发布时间:2010-01-31 17:32:03 文章类别:小技巧 原文地址:http://blog.sina.com.cn/s/blog_5e16f1770100h3cm.html
QQ群:91940767/145316219/141877998/80300084/194770436
淘宝店:http://latexstudio.taobao.com
其实现点如下:
%This is a 2-page sample illustrating how to use the
%multienum package
\documentclass{article}
\setlength{\textwidth}{6in}
\setlength{\textheight}{8.5in}
\setlength{\topmargin}{-0.5in}
\setlength{\oddsidemargin}{0.25in}
\usepackage{multicol,multienum}
\begin{document}
\begin{center}
{\Large\bf Sample formating using {\tt multienumerate}}
\end{center}
\bigskip
Sometimes we want to typeset the solutions to exercises. This
is easy to do using the {\tt multienumerate} environment.
\subsection*{Answers to All Exercises}
\begin{multienumerate}
\mitemxxxx{Not}{Linear}{Not}{Quadratic}
\mitemxxxo{Not}{Linear}{No; if $x=3$, then $y=-2$.}
\mitemxx{$(x_1,x_2)=(2 \frac{1}{3}t,t)$ or
$(s,3s-6)$}{$(x_1,x_2,x_3)=(2 \frac{5}{2}s-3t,s,t)$}
\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4} \frac{5}{4}s
\frac{3}{4}t-u,s,t,u)$
or $(s,t,u,\frac{1}{4}-s \frac{5}{4}t \frac{3}{4}u)$}
\mitemxxxx{$(2,-1,3)$}{None}{$(2,1,0,1)$}{$(0,0,0,0)$}
\end{multienumerate}
\bigskip
\hrule
\bigskip
We can also enumerate the items using an even-only or odd
only
counter.
\subsection*{Answers to Even-Numbered Exercises}
\begin{multienumerate}[evenlist]
\mitemxxxx{Not}{Linear}{Not}{Quadratic}
\mitemxxxo{Not}{Linear}{No; if $x=3$, then $y=-2$.}
\mitemxx{$(x_1,x_2)=(2 \frac{1}{3}t,t)$ or
$(s,3s-6)$}{$(x_1,x_2,x_3)=(2 \frac{5}{2}s-3t,s,t)$}
\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4} \frac{5}{4}s
\frac{3}{4}t-u,s,t,u)$
or $(s,t,u,\frac{1}{4}-s \frac{5}{4}t \frac{3}{4}u)$}
\mitemxxxx{$(2,-1,3)$}{None}{$(2,1,0,1)$}{$(0,0,0,0)$}
\end{multienumerate}
\hrule
\subsection*{Answers to Odd-Numbered Exercises}
\begin{multienumerate}[oddlist]
\mitemxxxx{Not}{Linear}{Not}{Quadratic}
\mitemxxxo{Not}{Linear}{No; if $x=3$, then $y=-2$.}
\mitemxx{$(x_1,x_2)=(2 \frac{1}{3}t,t)$ or
$(s,3s-6)$}{$(x_1,x_2,x_3)=(2 \frac{5}{2}s-3t,s,t)$}
\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4} \frac{5}{4}s
\frac{3}{4}t-u,s,t,u)$
or $(s,t,u,\frac{1}{4}-s \frac{5}{4}t \frac{3}{4}u)$}
\mitemxxxx{$(2,-1,3)$}{None}{$(2,1,0,1)$}{$(0,0,0,0)$}
\end{multienumerate}
\bigskip
\hrule
\bigskip
Sometimes we want to create sublists which are
enumerated using an alpha counter.
\begin{multienumerate}
\mitemx{Which of the following numbers is the solution of the
equation
$x 3=7$:}
\begin{multienumerate}
\mitemxxxx{1}{2}{3}{4}
\end{multienumerate}
\mitemx{The value of $\log_28$ is:}
\begin{multienumerate}
\mitemxxxx{1}{$-1$}{3}{$-3$}
\end{multienumerate}
\end{multienumerate}
\pagebreak
\begin{multicols}{2}
\subsection*{Answers to All Exercises}
\begin{multienumerate}
\mitemxx{Not}{Linear}
\mitemxx{Not}{Quadratic}
\mitemxx{Not}{Linear}
\mitemx{$(x_1,x_2)=(2 \frac{1}{3}t,t)$ or
$(s,3s-6)$}
\mitemx{$(x_1,x_2,x_3)=(2 \frac{5}{2}s-3t,s,t)$}
\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4} \frac{5}{4}s
\frac{3}{4}t-u,s,t,u)$
or $(s,t,u,\frac{1}{4}-s \frac{5}{4}t \frac{3}{4}u)$}
\mitemxx{$(2,-1,3)$}{None}
\mitemxx{$(2,1,0,1)$}{$(0,0,0,0)$}
\mitemxx{Not}{Linear}
\mitemxx{Not}{Quadratic}
\mitemxx{Not}{Linear}
\mitemx{$(x_1,x_2)=(2 \frac{1}{3}t,t)$ or
$(s,3s-6)$}
\mitemx{$(x_1,x_2,x_3)=(2 \frac{5}{2}s-3t,s,t)$}
\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4} \frac{5}{4}s
\frac{3}{4}t-u,s,t,u)$
or $(s,t,u,\frac{1}{4}-s \frac{5}{4}t \frac{3}{4}u)$}
\mitemxx{$(2,-1,3)$}{None}
\mitemxx{$(2,1,0,1)$}{$(0,0,0,0)$}
\mitemxx{Not}{Linear}
\mitemxx{Not}{Quadratic}
\mitemxx{Not}{Linear}
\mitemx{$(x_1,x_2)=(2 \frac{1}{3}t,t)$ or
$(s,3s-6)$}
\mitemx{$(x_1,x_2,x_3)=(2 \frac{5}{2}s-3t,s,t)$}
\mitemx{$(x_1,x_2,x_3,x_4)= (\frac{1}{4} \frac{5}{4}s
\frac{3}{4}t-u,s,t,u)$
or $(s,t,u,\frac{1}{4}-s \frac{5}{4}t \frac{3}{4}u)$}
\mitemxx{$(2,-1,3)$}{None}
\mitemxx{$(2,1,0,1)$}{$(0,0,0,0)$}
\end{multienumerate}
\subsection*{Multiple Choice}
\begin{multienumerate}
\mitemx{Which of the following numbers is the solution of the
equation
$x 3=7$:}
\begin{multienumerate}
\mitemxxxx{1}{2}{3}{4}
\end{multienumerate}
\mitemx{The value of $\log_28$ is:}
\begin{multienumerate}
\mitemxxxx{1}{$-1$}{3}{$-3$}
\end{multienumerate}
\mitemx{Which of the following numbers is the solution of the
equation
$x 3=7$:}
\begin{multienumerate}
\mitemxxxx{1}{2}{3}{4}
\end{multienumerate}
\mitemx{The value of $\log_28$ is:}
\begin{multienumerate}
\mitemxxxx{1}{$-1$}{3}{$-3$}
\end{multienumerate}
\mitemx{Which of the following numbers is the solution of the
equation
$x 3=7$:}
\begin{multienumerate}
\mitemxxxx{1}{2}{3}{4}
\end{multienumerate}
\mitemx{The value of $\log_28$ is:}
\begin{multienumerate}
\mitemxxxx{1}{$-1$}{3}{$-3$}
\end{multienumerate}
\end{multienumerate}
\end{multicols}
\end{document}


发表评论 取消回复