就像下面的命令
\newcommand\xx[2]{#1#2}
1
最佳答案
2
这是一个array
基于的解决方案。
\documentclass{article}
\usepackage{newtx} % optional (Times Roman text and math fonts)
\usepackage{array}
\newcommand\xx[2]{% feel free to choose a name other than 'xx'
\begingroup
\setlength\arraycolsep{1.5pt} % default: 5pt
\begin{array}{@{}r|l@{}}
\cline{2-2} #1 \\ \cline{1-1}
\end{array}
\endgroup}
\begin{document}
$b_1-\xx{1}{b_2}-\ldots-\xx{1}{b_s}$
\end{document}
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两个版本,一个带有tabularray
,第二个带有TiKz
:
\documentclass{article}
\usepackage{tabularray}
\usepackage{tikz}
\usetikzlibrary{calc,positioning}
\newcommand{\xx}[2]{\begin{tblr}{
colspec={r|l},
column{1} = {leftsep=0pt,rightsep=3pt},
column{2} = {leftsep=3pt,rightsep=0pt},
row{1} = {abovesep=1pt,belowsep=1pt},
}
\cline{2-2}
\ensuremath{#1} & \ensuremath{#2}\\
\cline{1-1}
\end{tblr}}
\newcommand{\xxx}[2]{\begin{tikzpicture}[baseline=(a.base),inner sep=2pt, outer sep=0pt, node distance=6pt]
\node (a) at (0,0) {\ensuremath{#1}};
\node [right=of a] (b) {\ensuremath{#2}};
\draw[very thin] ($(a.south west)-(2pt,0)$) -| ($(a)!.5!(b)$) |- ($(b.north east)+(2pt,0)$);
\end{tikzpicture}}
\begin{document}
\xx{a}{b}
\begin{equation}
b_1-\xx{1}{b_2}-\ldots-\xx{1}{b_s}
\end{equation}
\xxx{a}{b}
\begin{equation}
b_1-\xxx{1}{b_2}-\ldots-\xxx{1}{b_s}
\end{equation}
\end{document}
请随意调整参数;)
编辑
为了处理参数中的分数,我调整了 TiKz 版本:
\documentclass{article}
\usepackage{tabularray}
\usepackage{tikz}
\usetikzlibrary{calc,fit,positioning,tikzmark}
\newcommand{\xx}[2]{\begin{tblr}{
colspec={Q[r,m]|Q[l,m]},
column{1} = {leftsep=0pt,rightsep=3pt},
column{2} = {leftsep=3pt,rightsep=0pt},
row{1} = {abovesep=1pt,belowsep=1pt},
}
\cline{2-2}
\ensuremath{#1} & \ensuremath{#2}\\
\cline{1-1}
\end{tblr}}
\newcommand{\xxx}[2]{\begin{tikzpicture}[baseline=(O.base),inner sep=2pt, outer sep=0pt, node distance=6pt]
\coordinate (O) at (0,0);
\node [anchor=base] (a) at (O) {\ensuremath{#1}};
\node [right=of a, anchor=base] (b) {\ensuremath{#2}};
\draw[very thin] ($(a.south west)-(2pt,0)$) -| ($(a)!.5!(b)$) |- ($(b.north east)+(2pt,0)$);
\end{tikzpicture}}
\newcommand{\xxxx}[2]{\begingroup
\tikzmarknode{a}{#1}\,
\tikzmarknode{b}{#2}
\begin{tikzpicture}[remember picture, overlay,inner sep=2pt, outer sep=0pt, node distance=6pt]
\node (c) [fit=(a) (b)]{};
\draw[very thin] ($(c.south west)-(2pt,0)$) -| ($(a.east)!.5!(b.west)$) |- ($(c.north east)+(2pt,0)$);
\end{tikzpicture}
\endgroup}
\begin{document}
With \verb|tabularray|
\begin{equation}
b_1-\xx{1}{\frac{b_2}{\frac{b_1}{b_2}}}-\ldots-\xx{\frac{1}{2\frac{1}{2\frac{1}{2}}}}{b_s}
\end{equation}
With \verb|TiKz|
\begin{equation}
b_1-\xxx{\frac{1}{2}}{\frac{b_2}{\frac{b_1}{b_2}}}-\ldots-\xxx{1\frac{1}{2\frac{1}{2\frac{1}{2}}}}{b_s}
\end{equation}
Improving the \verb|TiKz| version
\begin{equation}
b_1-\xxxx{1}{\frac{b_2}{\frac{b_1}{b_2}}}-\ldots-\xxxx{1\frac{1}{2\frac{1}{2\frac{1}{2}}}}{b_s}
\end{equation}
\end{document}
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