at (2.1,-1.8) {$E$};
+ \node at (-1,0) {$\times$};
+ \end{tikzpicture}
+ \vfill
+\end{paracol}
+\vspace{-.75em}
+\begin{solution}*
+ $S_1MP$、$S_2MP$长度相等，但平面镜使在反射中一条光路发生半波损失，两条光路的相位差变化$\pi$，所以$P$点由原来的明纹变为暗纹.\sokka{B}
+\end{solution}
+
+\begin{choice}{A}[迈克尔逊干涉仪]
+ 如果使迈克尔逊干涉仪的动镜移动$0.233\mathrm{mm}$，观察到$792$个条纹的移动，则所用照明单色光源的波长是多少？
+\begin{tasks}(4)
+ \task $588\mathrm{nm}$
+ \task $294\mathrm{nm}$
+ \task $442\mathrm{nm}$
+ \task $552\mathrm{nm}$
+\end{tasks}
+\end{choice}
+\begin{solution}*
+ 移动带来的光程差满足$\delta=2d=N\lambda$，由此得$\lambda=\frac{2d}{N}=588.38\mathrm{mm}$.\sokka{A}
+\end{solution}
+
+\begin{choice}{B}[光栅]
+ 某元素的特征光谱中含有波长分别为$\lambda_1=450\mathrm{nm},\ \lambda_2=750\mathrm{nm}$的谱线，在光栅光谱中这两种波长的谱线有重合现象，重叠处$\lambda_2$的谱线级数将是
+\begin{tasks}(4)
+ \task $2,\ 4,\ 6,\ 8,\cdots$
+ \task $3,\ 6,\ 9,\ 12,\cdots$
+ \task $4,\ 8,\ 12,\ 16,\cdots$
+ \task $5,\ 10,\ 15,\ 20,\cdots$
+\end{tasks}
+\end{choice}
+\begin{solution}*
+ 由光栅方程$d\sin\theta=k_1\lambda_1=k_2\lambda_2$得$k_2=\frac{k_1\lambda_1}{\lambda_2}$，取整值得$k_2=3,6,9,12,\cdots$.
+\end{solution}
+
+\section{填空题（共18分）}
+\begin{problem}[弹簧振子][3]
+ 当弹簧振子以频率$f$做简谐振动时，它的动能的变化频率为\ans{$2f$}.
\end{problem}
\begin{solution}*
- $A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos{\Delta\varphi}}=2\sqrt{7}$.
+ 动能和势能变化趋势相反，所以二者变化频率相同. 势能$E_p\propto x^2$，由于$x$是周期为$T$的余弦函数，所以$x^2$的周期为$\frac T2$，即势能的变化频率等于动能的变化频率等于$2f$.
\end{solution}
-\begin{problem}[多普勒效应][6]
- 一辆运动的警车发出警铃声音频率为$f_s=1400\mathrm{Hz}$，从左向右移动，速度大小为$70\mathrm{m/s}$. 一个观察者从右向左移动，速度大小为$10\mathrm{m/s}$，空气中的声速$u=340\mathrm{m/s}$.观察者与警车相遇前后，听到警铃的频率分别为\ans{$1814.8\mathrm{Hz}$，$1126.8\mathrm{Hz}$}.
+\begin{problem}[驻波][3]
+ 在均匀介质中，一列余弦波沿$Ox$轴传播，波动方程为$y_1=A\cos\ab(2\pi t+\frac{2\pi x}3)$ (SI)，在$x=1\mathrm{m}$处反射，反射点为固定端，则反射波和入射波产生的驻波表达式为\ans{$2A\cos{\ab(2\pi t+\frac{7\pi}6)}\cos{\ab(\frac{2\pi x}{3}-\frac{7\pi}{6})}$}.
\end{problem}
+\begin{solution}
+\begin{itemize}
+ \item 考虑反射带来的半波损失，$x=1\mathrm{m}$处反射波的振动方程为$y_{10}=A\cos\ab(2\pi t+\frac{2\pi}{3}+\pi)$.
+ \item 反射后传播方向改变，考虑以$x=1$处为参考点需坐标变换$x'=x-1$，所以反射波的表达式为
+ \[y_2=A\cos\ab(2\pi t+\frac{2\pi}{3}-\frac{2\pi(x-1)}3+\pi)=A\cos\ab(2\pi t-\frac{2\pi x}{3}+\frac{7\pi}3)\]
+ \item 驻波表达式$y=y_1+y_2=2A\cos{\ab(2\pi t+\frac{7\pi}6)}\cos{\ab(\frac{2\pi x}{3}-\frac{7\pi}{6})}$.
+\end{itemize}
+\end{solution}
+
+\begin{problem}[双缝干涉][6]
+ 如图所示，在双缝干涉实验中，若把一厚度为$e$、折射率为$n$的薄云母片覆盖在$S_1$缝上，中央明条纹将向\ans{上}移动；覆盖云母片后，两束相干光至原中央明纹$O$处的光程差为\ans{$(n-1)e$}.
+\end{problem}
\begin{solution}*
- $f_1=\frac{340+10}{340-70}\times 1400\mathrm{Hz}\approx 1814.8\mathrm{Hz},\ f_2=\frac{340-10}{340+70}\times 1400\mathrm{Hz}\approx 1126.8\mathrm{Hz}$.
+ 覆盖云母片后，通过$S_1$的光路光程差变大，为抵消这一变化中央明纹需上移使通过$S_2$的光路变长；原光程差为零，现光程差即云母片带来的光程差$\delta=ne-e=(n-1)e$.
\end{solution}
-\section{计算题（共20分）}
-\begin{problem}[简谐振动][10]
- 一个质量为$1\mathrm{kg}$的物块沿$x$轴做简谐振动，振幅为$10\mathrm{cm}$，最大速度$2\times 10^{-2}\mathrm{m/s}$.在时间$t=4\mathrm{s}$时，物块位于$5\mathrm{cm}$.求
- \begin{enumerate}
- \item 简谐运动的周期和最大加速度.
- \item 简谐运动的位移方程.
- \end{enumerate}
+\begin{problem}[牛顿环][3]
+ 若把牛顿环装置（都是用折射率为$1.52$的玻璃制成的）由空气搬入折射率为$1.33$的水中，则干涉条纹\ans{变密}（变疏/变密）.
\end{problem}
+\begin{solution}*
+ 放入水中后每条条纹的光程差变大，为抵消这一变化条纹需向中心收缩，所以干涉条纹变密.
+\end{solution}
+
+\begin{problem}[弗琅禾费衍射][3]
+ 在单缝夫琅禾费衍射实验中，波长为$\lambda$的单色光垂直入射在宽度为$a=6\lambda$的单缝上，对应于衍射角为$30^\circ$的方向，单缝处波阵面可分成的半波带数目为\ans{6}.
+\end{problem}
+\begin{solution}*
+ 由衍射公式$a\sin{\theta}=k\lambda$得$k=3$，可分成的半波带数目为$2k=6$.
+\end{solution}
+
+\section{计算题（共46分）}
+\begin{paracol}{2}
+\begin{problem}[平面简谐波的波函数][10]
+ 一列平面简谐波在媒质中以波速$u=5\mathrm{m/s}$沿$x$轴正向传播，原点$O$处质元的振动曲线如图所示.
+\begin{enumerate}
+ \item 求解$x=25\mathrm{m}$处质元的振动方程并画出该点振动曲线.
+ \item 求解波动方程，并画出$t=3\mathrm{s}$时的波形曲线.
