如何用tcolorbox实现复杂样式

发布于 2021-03-22 09:51:15

请教各位大佬,如何用tcolorbox实现图中的样式?
微信图片_202103220946441.jpg
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微信图片_20210322094644.jpg
主要的问题是如何定位坐标?

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4 个回答
yuxtech.github.io
yuxtech.github.io 2021-03-24
\documentclass{ctexart}
\usepackage[many]{tcolorbox}
\newcounter{example}[section]

\tcbset{
  inbox/.style = {
    left = 0.5mm,
    right = 0.5mm,
    top = -0.5mm,
    bottom = -0.5mm,
    enhanced,
    boxrule = -1pt,
    colback = blue!80, 
    on line,
    overlay = {
      \draw [blue!80] (frame.west) -- ++ (-0.24cm,0);
    }
  },
  outbox/.style = {
    middle = 0mm,
    segmentation hidden,
    colback = white,
    arc = 0mm,
    enhanced,
    overlay = {
      \draw [blue!80] ([yshift=-1.358mm]frame.north west) -- (frame.south west)node[circle,draw,fill,inner sep=1.2pt,anchor=north]{};
    },
    boxrule = -0.1mm,
    enforce breakable,
    left = 1.5mm,
    top = -1.3mm,
    bottom = -1mm,
    right = 0mm
  }
}

\newcommand\example{
  \refstepcounter{example}
  \tcbox[inbox] { \color{white}例 \arabic{example} }\quad
  \label{exam\arabic{section}.\arabic{example}}
}

\newcommand\solution{
  \tcbox[inbox]{\color{white}解} \quad 
}

\usepackage[colorlinks]{hyperref}

\begin{document}
\section{第一节}

\begin{tcolorbox}[outbox]
\example 计算
  \[
    (a + b)(a - b) .
  \]
\tcblower
\solution 这是例 \ref{exam1.1} 的解答.
  \[
    (a + b)(a - b) = a^2 - b^2.
  \]
\end{tcolorbox}
\end{document} 

QQ截图20210324144441.png

LaTeXStudio
LaTeXStudio 2021-03-22

这是从一个站点挪来的代码,效果如下;
image.png

\documentclass{ctexart}

\usepackage{tcolorbox}
\tcbuselibrary{breakable,theorems,skins}
 
%\tcbuselibrary{breakable} %%を読み込んでいます。
\newcounter{reidaibangou} %%カウンタの定義
\newtcolorbox{reidai}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=44pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=blue!30!gray,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,blue!50!black,fill=blue!10!white]at ([xshift=22pt,yshift=-9pt]interior.north west) {\stepcounter{reidaibangou}\bfseries\sffamily 例題\thereidaibangou};},
    segmentation code={%
    \draw[dashed] (segmentation.west)--(segmentation.east);
    \node[inner sep=1pt,blue!50!black,fill=blue!10!white] at ([xshift=22pt,yshift=-8pt]segmentation.south west) {\bfseries\sffamily 解};},
    before upper={\setlength{\parindent}{1\ccwd}},
    before lower={\setlength{\parindent}{1\ccwd}},
}
%%%%%ここまでがreidai環境の定義。例えば本文中に以下のように記述してみよう。
\begin{document}
\begin{reidai}
次の問題に答えなさい。
\begin{enumerate}
    \item 8人を2つの組に分ける方法は何通りあるか。
    \item 6人を3つの部屋A,B,Cに入れる方法は何通りあるか。 {ただし}各部屋に少なくとも1人は入るものとする。
\end{enumerate}

\tcblower

区別があるかどうかを正しく考えます。
\begin{enumerate}
    \item なんだかんだで127通り
    \item なんだかんだで540通り
\end{enumerate}

\end{reidai}

\end{document}

地址:https://marukunalufd0123.hatenablog.com/entry/2019/03/15/071717

Banach spaces
Banach spaces 2021-03-22
像一个程序员那样去做排版,也许我真的是个程序员

这个可以修改一下王老师给的模板:

\documentclass{ctexart}

\usepackage{tcolorbox}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{mathrsfs}
\tcbuselibrary{breakable,theorems,skins}
\definecolor{Back}{RGB}{70,87,121}
\newcounter{reidaibangou} 
\newtcolorbox{reidai}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=2pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=white,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,white,fill=Back]at ([xshift=26pt,yshift=-8pt]interior.north west) {\stepcounter{reidaibangou}\sffamily 例~\thereidaibangou};
    \draw[Back,line width=1pt]([xshift=0pt,yshift=-8pt]interior.north west)--([xshift=-327pt,yshift=-8pt]interior.north east);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=2pt]interior.south west)--(interior.north west);},
    segmentation code={
   \draw[line width=1pt,Back] ([xshift=0pt,yshift=-14pt]segmentation.west)--([xshift=-325pt,yshift=-14pt]segmentation.east);
    \node[inner sep=1pt,white,fill=Back] at ([xshift=24pt,yshift=-8pt]segmentation.south west) {\sffamily 解};},
    before upper={\setlength{\parindent}{4\ccwd}},
    before lower={\setlength{\parindent}{4\ccwd}},
}

