答案1
如果您希望保留列间隙,这里有一种方法:
\documentclass{article}
\usepackage[TABcline]{tabstackengine}
\TABstackMath
\renewcommand\stackalignment{l}
\newcommand\lminus{\llap{$-$}}
{\catcode`-=\active
\gdef -{\lminus}}
\newcommand\minusactive{\phantom{-}\catcode`-=\active}
\setstacktabbedgap{3pt}
\begin{document}
$
\minusactive
\tabbedShortunderstack[r]{
&3&7&0&5&:5=741\\
-&3&5& &&\\
\TABcline{2,3}
& &2&0&&\\
&-&2&0&&\\
\TABcline{3,4}
& & & &5&\\
& & &-&5&\\
\TABcline{5}
& & & &0&
}
$
\end{document}
当然,该方法在内部起作用tabular
:
\documentclass{article}
\usepackage[TABcline]{tabstackengine}
\TABstackMath
\renewcommand\stackalignment{l}
\newcommand\lminus{\llap{$-$}}
{\catcode`-=\active
\gdef -{\lminus}}
\newcommand\minusactive{\phantom{-}\catcode`-=\active}
\setstacktabbedgap{3pt}
\begin{document}
\begin{tabular}{ll@{\hspace{10ex}}c}
1)&
$
\minusactive
\tabbedShortunderstack[r]{
&3&7&0&5&:5=741\\
-&3&5& &&\\
\TABcline{2,3}
& &2&0&&\\
&-&2&0&&\\
\TABcline{3,4}
& & & &5&\\
& & &-&5&\\
\TABcline{5}
& & & &0&
}
$
&
$741\cdot5 = 3705$\\
\end{tabular}
\end{document}