+\end{enumerate}
+\end{problem}
+\switchcolumn\centering
+ \vfill
+ \begin{tikzpicture}
+ \draw [->,thick] (-0.5,0)--(4,0) node [anchor=south] {$t(\mathrm{s})$} node [anchor=north] {\textcolor{H6}{$x(\mathrm{cm})$}};
+ \draw [->,thick] (0,-1.5)--(0,1.5) node [anchor=west] {$y(\mathrm{cm})$};
+ \draw[thick,H1] (0,0) sin (0.6,1) cos (1.2,0) sin (1.8,-1) cos (2.4,0) sin (3.0,1) cos (3.6,0);
+ \draw [densely dashed] (0,1)--(.6,1) node [anchor=east,at start] {$2$};
+ \node [anchor=north east] at (0,0) {$O$} node [anchor=south west] at (1.2,0) {$2$} node [anchor=south east] at (2.4,0) {$4$};
+ \draw [very thick,H2,dotted,line cap=round] (0,-1) cos (0.6,0) sin (1.2,1) cos (1.8,0) sin (2.4,-1) cos (3,0) sin (3.6,1);
+ \draw [very thick,H6,densely dashed,line cap=round] (-.8,0) sin (0,-1) cos (.8,0) sin (1.6,1) cos (2.4,0) sin (3.2,-1) cos (4,0);
+ \node [anchor=north] at (.8,0) {\textcolor{H6}{5}} node [anchor=north] at (2.4,0) {\textcolor{H6}{15}};
+ \end{tikzpicture}
+ \vfill
+\end{paracol}
+\vspace{-.75em}
\begin{solution}
\begin{enumerate}
- \item 由$v_{\max}=\omega,\ T=\frac{2\pi}{\omega}$得$T=10\pi\mathrm{s},\ \omega=0.2\mathrm{s}^{-1},\ a_{\max}=\omega^2A=4\times 10^{-3}\mathrm{m/s}^2$.\point{4}
- \item $t=4\mathrm{s}$时$x=\frac{1}{2}A,\ v<0$，所以此时相位$\varphi=\frac{\pi}{3}$.故位移方程为\point{2}
- \[x=0.1\cos{\ab[0.2\ab(t-4)]+\frac{\pi}{3}}=0.1\cos{\ab(0.2t+0.079\pi)}\eqno\point{4}\]
+ \item 由图知振幅$A=0.02\mathrm{m}$，角频率$\omega=\frac{2\pi}{T}=0.5\pi\mathrm{s}$. $t=0$时，$y=0,\ v>0$，初相$\varphi=-\frac\pi2$. 波动方程为\point{2}
+ \[y=0.02\cos{\ab[\frac\pi2\ab(t-\frac{x}{5})-\frac\pi2]}\eqno\point{2}\]
+ $x=25\mathrm{m}$处质元的振动方程为$y(x_0,t)=0.02\cos\ab(\frac\pi2t-\pi)$.\point{2}
+ \item 波动方程见上问. $t=3\mathrm{s}$时的波形方程为
+ \[y(x,t_0)=0.02\cos\ab(-\frac{\pi}{10}x+\pi)\eqno\point{2}\]
\end{enumerate}
\end{solution}
-\begin{problem}[牛顿环][10]
- 空气中，使用波长为$480\mathrm{nm}$平行单色光观察牛顿环.在反射光中测得某一明环的直径为$4.74\mathrm{mm}$，在它外面第$10$个明环的直径为$7.24\mathrm{mm}$.求
-\begin{enumerate}
- \item 平凸透镜的曲率半径.
- \item 直径为$4.74\mathrm{mm}$明环的条纹级数$k$.
- \item 假设把整个装置放入水中($n=1.33$)，原直径为$4.74\mathrm{mm}$明环的新直径.
-\end{enumerate}
+\vskip-2ex
+\begin{paracol}{2}
+\begin{problem}[光程和光程差][8]\footnote{\kaishu 选自37th CPhO预赛试题第10题，原题未并提供参考图片.}
+ 一艘船（如图中$S$）在$25\mathrm{m}$高的桅杆（$SS'$）上装有一天线（如图中$S''$），不断发射某种波长的无线电波，已知波长在$2-4\mathrm{m}$范围内，在高出海平面$150\mathrm{m}$的悬崖顶（$OP$）上有一接收站$P$能收到这无线电波，但当那艘船驶至离悬崖底部$OS=2\mathrm{km}$时，接收站就收不到无线电波.设海平面完全反射这无线电波，求所用无线电波的波长.
\end{problem}
+\switchcolumn\centering
+\vfill
+\begin{tikzpicture}[scale=0.8]
+ \coordinate [label=below left:{$S$}] (S) at (0,0);
+ \coordinate [label=above left:{$S'$}] (S') at (0,1);
+ \coordinate [label=below left:{$S''$}] (S'') at (0,-1);
+ \coordinate [label=below:{$M$}] (M) at (1.5,0);
+ \coordinate [label=above:{$P$}] (P) at (6,3);
+ \coordinate [label=below:{$O$}] (O) at (6,0);
+ \draw [thick,line join=round,line cap=round] (P) -- (O) -- (S) -- (S');
+ \draw [thick,dashed,line join=round] (P) -- (S) -- (S'') -- (M);
+ \draw [very thick,line cap=round,densely dashed,H1] (S') -- (P);
+ \draw [very thick,line cap=round,dotted,H1] (S') -- (M) -- (P);
+ \length{(-.25,0)}{(-.25,1)}{$a$}{(0,0.25)}
+ \pic ["$\theta$", draw, thick, angle radius=5mm, angle eccentricity=1.5] {angle=O--S--P};
+ \pic ["$\theta'$", draw, thick, angle radius=5mm, angle eccentricity=1.5] {angle=O--M--P};
+ \pic ["$\Delta\theta$", draw, thick, angle radius=10mm, angle eccentricity=1.5] {angle=S--P--M};
+ \draw [thick] (S') -- (0.75,-0.5) coordinate [label=below:$K$] (foot);
+ \pic ["$\theta$", draw, thick, angle radius=3mm, angle eccentricity=2] {angle=S''--S'--foot};
+ \length{($(S'')+({0.5*sin(atan(2/3))},{-0.5*cos(atan(2/3))})$)}{($(foot)+({0.5*sin(atan(2/3))},{-0.5*cos(atan(2/3))})$)}{$\scriptstyle 2a\sin\theta$}{({0.5*cos(atan(2/3))},{0.5*sin(atan(2/3))})}
+ \length{($(S)-(0,2)$)}{($(O)-(0,2)$)}{$2\mathrm{km}$}{(1.5,0)}
+ \length{($(O)+(0.5,0)$)}{($(P)+(0.5,0)$)}{$150\mathrm{m}$}{(0,0.5)}
+ \draw [dashed,H4] (S') --++ (6,0) -- (P);
+ \draw [dashed,H6] (S'') --++ (6,0) -- (P);
+\end{tikzpicture}
+\vfill
+\end{paracol}
+\vspace{-.75em}
\begin{solution}
-\begin{enumerate}
- \item $R=\ab(r_{k+10}^2-r_k^2)/10\lambda=1.56\mathrm{m}$.\point{4}
- \item 由$r_k=\sqrt{\frac{2k-1}{2}\lambda R}$得$k=\frac{r_k^2}{\lambda R}+\frac{1}{2}=8$.\point{3}
- \item 由于光在介质中的波长与折射率成反比，所以此时$d_k^{\prime}=\frac{d_k}{\sqrt{n}}=4.11\mathrm{mm}$.\point{3}
-\end{enumerate}
+ 考虑半波损失，经海平面反射的光波与直达$P$点的光波之间的光程差为
+ \[\delta=2a\sin\theta+\frac\lambda2\eqno\point{3}\]
+ 由几何关系$\sin\theta=\sin\arctan\ab(\frac{150}{2000})=0.075$. 利用干涉相消条件得无线电波长为\point{2}
+ \[\delta=\frac{2k+1}{2}\lambda,\ \lambda=\frac{2a\sin\theta}{k}\xlongequal[2<\lambda<4]{k=1}3.74\mathrm{m}\eqno\point{3}\]
+ \textbf{本题中$\Delta\theta$不可忽略，不得认为$\theta=\angle PMO$! 否则会得到$k=2.18$.}
+ \vskip1ex\hrule\vskip1ex
+ 另一种更精确的解法是直接计算两条光路的长度之差\footnote{值得一提的是，因上一种解法近似认为$S'K\bot S''K$，相对于这种解法有$7.73\ee{-3}\%$的误差（小数点第四位及以后不同）.}
+ \[\begin{aligned}
+ \delta&=\overline{S''P}-\overline{S'P}+\frac\lambda2=\sqrt{\ab(\overline{SO})^2+\ab(\overline{OP}+\overline{SS''})^2}-\sqrt{\ab(\overline{SO})^2+\ab(\overline{OP}-\overline{SS''})^2}+\frac\lambda2\\
+ &=\sqrt{2000^2+(150+25)^2}-\sqrt{2000^2+(150-25)^2}+\frac\lambda2=\frac{2k+1}{2}\lambda
+ \end{aligned}\]
+
+ 解得$\lambda\xlongequal{k=1}3.74\mathrm{m}$.
\end{solution}
-\section{计算题\! $^\dagger$（共15分）}
+\begin{problem}[薄膜干涉][10]
+波长为$\lambda=500\mathrm{nm}$的单色光垂直入射到\textbf{置于空气中}的上下表面平行的薄膜上，已知膜的折射率$n=1.25$，求反射光、透射光最强时膜的最小厚度.
+\end{problem}
+\begin{solution}*
+ 两个表面反射光光程差为$\delta=2ne+\frac\lambda2$. 分别由反射光干涉相长$\delta=k\lambda$和相消$\delta=\frac{2k+1}{2}\lambda$得\point{4}
+ \[e_{\min_1}=\frac{\lambda}{4n}=100\mathrm{nm},\ e_{\min_2}=\frac{\lambda}{2n}=200\mathrm{nm}\eqno\point{4}\]
+\begin{itemize}
+ \item 反射光干涉相长时，反射光最强，膜的最小厚度为$e_{\min_1}=100\mathrm{nm}$.\point{1}
+ \item 反射光干涉相消时，透射光最强，膜的最小厚度为$e_{\min_2}=200\mathrm{nm}$.\point{1}
+\end{itemize}
+\end{solution}
-\begin{problem}[弹簧振子][15]
- 一个质量为$m=1\mathrm{kg}$的盘子刚性连接竖直悬挂的轻弹簧下端，弹簧的劲度系数为$k=90\mathrm{N/m}$.盘子在竖直方向做简谐运动，振幅$A=10.0\mathrm{cm}$.现有一个质量$m_2=1\mathrm{kg}$的物体自由落下掉在盘上，没有反弹.当\textbf{盘子位于向上最大位移处}，盘子与物体发生碰撞，物体从开始自由自由落体发生处，距离为$h=20\mathrm{cm}$，假设碰撞瞬间完成.求碰撞后盘子和物体组成的系统，它的振动周期、振幅和振动能量.