\newcounter{reidaibangouA} 
\newtcolorbox{reidaiA}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=2pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=white,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,white,fill=Back]at ([xshift=26pt,yshift=-8pt]interior.north west) {\stepcounter{reidaibangou}\sffamily 例~\thereidaibangou};
    \draw[Back,line width=1pt]([xshift=0pt,yshift=-8pt]interior.north west)--([xshift=-327pt,yshift=-8pt]interior.north east);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=2pt]interior.south west)--([xshift=0pt,yshift=-8pt]interior.north west);},
    segmentation code={
   \draw[line width=1pt,Back] ([xshift=0pt,yshift=-14pt]segmentation.west)--([xshift=-325pt,yshift=-14pt]segmentation.east);
    \node[inner sep=1pt,white,fill=Back] at ([xshift=24pt,yshift=-8pt]segmentation.south west) {\sffamily 解};},
    before upper={\setlength{\parindent}{4\ccwd}},
    before lower={\setlength{\parindent}{4\ccwd}},
}

\newcounter{reidaibangouB}
\newtcolorbox{reidaiB}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=2pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=white,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,white,fill=Back]at ([xshift=26pt,yshift=-8pt]interior.north west) {\stepcounter{reidaibangou}\sffamily 例~\thereidaibangou};
    \draw[Back,line width=1pt]([xshift=0pt,yshift=-8pt]interior.north west)--([xshift=-327pt,yshift=-8pt]interior.north east);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=0pt]interior.south west)--([xshift=0pt,yshift=-8pt]interior.north west);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=0pt]interior.south west)--([xshift=-327pt,yshift=0pt]interior.south east);},
    segmentation code={
   \draw[line width=1pt,Back] ([xshift=0pt,yshift=-14pt]segmentation.west)--([xshift=-325pt,yshift=-14pt]segmentation.east);
    \node[inner sep=1pt,white,fill=Back] at ([xshift=24pt,yshift=-8pt]segmentation.south west) {\sffamily 解};},
    before upper={\setlength{\parindent}{4\ccwd}},
    before lower={\setlength{\parindent}{4\ccwd}},
}


\begin{document}
\begin{reidai}
求 $\dfrac{(1+i)^3(\sqrt{3}-1\i)}{1+\sqrt{3}i}$
\tcblower 因为
\[
1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)
\]
\[
\sqrt{3}-i=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)
\]
\[
1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)
\]
所以 

$\begin{array}{l}
          \text{原式}=\frac{(\sqrt{2})^3\times 2}{2}\left[\cos\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)\right]  \\
\hspace{0.85cm}=2\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
\hspace{0.85cm}=2+2i
\end{array}$
\end{reidai}

\begin{reidaiA}
求 $\dfrac{(1+i)^3(\sqrt{3}-1\i)}{1+\sqrt{3}i}$
\tcblower 因为
\[
1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)
\]
\[
\sqrt{3}-i=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)
\]
\[
1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)
\]
所以

$\begin{array}{l}
          \text{原式}=\frac{(\sqrt{2})^3\times 2}{2}\left[\cos\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)\right]  \\
\hspace{0.85cm}=2\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
\hspace{0.85cm}=2+2i
\end{array}$
\end{reidaiA}


\begin{reidaiB}
求 $\dfrac{(1+i)^3(\sqrt{3}-1\i)}{1+\sqrt{3}i}$
\tcblower 因为
\[
1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)
\]
\[
\sqrt{3}-i=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)
\]
\[
1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)
\]
所以

$\begin{array}{l}
          \text{原式}=\frac{(\sqrt{2})^3\times 2}{2}\left[\cos\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)\right]  \\
\hspace{0.85cm}=2\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
\hspace{0.85cm}=2+2i
\end{array}$
\end{reidaiB}

\end{document} 

最后我们得到如下效果,当然也可以根据自己的要求随意修改。
fig1.jpg
fig2.jpg

稍作了修正:

\documentclass{ctexart}

\usepackage{tcolorbox}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{mathrsfs}
\tcbuselibrary{breakable,theorems,skins}
\definecolor{Back}{RGB}{70,87,121}
\newcounter{reidaibangou} 
\newtcolorbox{reidai}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=2pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=white,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,white,fill=Back]at ([xshift=26pt,yshift=-16pt]interior.north west) {\stepcounter{reidaibangou}\sffamily 例~\thereidaibangou};
    \draw[Back,line width=1pt]([xshift=0pt,yshift=-16pt]interior.north west)--([xshift=-327pt,yshift=-16pt]interior.north east);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=2pt]interior.south west)--(interior.north west);},
    segmentation code={
   \draw[line width=1pt,Back] ([xshift=0pt,yshift=-14pt]segmentation.west)--([xshift=-325pt,yshift=-14pt]segmentation.east);
    \node[inner sep=1pt,white,fill=Back] at ([xshift=24pt,yshift=-8pt]segmentation.south west) {\sffamily 解};},
    before upper={\setlength{\parindent}{3.5\ccwd}},
    before lower={\setlength{\parindent}{3.5\ccwd}},
}