+\vskip-2ex
+\begin{paracol}{2}
+\begin{problem}[光栅][12]
+ 如右图所示，$AB$之间的虚线为一透射式光栅，该光栅在$1\mathrm{mm}$内刻画有$500$条狭缝，单条狭缝的缝宽为$a=0.5\mu\mathrm{m}$，一波长为$\lambda=500\mathrm{nm}$的单色平行光斜入射在该光栅上，入射角$\theta=30^\circ$（从光栅光轴下方入射），在光栅后放置凸透镜和观察屏（屏位于透镜的焦平面处），问屏上能看到哪几级谱线？
\end{problem}
+\switchcolumn\centering
+\vfill
+\begin{tikzpicture}[decoration={markings,mark=between positions .25 and .8 step 20mm with {\arrow{stealth}}}]
+ \filldraw [thick,fill=gray] (0,1.6) rectangle (0.15,0.5);
+ \draw [densely dashed,thick] (0,0.5)--(0,-0.5);
+ \filldraw [thick,fill=gray] (0,-1.6) rectangle (0.15,-0.5);
+ \draw [densely dashed,thick] (-1.5,0)--(2,0);
+ \coordinate (b) at (-1.5,0);
+ \coordinate (a) at (0,0);
+ \coordinate (c) at (-1,-0.4);
+ \coordinate (d) at (1.5,0.2);
+ \coordinate (e) at (2,0);
+ \foreach \a in {-0.5,-0.25,0,0.25,0.5}
+ \draw [yshift=\a cm,thick,postaction=decorate,->] (-1,-0.4)--(0,0)--(1.5,0.2);
+ \pic["$\theta$", draw=black, very thick, angle
+ eccentricity=1.35, angle radius=24]{angle=b--a--c};
+ \pic["$\varphi$", draw=black, very thick, angle
+ eccentricity=1.35, angle radius=36]{angle=e--a--d};
+ \node[anchor=east] at (0,0.6) {$A$} node[anchor=east] at (0,-0.6) {$B$};
+\end{tikzpicture}
+\vfill
+\end{paracol}
+\vspace{-.75em}
\begin{solution}*
- Answer omitted.
+ 光栅常数$d=\frac{1\ee{-3}}{500}=2\mu\mathrm{m}$. 由于光栅方程\point{2}
+ \[d(\sin\varphi-\sin\theta)=k\lambda\eqno\point{2}\]
+\begin{itemize}
+ \item 令$\varphi=\pm 90^\circ$得$k_{\min}=-6,\ k_{\max}=2$. 由缺级条件得$k'=\frac{d}{a}=\pm 4$，第$-4$级缺级.\point{4}
+ \item 由于$\varphi$无法取到$\pm 90^\circ$，所以屏上可见主极大级次为$k=0,\ \pm 1,\ -2,\ -3,\ -5$.\point{4}
+\end{itemize}
+\end{solution}
+
+\begin{problem}[驻波][6]
+ 由振动频率为$400\mathrm{Hz}$的音叉在两端固定拉紧的弦线上建立驻波.这个驻波共有三个波腹，其振幅为$0.30\mathrm{cm}$，波在弦上的速度为$320\mathrm{m/s}$.
+\begin{enumerate}
+ \item 求此弦线的长度.
+ \item 若以弦的中点为坐标原点，试写出弦线上驻波的表达式.
+\end{enumerate}
+\end{problem}
+\begin{solution}
+\begin{enumerate}
+ \item 由题意得弦长为$1.5$个波长，即$l=1.5\lambda=1.5\frac{u}{f}=1.2\mathrm{m}$.\point{2}
+ \item 驻波的角频率$\omega=2\pi f=800\pi$，波矢$k=\frac{2\pi}{\lambda}=\frac{5}{2}\pi$. 设驻波的表达式为
+ \[y=3\ee{-3}\cos\ab(800\pi t+\phi)\cos\ab(\frac52\pi x+\varphi)\eqno\point{2}\]
+ 中点$x=0$处是波腹，所以$\cos\varphi=1,\ \varphi=0\ \text{or}\ \pi$. 所以驻波的表达式为
+ \[y=\pm 3\ee{-3}\cos\ab(800\pi t+\phi)\cos\ab(\frac52\pi x)\eqno\point{2}\]
+ 符号$\pm$对应$\varphi$的两个解，$\phi$由初始条件决定.
+\end{enumerate}
\end{solution}
\ No newline at end of file
Modified: trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-doc.tex
===================================================================
--- trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-doc.tex 2023-12-13 21:23:15 UTC (rev 69114)
+++ trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-doc.tex 2023-12-13 21:23:27 UTC (rev 69115)
@@ -3,9 +3,9 @@
\fancyhead[R]{\color{H7}\rightmark\,}
\centerline{Xia Mingyu, \href{https://www.hdu.edu.cn}{Hangzhou Dianzi University}}
-\ddmmyyyydate
+\yyyymmdddate
\centerline{\mailto{xiamyphys at gmail.com}}
-\centerline{\today,\quad Version 1.0a}
+\centerline{\today,\quad Version 1.2a}
This is the document for \pkg{LiteSolution} template, which provides a lite design of the solution of test paper.
@@ -13,7 +13,7 @@
\section{Introduction}
\subsection{The purpose of this template}
-This template provides a lite and fresh template, and mainly used for typesetting solutions of final, textbooks' and other exercises. This template is developed based on \href{https://www.ctan.org/pkg/elegantbook}{\pkg*{ElegantBook}} and \href{https://github.com/Azure1210/VividBooK}{\pkg*{VividBooK}}, ported and improved the chapter design module code of \href{https://www.overleaf.com/latex/templates/the-legrand-orange-book-template-english/jtctyfmnpppc}{The Legrand Orange Book}. I'd like to express my gratitude to the template authors above for their previous work.
+This template provides a lite and fresh template, and mainly used for typesetting solutions of mid-term or final exam, textbooks and other exercises. This template is developed based on \pkg{ElegantBook} and \href{https://github.com/Azure1210/VividBooK}{\pkg*{VividBooK}}, ported and improved the chapter design module code of \href{https://www.overleaf.com/latex/templates/the-legrand-orange-book-template-english/jtctyfmnpppc}{\pkg*{The Legrand Orange Book}}. I'd like to express my gratitude to the template authors above for their previous work.
If you meet bugs when using this template, or you have better suggestions or ideas, or you want to participate in the development of the template or other templates by me, welcome to contact via email \href{mailto:xiamyphys at gmail.com}{xiamyphys at gmail.com}.
@@ -21,10 +21,8 @@
\subsection{Loading \pkg{LiteSolution} and its modes}
-Update all the packages to the latest version or switch to portable version instead by implementing the commands
+You should update all the packages to the latest version or switch to portable version instead by implementing the commands
\begin{verbatim}
- tlmgr update --self
- tlmgr update --all
tlmgr update --self --all --reinstall-forcibly-removed
\end{verbatim}
@@ -36,7 +34,7 @@
\section{Modes of \pkg{LiteSolution}}
\subsection{The \mode{answer} mode}
-This mode has two options, \mode{ans} and \mode{noans}, which can show and hide answers respectively. After you choose the \mode{noans}, the contents in the environment \cmd{solution}, the command \cmd{ans} and the answers in the multiple choice questions will all become the same color as the pagecolor. So the area that originally contained the answer will be replaced by an area of the same blank size. You can generate exams without answers and solutions by enabling \mode{noans}.
+This mode has two options, \mode{ans} and \mode{noans}, which can show and hide answers respectively. After you choose the \mode{noans}, the contents in the environment \cmd{solution}, the command \cmd{ans} and the answers in the multiple choice questions will all disappear. So the area that originally contained the answer will be replaced by an area of the same blank size. You can generate exams without answers and solutions by enabling \mode{noans}.
\subsection{The \mode{mtpro2} mode}
If you've installed the \emph{Mathtime Pro 2 Lite} font in your computer, then you can use this mode to change the math font.
@@ -44,22 +42,35 @@
\subsection{The \mode{counter}}
This mode has two options, \mode{separate} and \mode{continuous}, which can make the page counter between chapters be remaked or continuous.The page numbers between each test question will be continuous when you use the \mode{continuous} mode or the page number of each test question will start from 1 when you use the \mode{separate}.
-
\section{Commands of \pkg{LiteSolution}}
\subsection{The \cmd{chapterimage} command}
\begin{verbatim}
\chapterimage{cover1.png}
\end{verbatim}
+
This command can assign the title background image for each subsequent chapter.
\subsection{The \cmd{chapterfont} command}
+\begin{verbatim}
+ \chapterfont{PingFang HK} \chapterfont*{PingFang HK}
+\end{verbatim}
+
This command can assign the title font for each subsequent chapter, if you do not use this command, the title font will be \emph{songti} in Chinese and \emph{Libertinus} in English.
+If a star (*) is added after the \verb|\chapterfont|, then \TeX Live will call the font file from the current path, note from the system. And the file in the current path only support the \verb|.ttf| format.
+
\subsection{The \cmd{ans} command}
-This command can underlines the answer and changes the color of the answer to \textcolor{1号色}{Blue Sapphire}.
+This command can underlines the answer and changes the color of the answer to \textcolor{H5}{Blue Sapphire}.
-\paragraph{If mode \mode{noans} is enabled, the answer will disappear, leaving only a horizontal line the same width as the answer.}
+If mode \mode{noans} is enabled, the answer will disappear, leaving only a horizontal line the same width as the answer.
+\subsection{The \cmd{solute} command}
+\begin{verbatim}
+ \solute{3} % create a solution box with the height of 3em.
+\end{verbatim}
+
+This command can create a fixable solution box when the mode \mode{noans} is enabled.
+
\subsection{The \cmd{watermark} command}
\begin{verbatim}
\watermark{ctanlion.pdf}
@@ -78,6 +89,10 @@
\verb|\titlelogo{#1}{#2}| & Add emoji with link in text & \verb|\point{#1}| & Add score & \verb|\i| & $\mathrm{i}$\\
\hline
\verb|\sokka{#1}| & 故本题选择\#1项 & \verb|\d| & $\mathrm{d}$ & \verb|\e| & $\mathrm{e}$\\
+ \hline
+ \verb|\xSim[#1]{#2}| & $\xSim[r_1\times2]{r_2+r_1}$ & \verb|\ee{#1}| & $\ee{-34}$ & \verb|\mat{#1}^\T| & $\mat{A}^\T$\\
+ \hline
+ \verb|\rank{#1}| & $\rank{\mat{AB}}$ & \verb|\QED| & \QED & \verb|$5\unit{kg}$| & $5\unit{kg}$\\
\bottomrule
\end{tabular}
\end{center}
@@ -94,12 +109,6 @@
\task This is choice C \task This is choice D
\end{tasks}
\end{choice}
-\begin{choice}{D}
-If you want to add choice only.
-\begin{tasks}(4) % 4 choices per line
- \task Chc A \task Chc B \task Chc C \task Chc D
-\end{tasks}
-\end{choice}
\end{tcblisting}
\begin{paracol}{2}
@@ -125,6 +134,7 @@
\end{tikzpicture}
\vfill
\end{paracol}
+\begin{tcolorbox}
\begin{verbatim}
\begin{paracol}{2}
\begin{choice}{D}[Gaussian theory]
@@ -137,6 +147,7 @@
\switchcolumn\centering\vfill\tikz{...}\vfill
\end{paracol}
\end{verbatim}
+\end{tcolorbox}
\subsection{The \cmd{problem} environment}
Sightly different from the cmd{choice} environment: the two variables are points and keywords, and the question number counter is shared with the multiple-choice question number counter.
@@ -157,7 +168,7 @@
\begin{paracol}{2}
\begin{problem}[Gaussian theory \& Field strength][6]
- 一均匀带电直导线长为$d$，电荷线密度为$+\lambda$.过导线中点$O$作一半径为$R$（$R>\frac{d}{2}$）的球面$S$，$P$为带电直导线的延长线与球面$S$的交点. 则通过该球面的电场强度通量$\Phi_e=$\ans{$\frac{\lambda d}{\varepsilon_0}$}，带电直线的延长线与球面交点$P$处的电场强度的大小为\ans{$\frac{\lambda d}{4\pi\varepsilon_0(R^2-d^2/4)}$}，方向\ans{沿矢径$\boldsymbol{OP}$}.
+ 一均匀带电直导线长为$d$，电荷线密度为$+\lambda$.过导线中点$O$作一半径为$R$（$R>\frac{d}{2}$）的球面$S$，$P$为带电直导线的延长线与球面$S$的交点. 则通过该球面的电场强度通量$\Phi_e=$\ans{$\frac{\lambda d}{\varepsilon_0}$}，带电直线的延长线与球面交点$P$处的电场强度的大小为\ans{$\frac{\lambda d}{4\pi\varepsilon_0(R^2-d^2/4)}$}，方向\ans{沿矢径$\vv{OP}$}.
\end{problem}
\switchcolumn\centering
\vfill
@@ -174,6 +185,7 @@
\end{tikzpicture}
\vfill
\end{paracol}
+\begin{tcolorbox}
\begin{verbatim}
\begin{paracol}{2}
\begin{problem}[Gaussian theory \& Field strength][6]
@@ -183,6 +195,7 @@
\switchcolumn\centering\vfill\tikz{...}\vfill
\end{paracol}
\end{verbatim}
+\end{tcolorbox}
\subsection{The \cmd{note} environment}
\begin{tcblisting}{sidebyside}
@@ -191,6 +204,20 @@
\end{note}
\end{tcblisting}
+\subsection{The \cmd{proof} environment}
+\begin{tcblisting}{sidebyside}
+\begin{proof}
+ Due to \emph{Langrange's Thorem}...
+\end{proof}
+\end{tcblisting}
+
+If a star (*) is added after the \verb|\begin{solution}|, then the content will follow the
+\begin{tcblisting}{sidebyside}
+\begin{proof}*
+ Due to \emph{Langrange's Thorem}...
+\end{proof}
+\end{tcblisting}
+
\subsection{The \cmd{solution} environment}
\begin{tcblisting}{sidebyside}
\begin{solution}
@@ -205,11 +232,13 @@
\end{solution}
\end{tcblisting}
-\paragraph{If mode \mode{noans} is enabled, the solution will disappear, leaving only a blank box with the same height as the solution, and the name of the box will change to \emph{\textcolor{1号色}{\textbf{\faIcon{ban} 答案隐藏}}}.}
+If mode \mode{noans} is enabled, the solution will disappear, leaving only a blank box with the same height as the solution, and the name of the box will change to \emph{\textcolor{1号色}{\textbf{\faIcon{pen-square} 答题区域}}}.
-\newpage
+\section{Known Issues}
+\TeX Live will return errors when you enable the mode \mode{noans} and use the \cmd{solution} environment in \cmd{paracol} environment.
+
\section{Version History}
-This template is used to type the mid-term and final exam solutions of \emph{College Physics}. Initially, I used the \href{https://www.ctan.org/pkg/elegantbook}{ElegantBook} template for layout, however, it's no longer be maintained since January 1st, 2023, so I trun to use the \href{https://github.com/Azure1210/VividBooK}{\pkg*{VividBooK}} instead. But this template is too bloated and some functions \& designs need to be redesigned, so I started developing \pkg{LiteSolution}.
+This template is used to type the mid-term and final exam solutions of \emph{College Physics}. Initially, I used the \href{https://www.ctan.org/pkg/elegantbook}{ElegantBook} template for layout, however, it's no longer be maintained since January 1st, 2023, so I turn to use the \href{https://github.com/Azure1210/VividBooK}{\pkg*{VividBooK}} instead. But this template is too bloated and some functions \& designs need to be redesigned, so I started developing \pkg{LiteSolution}.
\textsf{\bfseries Version 0.1a} was finished developing on 29 June, 2023 and released on \href{https://mp.weixin.qq.com/s/kd4StYk3XybhNQZkAfoY6A}{\faIcon{weixin} WeChat Public Account: 物理问题作} with the name \emph{FreshSolution}. This version redesigned the \cmd{exercise} environment and the \cmd{solution} environment in terms of designs and functions, and improvements have been made to the design of the chapterimage part.
@@ -217,13 +246,17 @@
\textsf{\bfseries Version 1.0a} was finished developing on 15 November, 2023. This version has redesigned the \cmd{chapterimage} part, \cmd{choice} and replaced the \cmd{exercise} environment with the \cmd{problem} environment.
-\datechange{06/07/2023}{Version 0.1b}
+\textsf{\bfseries Version 1.2a} was finished developing on \textcolor{H1}{13 November, 2023}. This version has integrated the chapter design of \href{https://www.ctan.org/pkg/elegantbook}{\pkg*{ElegantBook}} into the star (*) key value of this template \cmd{chapter}. And some commands friendly for matrices typesetting were added in. The command \cmd{chapterfont} was redesigned that it can call the font in the local path or call system font with(out) a star (*) after it. Also, the environment \cmd{proof} was redesigned, and the mode \mode{noans} in this version supports hide the solution box and replace it with a fixable solution box. Today (\textcolor{H1}{2023/12/13}) is \emph{The National public memorial day of Nanjing Massacre}. Hope for world peace.
+
+\newpage
+\subsection*{Update Announcements}
+\datechange{2023/07/06}{Version 0.1b}
\begin{itemize}
\item Support page number remaking between chapters.
\item Added \cmd{watermark} command.
\end{itemize}
-\datechange{15/11/2023}{Version 1.0a}
+\datechange{2023/11/15}{Version 1.0a}
\begin{itemize}
\item Redesigned the \cmd{chapterimage} part, include the layout and the code.
\item Redesigned the \cmd{choice} and \cmd{solution} environment, keywords become optional and supports star (*) key.
@@ -231,10 +264,19 @@
\item Added the \cmd{note} environment and some customer commands.