\newcounter{reidaibangouA} 
\newtcolorbox{reidaiA}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=2pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=white,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,white,fill=Back]at ([xshift=26pt,yshift=-16pt]interior.north west) {\stepcounter{reidaibangou}\sffamily 例~\thereidaibangou};
    \draw[Back,line width=1pt]([xshift=0pt,yshift=-16pt]interior.north west)--([xshift=-327pt,yshift=-16pt]interior.north east);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=2pt]interior.south west)--([xshift=0pt,yshift=-16pt]interior.north west);},
    segmentation code={
   \draw[line width=1pt,Back] ([xshift=0pt,yshift=-14pt]segmentation.west)--([xshift=-325pt,yshift=-14pt]segmentation.east);
    \node[inner sep=1pt,white,fill=Back] at ([xshift=24pt,yshift=-8pt]segmentation.south west) {\sffamily 解};},
    before upper={\setlength{\parindent}{3.5\ccwd}},
    before lower={\setlength{\parindent}{3.5\ccwd}},
}

\newcounter{reidaibangouB}
\newtcolorbox{reidaiB}[1][]{enhanced,boxrule=0.5mm,
    top=2pt,left=2pt,right=4pt,bottom=2pt,arc=0mm,
    colframe=white,
    boxrule=1pt,
    underlay={
    \node[inner sep=1pt,white,fill=Back]at ([xshift=26pt,yshift=-16pt]interior.north west) {\stepcounter{reidaibangou}\sffamily 例~\thereidaibangou};
    \draw[Back,line width=1pt]([xshift=0pt,yshift=-16pt]interior.north west)--([xshift=-327pt,yshift=-16pt]interior.north east);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=0pt]interior.south west)--([xshift=0pt,yshift=-16pt]interior.north west);
    \draw[Back,line width=1pt]([xshift=0pt,yshift=0pt]interior.south west)--([xshift=-327pt,yshift=0pt]interior.south east);},
    segmentation code={
   \draw[line width=1pt,Back] ([xshift=0pt,yshift=-14pt]segmentation.west)--([xshift=-325pt,yshift=-14pt]segmentation.east);
    \node[inner sep=1pt,white,fill=Back] at ([xshift=24pt,yshift=-8pt]segmentation.south west) {\sffamily 解};},
    before upper={\setlength{\parindent}{3.5\ccwd}},
    before lower={\setlength{\parindent}{3.5\ccwd}},
}


\begin{document}
\begin{reidai}
求 $\dfrac{(1+i)^3(\sqrt{3}-1\i)}{1+\sqrt{3}i}$,文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,
\tcblower 因为文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,
\[
1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)
\]
\[
\sqrt{3}-i=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)
\]
\[
1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)
\]
所以 文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,文字测试内容文字测试内容 $1+\sqrt{3}i$,

$\begin{array}{l}
          \text{原式}=\frac{(\sqrt{2})^3\times 2}{2}\left[\cos\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)\right]  \\
\hspace{0.85cm}=2\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
\hspace{0.85cm}=2+2i
\end{array}$
\end{reidai}

\begin{reidaiA}
求 $\dfrac{(1+i)^3(\sqrt{3}-1\i)}{1+\sqrt{3}i}$
\tcblower 因为
\[
1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)
\]
\[
\sqrt{3}-i=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)
\]
\[
1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)
\]
所以

$\begin{array}{l}
          \text{原式}=\frac{(\sqrt{2})^3\times 2}{2}\left[\cos\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)\right]  \\
\hspace{0.85cm}=2\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
\hspace{0.85cm}=2+2i
\end{array}$
\end{reidaiA}


\begin{reidaiB}
求 $\dfrac{(1+i)^3(\sqrt{3}-1\i)}{1+\sqrt{3}i}$
\tcblower 因为
\[
1+i=\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4}\right)
\]
\[
\sqrt{3}-i=2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin \left(-\frac{\pi}{6}\right)\right)
\]
\[
1+\sqrt{3}i=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)
\]
所以

$\begin{array}{l}
          \text{原式}=\frac{(\sqrt{2})^3\times 2}{2}\left[\cos\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{4}\times 3-\frac{\pi}{6}-
\frac{\pi}{3}\right)\right]  \\
\hspace{0.85cm}=2\sqrt{2}(\cos \frac{\pi}{4}+i\sin \frac{\pi}{4})\\
\hspace{0.85cm}=2+2i
\end{array}$
\end{reidaiB}


\end{document} 

fig4.jpg
fig5.jpg

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