\end{itemize}
+\datechange{\today}{Version 1.2a}
+\begin{itemize}
+ \item Fixed the bug that the gap around the chapter image.
+ \item Added some commands for matrices.
+ \item Redesigned the \cmd{chapterfont} command.
+ \item Redesigned the \cmd{proof} environment.
+ \item Supports to adjust the height of solution box when output the exam paper without answer.
+ \item Add the \emph{排版规范} in \emph{中文（简体）} after the package document.
+ \item Fixed the bug that warnings of the packages \pkg{xeCJK} and \pkg{fontspec}.
+\end{itemize}
+
\subsection*{Future Plans}
\begin{itemize}
- \item Plan to integrate the color management of \href{https://www.ctan.org/pkg/elegantbook}{\pkg*{ElegantBook}}.
- \item Plan to integrate the chapter design of \href{https://www.ctan.org/pkg/elegantbook}{\pkg*{ElegantBook}} into the star (*) key value of this template \cmd{chapter}.
\item Plan to add the \mode{dark} mode to this template to make the text color light while make the page color dark to protect eyesight.
\item Plan to change the \verb|*.cls| file to a block code design to make it easier for subsequent developers to maintain or modify.
\item ......
Added: trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-gauge.tex
===================================================================
--- trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-gauge.tex (rev 0)
+++ trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-gauge.tex 2023-12-13 21:23:27 UTC (rev 69115)
@@ -0,0 +1,65 @@
+\chapter*{\pkg{LiteSolution}~试卷解析模板排版规范}% standard
+\fancyhead[L]{\,\color{1号色}\kaishu\href{http://weixin.qq.com/r/hR1SSofEIdpercMp90iX}{物理問題作}}
+\fancyhead[R]{\color{1号色}\ttfamily\mailto{xiamyphys at gmail.com}\,}
+
+
+\centerline{夏明宇, \href{https://www.hdu.edu.cn}{杭州电子科技大学}}
+\yyyymmdddate
+\centerline{\mailto{xiamyphys at gmail.com}}
+
+\section{模板介绍}
+本模板用于高校期中 \& 期末试卷排版，支持一键输出无答案试卷.
+
+\section{数学公式}
+\subsection{正体问题}
+\begin{enumerate}
+ \item 微分算符要使用正体：\verb|\d|.
+ \item 自然对数要使用正体：\verb|\e|.
+ \item 三角函数、exp正体：\verb|\function|
+ \item 撇\verb|'|和\verb|^\prime|的作用相同：\href{https://tex.stackexchange.com/a/538413/299948}{TeX Stackexchange}.
+\end{enumerate}
+
+\subsection{行内 \& 行间公式}
+\begin{enumerate}
+ \item 行内公式：要\verb|$ ... $|不要\verb|\( ... \)|
+ \item 行间公式：要\verb|\[ ... \]|不要\verb|$$ ... $$|
+\end{enumerate}
+至于为何，“不要”的命令在\LaTeX 中属于「脆弱」命令，这里不做详解.
+
+\href{https://www.zhihu.com/question/27589739}{\LaTeX 中的「要」和「不要」 - 知乎}
+
+\subsection{选填题}
+\begin{enumerate}
+ \item 使用tasks环境判断高度
+ \begin{verbatim}
+ \begin{tasks}(一行选项个数，自己掂量)
+ \task 选项1 \task 选项2
+ \task 选项3 \task 选项4
+ \end{tasks}
+ \end{verbatim}
+
+ 不要使用\verb|tabular|制作选择题选项，选项不是画表格！
+ \item 答案不要擅自使用\verb|\underline{}|命令，而是使用\verb|\ans{}|命令，否则制作空白无答案版时答案将无法隐藏.
+\end{enumerate}
+
+\subsection{等高括号}
+\begin{enumerate}
+ \item 凡是你觉得很高的都需做等高括号处理：\verb|\ab( ), \ab[ ], \ab\{ \}|
+ \item 但也不要所有括号都做等高处理，比如\verb|f\ab(x)|和\verb|f(x)|的效果是一样的，多一个\verb|\ab|判断高度只会徒增编译时间.
+\end{enumerate}
+
+\section{偷懒技巧}
+以\verb|\frac{}{}|为例
+\begin{enumerate}
+ \item \verb|\frac12 == \frac{1}{2}, \frac\pi2 == \frac{\pi}{2}, \frac yx == \frac{y}{x}|
+ \item \verb|frac{}{}|后跟两个键值，如果未使用\verb|{}|界定作用域，则默认scan到第1个char或宏（如\verb|\alpha|）时将其放入第一个键值，scan到第2个char或宏（如\verb|\alpha|）时将其放入第2个键值.
+ \item 不会用的话别乱用，老实儿使用\verb|{}|界定作用域，不要出现诸如\verb|\frac x'2|: $\frac x'2$（事实你要写的是$\frac{x'}{2}$）.
+\end{enumerate}
+
+\section{排版美观}
+\begin{enumerate}
+ \item 尽量不要（甚至杜绝）题目在上一页，答案在下一页的情况（\emph{\textbf{六级完形文章和题目不在一页的痛你应该经历过}}）.
+ \item 杜绝出现题目/解答盒子断开跨页的情况，这很不美观. 如果一个题目解析实在过长，请尝试缩句，或者使用分栏\verb|multicols|环境以节省空间.
+ \item 如果因第$n+1$页开头的题目盒子高度小于第$n$页末尾剩余了空间而导致了空间浪费，请试着将后文中高度与第$n$页最后一道题目+剩余空间相当的题目位置与其交换.
+ \item 最完美的排版：除非遇到下一个\verb|section|，每一页从头到尾都恰好填满了题目与解析.
+\end{enumerate}
\ No newline at end of file
Property changes on: trunk/Master/texmf-dist/doc/latex/litesolution/litesolution-gauge.tex
___________________________________________________________________
Added: svn:eol-style
## -0,0 +1 ##
+native
\ No newline at end of property
Modified: trunk/Master/texmf-dist/doc/latex/litesolution/litesolution.pdf
===================================================================
(Binary files differ)
Modified: trunk/Master/texmf-dist/doc/latex/litesolution/litesolution.tex
===================================================================
--- trunk/Master/texmf-dist/doc/latex/litesolution/litesolution.tex 2023-12-13 21:23:15 UTC (rev 69114)
+++ trunk/Master/texmf-dist/doc/latex/litesolution/litesolution.tex 2023-12-13 21:23:27 UTC (rev 69115)
@@ -1,29 +1,11 @@
-\documentclass{litesolution}
-\usepackage[level]{datetime}
+\documentclass[ans,continuous]{litesolution}
-% document setup
-\makeatletter
-\def\@pkg#1{\texorpdfstring{\href{https://www.ctan.org/pkg/#1}{\textcolor{pkgcolor}{\textsf{#1}}}}{“#1”}}
-\def\s at pkg#1{\texorpdfstring{\textcolor{pkgcolor}{\textsf{#1}}}{“#1”}}
-\DeclareRobustCommand\pkg{\@ifstar\s at pkg\@pkg}
-\def\mode#1{\texorpdfstring{\textcolor{moducolor}{\textsf{#1}}}{“#1”}}
-\def\cmd#1{\texorpdfstring{\textcolor{cmdcolor}{\textsf{#1}}}{“#1”}}
-\def\datechange#1#2{%
- \noindent{\makebox[\textwidth][r]{\color{H7}\rule{1.15\textwidth}{.4pt}}}
- \noindent\makebox[0pt][r]{\makebox[-3em][r]{\small\textbf{\textcolor{H7}{#1}}}\;\;}{\sffamily Update: \ignorespaces#2}}
-\makeatother
-\setmainfont{Libertinus Serif}
-\setsansfont{Libertinus Sans}
-\setmonofont{NotoSansMono}[
- Scale=MatchLowercase,
- Extension=.ttf,
- UprightFont=*-Light,
- BoldFont=*-Medium
-]
-
\begin{document}
\chapterimage{cover1.png}
\input{litesolution-doc.tex}
\input{litesolution-demo.tex}
+\chapterfont{PingFang HK}
+\input{litesolution-gauge.tex}
+
\end{document}
\ No newline at end of file
Modified: trunk/Master/texmf-dist/tex/latex/litesolution/litesolution.cls
===================================================================
--- trunk/Master/texmf-dist/tex/latex/litesolution/litesolution.cls 2023-12-13 21:23:15 UTC (rev 69114)
+++ trunk/Master/texmf-dist/tex/latex/litesolution/litesolution.cls 2023-12-13 21:23:27 UTC (rev 69115)
@@ -14,15 +14,21 @@
%% ************************************************************************************************************
\NeedsTeXFormat{LaTeX2e}
-\ProvidesClass{litesolution}[15/11/2023 v1.0a LiteSolution document class]
+\PassOptionsToPackage{quiet}{xeCJK}
+\PassOptionsToPackage{quiet}{fontspec}
+\PassOptionsToPackage{no-math}{fontspec}
+
+\ProvidesClass{litesolution}[2023/12/13 v1.2a LiteSolution document class]
\RequirePackage{kvoptions}
\RequirePackage{etoolbox}
\SetupKeyvalOptions{family=LITE, prefix=LITE@, setkeys=\kvsetkeys}
\newcommand{\ekv}[1]{\kvsetkeys{LITE}{#1}}
+
% ---- 全局选项
\DeclareStringOption[ans]{answer}
\DeclareVoidOption{ans}{\ekv{answer=ans}}
\DeclareVoidOption{noans}{\ekv{answer=noans}}
+\DeclareVoidOption{preprint}{\ekv{answer=preprint}}
\DeclareStringOption[separate]{counter}
\DeclareVoidOption{separate}{\ekv{counter=separate}}
@@ -34,24 +40,31 @@
% ----- Default Options -----
\DeclareDefaultOption{\PassOptionsToClass{\CurrentOption}{book}}
\ProcessKeyvalOptions*\relax
-% \ProcessOptions*\relax
\LoadClass[a4paper,oneside]{book}
+\RequirePackage[letterpaper,margin=0.75in,headheight=0.25in,headsep=0.2in,footskip=0.2in]{geometry}
+\usepackage[level]{datetime}
\RequirePackage{extarrows}% 长等号
-\RequirePackage{fixdif,derivative}% 微分算子修复
\RequirePackage{physics2}
\usephysicsmodule{ab,ab.legacy,braket,nabla.legacy}
-\RequirePackage{qrcode}
-\qrset{link, height=4em}
-
+\RequirePackage{amsmath,amssymb}
+\usepackage{nicematrix}
+\RequirePackage{bm}
+\RequirePackage{esint}
\RequirePackage[e]{esvect}% 矢量箭头
-\RequirePackage{setspace}
\RequirePackage{bbm}% 花体字
\RequirePackage{dutchcal}% 花体字2
\RequirePackage{mathrsfs}% 花体字3
\RequirePackage{csquotes}% 引号
\RequirePackage{tasks}% 选择题
-\settasks{label=\color{normalcolor}\Alph*.}
+\settasks{label=\Alph*.}
+\RequirePackage{setspace}
+\setstretch{1.5}
+\AtBeginDocument{
+ \everymath{\displaystyle}
+ \setlength{\abovedisplayskip}{3pt}
+ \setlength{\belowdisplayskip}{3pt}
+}
\definecolor{pkgcolor}{Hsb}{103,.8,.5}
\definecolor{moducolor}{Hsb}{290,.8,.5}
@@ -67,8 +80,6 @@
\RequirePackage{hyperref}
\hypersetup{colorlinks,urlcolor=H1,linkcolor=H2,filecolor=filecolor,pdfstartview=FitH,pdfview=FitH,pdfcreator=XeTeX output}
-\RequirePackage[letterpaper,margin=0.75in,headheight=0.25in,headsep=0.2in,footskip=0.2in]{geometry}
-
\RequirePackage{indentfirst,comment}
% fontsetting
\ifdefstring{\LITE at math}{mtpro2}{
@@ -75,84 +86,65 @@
\let\Bbbk\relax
\RequirePackage[lite]{mtpro2}
}{\relax}
-
\RequirePackage[UTF8,scheme=plain]{ctex}
-\def\chapterfont#1{
- \setCJKfamilyfont{chapterfont}{#1}[Extension =.ttf]
- \gdef\@chapterfont{#1}
+\RequirePackage{xeCJK}
+\RequirePackage[no-math]{fontspec}
+\RequirePackage{fixdif,derivative}
+\NewDocumentCommand\chapterfont{s m}{
+ \IfBooleanTF{#1}{
+ \setCJKfamilyfont{chapterfont}{#2}[Extension =.ttf]
+ }
+ {
+ \setCJKfamilyfont{chapterfont}{#2}
+ }
+ \gdef\@chapterfont{#2}
}
+\setmainfont{Libertinus Serif}
+\setsansfont{Libertinus Sans}
\setcounter{tocdepth}{1}
-
-\PassOptionsToPackage{no-math}{fontspec}
-\PassOptionsToPackage{quiet}{fontspec}
-\RequirePackage[no-math]{fontspec}
-
\AfterEndPreamble{
- \setlength\parindent{2\ccwd}}
+ \setlength\parindent{2\ccwd}}{\relax}
-%% 章节以及页脚图形
+%% 章节
\RequirePackage{xcolor}
\RequirePackage{silence}
+\RequirePackage[center,pagestyles]{titlesec}
+\RequirePackage[title,titletoc,header]{appendix}
- \definecolor{backgroundcolor}{HTML}{F8F7E9}
- \definecolor{headrulecolor}{HTML}{640125}
- \pagecolor{backgroundcolor!20}
- \definecolor{1号色}{HTML}{4D5AAF} %
- \definecolor{2号色}{HTML}{006E54} %
+% ----- box -----
+\definecolor{headrulecolor}{HTML}{640125}
+\definecolor{1号色}{HTML}{4D5AAF} %
+\definecolor{2号色}{HTML}{006E54} %
- \ifdefstring{\LITE at answer}{ans}{
+\ifdefstring{\LITE at answer}{ans}{
\definecolor{solutioncolor}{HTML}{0F2350} %
- \newcommand{\solutionname}{\faIcon{check-square} 分析与解}
+ \def\solutionname{\faIcon{check-square} 分析与解}
+ \def\ans#1{\underline{\color{solutioncolor}\ #1\ }}
+ \def\choiceans#1{{\color{solutioncolor}\ #1\ }}
+ \definecolor{backgroundcolor}{HTML}{F8F7E9}
+ \def\solute#1{\relax}
+ \def\hidesolution{\relax}
}{\relax}
+
\ifdefstring{\LITE at answer}{noans}{
\definecolor{solutioncolor}{HTML}{FEFEFB} %
- \definecolor{solutionbg}{RGB}{130,73,71} %
- \newcommand{\solutionname}{\faIcon{ban} 答案隐藏}
+ \def\solutionname{\faIcon{pen-square} 答题区域}
+ \def\ans#1{\underline{\phantom{\color{solutioncolor}#1}}}
+ \def\choiceans#1{\phantom{\color{solutioncolor}}}
+ \definecolor{backgroundcolor}{HTML}{FFFFFF}
+ \def\solute#1{\begin{draft}\vspace{#1em}\end{draft}}
+ \def\hidesolution{
+ \excludecomment{solution}
+ \let\endsolution\relax
+ }
}{\relax}
- \definecolor{normalcolor}{HTML}{000000} %
+\pagecolor{backgroundcolor!20}
+\def\notename{\faIcon{info-circle} 注意}
+\def\proofname{证明.}
+\def\problemname{\faIcon{pen-square} 题目} % 题目=习题
-% ----- box -----
-%% 章节设置
-\RequirePackage[center,pagestyles]{titlesec}
-\RequirePackage[title,titletoc,header]{appendix}
-
-\newcommand{\notename}{\faIcon{snowflake} 注意}
-\renewcommand*{\proofname}{证明}
-\newcommand{\problemname}{\faIcon{pen-square} 题目} % 题目=习题
-
-\newcommand{\remarkname}{注}
-\renewcommand*{\date}[8]{
- \vspace{-0.5em}\color{normalcolor}
- \begin{minipage}{0.64\textwidth}
- \begin{multicols}{2}
- \textbf{考试时间}：\kaishu #1
-
- \textbf{\songti 课程编号}：\kaishu #3
-
- \textbf{\songti 任课教师}：\kaishu #2
-
- \textbf{\songti 解析制作}：\kaishu #4
- \end{multicols}
- \end{minipage}
- \hfill
- \begin{minipage}{0.15\textwidth}
- \centering
- #5
-
- \vskip1ex\scriptsize\textbf{#6}
- \end{minipage}
- \hfill
- \begin{minipage}{0.15\textwidth}
- \centering
- #7
-
- \vskip1ex\scriptsize\textbf{#8}
- \end{minipage}
- \vspace{-0.5em}
-}
-
\RequirePackage[most]{tcolorbox}
\tcbuselibrary{breakable}
\tcbset{
@@ -176,22 +168,18 @@
\begin{tcolorbox}[
common,
borderline west={1.5pt}{0mm}{2号色},
- title=\bfseries\color{2号色}{\problemname}\theprob\hfill【\quad\color{solutioncolor} #1 \color{2号色}\quad 】,
- coltitle=2号色,
- colupper=normalcolor,
- collower=normalcolor
+ title=\color{2号色}{\problemname}\theprob\hfill【\quad \choiceans#1 \color{2号色}\quad 】,
+ coltitle=2号色
]
}{
\begin{tcolorbox}[
common,
borderline west={1.5pt}{0mm}{2号色},
- title=\bfseries\color{2号色}{\problemname}\theprob\hfill\color{2号色}\faIcon{tag}~#2\qquad【\quad\color{solutioncolor} #1 \color{2号色}\quad 】,
- coltitle=2号色,
- colupper=normalcolor,
- collower=normalcolor
+ title=\color{2号色}{\problemname}\theprob\hfill\color{2号色}\faIcon{tag}~#2\qquad【\quad \choiceans#1 \color{2号色}\quad 】,
+ coltitle=2号色
]
}
- \rmfamily\color{normalcolor}}{\end{tcolorbox}}
+ \rmfamily}{\end{tcolorbox}}
\NewDocumentEnvironment{problem}{s o o}{
\refstepcounter{prob}
@@ -207,8 +195,6 @@
borderline west={1.5pt}{0mm}{2号色},
title=\color{2号色}{\problemname} \theprob,
coltitle=2号色,
- colupper=normalcolor,
- collower=normalcolor
]
}{
\begin{tcolorbox}[
@@ -216,8 +202,6 @@
borderline west={1.5pt}{0mm}{2号色},
title=\color{2号色}{\problemname} \theprob\tagname~#2,
coltitle=2号色,
- colupper=normalcolor,
- collower=normalcolor
]
}
}{
@@ -227,8 +211,6 @@
borderline west={1.5pt}{0mm}{2号色},
title=\color{2号色}{\problemname} \theprob（本题#3分）,
coltitle=2号色,
- colupper=normalcolor,
- collower=normalcolor
]
}{
\begin{tcolorbox}[
@@ -236,37 +218,24 @@
borderline west={1.5pt}{0mm}{2号色},
title=\color{2号色}{\problemname} \theprob（本题#3分）\tagname~#2,
coltitle=2号色,
- colupper=normalcolor,
- collower=normalcolor
]
}
}
- \rmfamily\color{normalcolor}}{\end{tcolorbox}}
+ \rmfamily}{\end{tcolorbox}}
-\newcommand{\noting}[1]{
- \begin{tcolorbox}[
- common,
- borderline west={1.5pt}{0mm}{orange},
- coltitle=orange,
- ]%
- {\textbf{\color{orange}\notename}}
- \kaishu\color{normalcolor}#1\end{tcolorbox}}
-
\newenvironment{note}{
\begin{tcolorbox}[
common,
- borderline west={1.5pt}{0mm}{orange},
- title=\textbf{\color{orange}\notename},
- coltitle=2号色,
+ borderline west={1.5pt}{0mm}{H1},
+ title=\color{H1}\notename,
+ coltitle=H1
]
- \kaishu\color{normalcolor}}{\end{tcolorbox}}
+ \kaishu}{\end{tcolorbox}}
-\definecolor{second}{RGB}{226,106,106} %
-\newenvironment{proof}{
- \par\noindent\bfseries\songti{\color{second}\proofname\;}
- \color{black!90}\fangsong}{
- \hfill\ensuremath{\square}
- \par}
+\NewDocumentEnvironment{proof}{s}{
+ \par\noindent\songti{\color{H1}\proofname\;}
+ \color{solutioncolor!90}\fangsong
+}{\IfBooleanTF{#1}{\relax}{\QED}}
\NewDocumentEnvironment{solution}{s}{
\IfBooleanTF{#1}{
@@ -275,13 +244,13 @@
borderline west={1.5pt}{0mm}{1号色},
coltitle=1号色
]
- \textcolor{1号色}{\textbf{\solutionname}}
+ \color{1号色}\textbf{\solutionname}
}
{
\begin{tcolorbox}[
common,
borderline west={1.5pt}{0mm}{1号色},
- title=\textcolor{1号色}{\textbf{\solutionname}},
+ title=\color{1号色}\solutionname,
coltitle=1号色,
colupper=solutioncolor,
collower=solutioncolor
@@ -288,6 +257,16 @@
]}%
\kaishu\color{solutioncolor}}{\end{tcolorbox}}
+\newenvironment{draft}{
+\begin{tcolorbox}[
+ common,
+ borderline west={1.5pt}{0mm}{1号色},
+ coltitle=1号色
+ ]
+ \color{1号色}\bfseries\solutionname
+ }
+{\end{tcolorbox}}
+
\RequirePackage{enumerate}
% list/itemize/enumerate setting
\RequirePackage[shortlabels,inline]{enumitem}
@@ -295,17 +274,13 @@
\RequirePackage{graphicx}
\graphicspath{{./figure/}{./figures/}{./image/}{./images/}{./graphics/}{./graphic/}{./pictures/}{./picture/}}
-\RequirePackage{bm}% 数学加粗
\RequirePackage{pdfpages}
-\RequirePackage{esint}% 面积分
\RequirePackage{wrapstuff}% 图文绕排
\RequirePackage{booktabs}
\RequirePackage{paracol}
\columnratio{0.67}
\RequirePackage{multicol,multirow}
-
\RequirePackage{fancyvrb}
-%%中文结构名字
%绘图
\RequirePackage{tikz} %% load tikz without tikz
@@ -313,7 +288,6 @@
\tikzset{>=stealth,line cap=round,line join=round}
\RequirePackage{pgfplots}
\pgfplotsset{compat=1.9}
-%电路图绘制
\RequirePackage{circuitikz}
\ctikzset{logic ports=ieee,logic ports/scale=0.6,fill=backgroundcolor}
@@ -328,6 +302,9 @@
\setlist[itemize,2]{label={\eitemii}}
\setlist[itemize,3]{label={\eitemiii}}
+\titleformat{\chapter}[hang]{\bfseries}{
+ \filcenter\LARGE\enspace\bfseries}{1pt}{
+ \LARGE\bfseries\color{1号色}\filcenter}[]
\def\thesection{\arabic{section}}
\titleformat{\section}[hang]{\bfseries}{
\Large\bfseries\heiti{\color{1号色}\thesection.}\enspace}{1pt}{%
@@ -344,18 +321,17 @@
\def\chapterimage{\gdef\@chapterimage}
\def\@makechapterhead#1{%
\begin{tikzpicture}[remember picture,overlay]
- \fill [fill=backgroundcolor,opacity=0.2] (current page.north west) rectangle +(\paperwidth,-\paperwidth/13*6);
- \node[scope fading=south,anchor=north,inner sep=0pt] at (current page.north) {\includegraphics[width=\paperwidth]{\@chapterimage}};
- \draw [line width=2pt,2号色!40] ($(current page.north west)-(0,3.5/13*\paperwidth)$) -- ($(current page.north east)-(0,3.5/13*\paperwidth)$);
- \draw [line width=2pt,2号色!40] ($(current page.north west)-(0,4.5/13*\paperwidth)$) -- ($(current page.north east)-(0,4.5/13*\paperwidth)$);
- \fill [backgroundcolor!40,opacity=0.5] ($(current page.north west)-(0,3.5/13*\paperwidth)$) rectangle +(\paperwidth,-\paperwidth/13);
- \node at ($(current page.north west)-(-0.5*\paperwidth,4/13*\paperwidth)$) {\ifcsname @chapterfont\endcsname\CJKfamily{chapterfont}\fi\huge\bfseries\color{normalcolor}#1~\color{2号色}\faIcon{feather-alt}};
+ \node[scope fading=south,anchor=north,inner sep=0pt,outer sep=0pt] at (current page.north) {\includegraphics[width=\paperwidth]{\@chapterimage}}; % 图片
+ \draw [line width=2pt,2号色!40] ($(current page.north west)-(0,3.5/13*\paperwidth)$) -- ($(current page.north east)-(0,3.5/13*\paperwidth)$); % 上线
+ \draw [line width=2pt,2号色!40] ($(current page.north west)-(0,4.5/13*\paperwidth)$) -- ($(current page.north east)-(0,4.5/13*\paperwidth)$); % 下线
+ \fill [backgroundcolor!40,opacity=0.5] ($(current page.north west)-(0,3.5/13*\paperwidth)$) rectangle +(\paperwidth,-\paperwidth/13); % 中间
+ \node at ($(current page.north west)-(-0.5*\paperwidth,4/13*\paperwidth)$) {\ifcsname @chapterfont\endcsname\CJKfamily{chapterfont}\fi\huge\bfseries#1~\color{2号色}\faIcon{feather-alt}}; % 标题
\end{tikzpicture}\par
\vspace{3.5\paperwidth/13}
\thispagestyle{plain}
\fancypagestyle{plain}{
\fancyhead{} % 页眉清空
- \def\headrulewidth{0pt} % 去页眉线
+ \renewcommand{\headrule}{} % 首页去页眉线
}
\ifdefstring{\LITE at counter}{separate}{
\setcounter{page}{1}
@@ -362,60 +338,41 @@
}{\relax}
}
\def\@makeschapterhead#1{
- \centering
- \ifcsname @chapterfont\endcsname\CJKfamily{chapterfont}\fi\huge\bfseries\color{normalcolor}#1\normalsize
- \thispagestyle{plain}
+ {\parindent \z@ \raggedright
+ \normalfont
+ \interlinepenalty\@M
+ \centering \ifcsname @chapterfont\endcsname\CJKfamily{chapterfont}\fi \huge \bfseries \color{1号色!80!2号色!80!black} #1\par\nobreak
+ \vspace*{1em}
+ }
+ \thispagestyle{plain}
\fancypagestyle{plain}{
\fancyhead{} % 页眉清空
- \def\headrulewidth{0pt} % 去页眉线
+ \renewcommand{\headrule}{} % 首页去页眉线
}
\ifdefstring{\LITE at counter}{separate}{
\setcounter{page}{1}
}{\relax}
+ \setcounter{section}{0}
+ \setcounter{prob}{0}
}
%%%%%%%%%%%chapter图片定制结束%%%%%%%%%%%%%%%
-\newcommand{\watermark}[1]{\RequirePackage{wallpaper}\CenterWallPaper{0.5}{#1}}
+\def\watermark#1{\RequirePackage{wallpaper}\CenterWallPaper{0.5}{#1}}
%%%%%%%%%%%%%%%%%%%%%%%
\RequirePackage{ulem}
\RequirePackage{fancyhdr}
\RequirePackage{fontawesome5}
-
\fancyhf{}
-
-\fancyfoot[C]{\color{1号色}\sout{\hbox to 0.4\textwidth{}}\quad\faIcon{chevron-left}\;第~\color{normalcolor}{\sffamily\thepage}\color{1号色}~页\;\faIcon{chevron-right}\quad\sout{\hbox to 0.4\textwidth{}}}
+\fancyfoot[C]{\color{1号色}\sout{\hbox to 0.4\textwidth{}}\quad\faIcon{chevron-left}\;第~{\sffamily\thepage}\color{1号色}~页\;\faIcon{chevron-right}\quad\sout{\hbox to 0.4\textwidth{}}}
\fancyfoot[L]{\color{1号色}{$\boldsymbol{\cdot}$}}
\fancyfoot[R]{\color{1号色}{$\boldsymbol{\cdot}$}}
+\renewcommand{\headrule}{\color{headrulecolor}\hrule width\textwidth}
-\renewcommand{\headrule}{\color{headrulecolor}\hrule width\textwidth}
\pagestyle{fancy}
-
\renewcommand{\sectionmark}[1]{\markright{#1}{}}
\renewcommand{\chaptermark}[1]{\markboth{#1}{}}
%%%%%%%%%%%%%%%%%%%%%%%%%
-%%%%%%%%%%%%%%%%[第三方定制]%%%%%%%%%%%%%%%%%%%%%
-\def\e{\mathrm{e}}
-\def\i{\mathrm{i}}
-\newcommand\mailto[1]{\href{mailto:#1}{\nolinkurl{#1}}}
-\newcommand\sokka[1]{故本题选择 \textbf{#1} 项.}
-\newcommand\ans[1]{\underline{\color{solutioncolor}\ #1\ }}
-\newcommand\length[4]{
-\node (a) at ($0.5*($#2+#1$)$) {#3};
-\draw [->|] ($0.5*($#2+#1+#4$)$) -- #2;
-\draw [->|] ($0.5*($#2+#1-#4$)$) -- #1;}
-\newcommand\cdotfill{\leavevmode\cleaders\hb at xt@.44em{\hss$\cdot$\hss}\hfill\kern\z@}
-\newcommand{\point}[1]{\cdotfill\texttt{(#1pt)}}
-\setstretch{1.5}
-\AtBeginDocument{\everymath{\displaystyle}}
-
-\makeatletter
-\protected\def\titlelogo#1#2{
- \leavevmode at ifvmode
- \lower\dimexpr\f at size\p@*1/10\hbox{\href{#1}{\includegraphics[height={\f at size\p@}]{#2}}}
- }
-\makeatother
-
\usepackage{listings,dirtree}
\lstdefinestyle{TeX}{
language = [LaTeX]TeX,
@@ -429,4 +386,115 @@
commentstyle = \color{darkgray},
tabsize = 2,
keywords = {chapter,choice,problem,solution,enumerate}
-}
\ No newline at end of file
+}
+
+% document setup
+\def\e{\mathrm{e}}
+\def\i{\mathrm{i}}
+\def\ee#1{\times 10^{#1}}
+\def\T{\mathsf{T}}
+\def\unit#1{\,\mathrm{#1}}
+\def\mat#1{\mathbf{#1}}
+\def\rank#1{\text{R}({#1})}
+\def\QED{\hfill\sffamily Q.E.D. }
+\def\mailto#1{\href{mailto:#1}{\nolinkurl{#1}}}
+\def\sokka#1{故本题选择 \textbf{#1} 项.}
+\def\length#1#2#3#4{
+\node (a) at ($0.5*($#2+#1$)$) {#3};
+\draw [->|] ($0.5*($#2+#1+#4$)$) -- #2;
+\draw [->|] ($0.5*($#2+#1-#4$)$) -- #1;
+}
+\def\point#1{\leavevmode\cleaders\hb at xt@.44em{\hss$\cdot$\hss}\hfill\kern\z@\texttt{(#1pt)}}
+\makeatletter
+\pdfstringdefDisableCommands{%
+ \def\pkg#1{<#1>}%
+ \def\mode#1{<#1>}%
+ \def\titlelogo#1#2{<#1#2>}%
+}
+\def\@pkg#1{\texorpdfstring{\href{https://www.ctan.org/pkg/#1}{\textcolor{pkgcolor}{\textsf{#1}}}}{“#1”}}
+\def\s at pkg#1{\texorpdfstring{\textcolor{pkgcolor}{\textsf{#1}}}{“#1”}}
+\DeclareRobustCommand\pkg{\@ifstar\s at pkg\@pkg}
+\def\mode#1{\texorpdfstring{\textcolor{moducolor}{\textsf{#1}}}{“#1”}}
+\def\cmd#1{\texorpdfstring{\textcolor{cmdcolor}{\textsf{#1}}}{“#1”}}
+\def\datechange#1#2{%
+ \noindent{\makebox[\textwidth][r]{\color{H7}\rule{1.15\textwidth}{.4pt}}}
+ \noindent\makebox[0pt][r]{\makebox[-3em][r]{\small\textbf{\textcolor{H7}{#1}}}\;\;}{\sffamily Update: \ignorespaces#2}}
+\protected\def\@titlelogo#1#2{
+ \leavevmode at ifvmode
+ \lower\dimexpr\f at size\p@*1/10\hbox{\href{#1}{\includegraphics[height={\f at size\p@}]{#2}}}
+ }
+\protected\def\s at titlelogo#1#2{
+ \leavevmode at ifvmode
+ \lower\dimexpr\f at size\p@*1/10\hbox{\includegraphics[height={\f at size\p@}]{#1}}
+ }
+\DeclareRobustCommand\titlelogo{\@ifstar\s at titlelogo\@titlelogo}
+
+\RequirePackage{qrcode}
+\qrset{link, height=4em}
+
+\renewcommand*{\date}[8]{
+ \vspace{-0.5em}
+ \begin{minipage}{0.64\textwidth}
+ \begin{multicols}{2}
+ \textbf{考试时间}：\kaishu #1
+
+ \textbf{\songti 课程编号}：\kaishu #3
+
+ \textbf{\songti 任课教师}：\kaishu #2
+
+ \textbf{\songti 解析制作}：\kaishu #4
+ \end{multicols}
+ \end{minipage}
+ \hfill
+ \begin{minipage}{0.15\textwidth}
+ \centering
+ \qrcode{#5}
+
+ \vskip1ex\scriptsize\textbf{\href{#5}{#6}}
+ \end{minipage}
+ \hfill
+ \begin{minipage}{0.15\textwidth}
+ \centering
+ \qrcode{#7}
+
+ \vskip1ex\scriptsize\textbf{\href{#7}{#8}}
+ \end{minipage}
+ \vspace{-0.5em}
+}
+
+\def\sim at x@scale{.15}
+\def\sim at y@scale{.05}
+\def\sim at y@thick{.02}
+\newsavebox\sim at upper
+\newsavebox\sim at lower
+% extensible sim symbol
+\NewDocumentCommand{\xSim}{ O{} m }{%
+ \TextOrMath{%
+ \PackageError{TEST}{`\string\xSim` is valid in math mode only.}{}%
+ }{
+ % math mode only, hence no need to eliminate spaces
+ \sbox\sim at upper{$\scriptstyle #2$}
+ \sbox\sim at lower{$\scriptstyle #1$}
+ \pgfmathparse{min(max(\wd\sim at upper/1em, \wd\sim at lower/1em, 1.0), 1.5)}
+ \edef\sim at ratio{\pgfmathresult}
+ \def\sim at x {\sim at x@scale * \sim at ratio}
+ \def\sim at y {\sim at y@scale * \sim at ratio}
+ \def\sim@@y{\sim at y@thick * \sim at ratio}
+ \pgfmathparse{floor(max(\wd\sim at upper/1em, \wd\sim at lower/1em)) + 1}
+ \edef\sim at wd{\pgfmathresult em}
+ \mathrel{
+ \begin{tikzpicture}[baseline=-.7ex]
+ \filldraw[line width=.2pt]
+ (0, 0)
+ .. controls +(\sim at x, \sim at y+\sim@@y) and +(-\sim at x, -\sim at y) ..
+ +(\sim at wd, 0)
+ node[midway, above] {\usebox\sim at upper}
+ node[midway, below] {\usebox\sim at lower}
+ .. controls +(-\sim at x, -\sim at y-\sim@@y) and +(\sim at x, \sim at y) ..
+ (0, 0);
+ \end{tikzpicture}
+ }
+ }%
+}
+\makeatother
+\hidesolution
\ No newline at end of file
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