重复除法 - 从十进制转换为其他进制

Question 1

编辑：这个答案有 3 个版本，第一个版本基于我对上述方法的误解，第二个版本（我希望）是更正确的解释。第三个版本允许最多 36 个基数的转换（但仅限于引擎的数字范围pgfmath，即 0-16383）。

第一个：

\documentclass{standalone}
\usepackage{tikz}
\begin{document}

\newcount\total
\newcount\lasttotal
\newcount\targetbase

\def\basetenconversiontable#1#2{%
    \begin{tikzpicture}[every node/.style={minimum width=1cm, minimum height=0.5cm}, x=1cm,y=0.5cm]
    %
    \total=#1%
    \targetbase=#2
    \def\newnumber{}
    %
    \pgfmathloop
    \ifnum\total<1
    \else
        %
        \ifnum\pgfmathcounter>1
            \node at (\pgfmathcounter, -\pgfmathcounter+1) (tmp) {\the\targetbase};
            \draw (tmp.north west) |- (tmp.south east);
            %
            \node at (\pgfmathcounter-1, -\pgfmathcounter) (tmp) {\pgfmathparse{int(\total*\targetbase)}\pgfmathresult};
            \draw (tmp.south west) -- (tmp.south east);
            %
            \pgfmathparse{int(\lasttotal-\total*\targetbase)}%
            \let\digit=\pgfmathresult
            \node at (\pgfmathcounter-1, -\pgfmathcounter-1) [text=red] {\digit};
            \edef\newnumber{\digit\newnumber}
        \fi
        %
        \ifnum\total<\targetbase
                \edef\newnumber{\the\total\newnumber}
            \node at (\pgfmathcounter, -\pgfmathcounter) [text=red]  {\the\total};
        \else
            \node at (\pgfmathcounter, -\pgfmathcounter) {\the\total};
        \fi
        \lasttotal=\total
        \divide\total by\targetbase
    \repeatpgfmathloop    
    \draw [->] (\pgfmathcounter-1,-\pgfmathcounter-1) -- ++(-0.5,0); 
    \node [anchor=west] at (1, -\pgfmathcounter-2) {$#1=\newnumber_{\the\targetbase}$};
    \end{tikzpicture}   
}

\begin{tabular}{c}
\basetenconversiontable{478}{7} \\ 
\basetenconversiontable{1362}{5} \\
\basetenconversiontable{365}{3} \\
\basetenconversiontable{637}{2}
\end{tabular}

\end{document}

在此处输入图片描述

第二个：

\documentclass{standalone}
\usepackage{tikz}
\begin{document}

\newcount\columntotal
\newcount\nextcolumntotal
\newcount\tmptotal
\newcount\tmptmptotal

\newcount\targetbase
\newcount\digitcount


\def\getonedigit#1#2;{#1}
\def\getndigits#1{%
    \begingroup%
        \tmptotal=#1
        \pgfmathloop%
        \ifnum\tmptotal<10%
        \else%
            \divide\tmptotal by10%
        \repeatpgfmathloop%
        \pgfmathsmuggle\pgfmathcounter%
    \endgroup%
    \edef\ndigits{\pgfmathcounter}%
}

\def\baseconversiontable#1#2{%
    \begingroup%
    \getndigits{#1}%
    \pgfmathsetlengthmacro\tablecolumnwidth{\ndigits*width("$4$")}%
    \def\convertednumber{}%
    \begin{tikzpicture}
    \node (base conversion table) {%
        \begin{tikzpicture}
        [
            table node/.style={
                text width=\tablecolumnwidth, 
                inner sep=0pt,
                align=right, 
                minimum height=0.5cm, 
                minimum width=\tablecolumnwidth+5pt
            }, 
            x=\tablecolumnwidth+5pt,
            y=0.5cm
        ]
            \targetbase=#2
            \dobaseconversiontable{#1}%
        \end{tikzpicture}%
    };%
    \draw [->](base conversion table.south east) -- ++(-0.5,0);
    \node [below, anchor=north west] at (base conversion table.south west) {$#1=\convertednumber_{#2}$};    
    \end{tikzpicture}
    \endgroup%
}


\def\dobaseconversiontable#1{%
    \columntotal=#1%
    \getndigits{\columntotal}%
    \tmptotal=#1%
    \divide\tmptotal by\targetbase
    \nextcolumntotal=\tmptotal
    %
    \ifnum\columntotal<\targetbase
        \node at (0, 0) [table node, red] (@) {$\the\columntotal$};%
        \xdef\convertednumber{\the\columntotal\convertednumber}%
    \else%
        \node at (1, 0) [table node] (@) {$\the\targetbase$};%
        \draw (@.north west) |- (@.south east);
        %
        \getndigits{\tmptotal}%
        \digitcount=\ndigits%
        %   
        \node at (0,0) [table node] {$\the\columntotal$};%
        \tmptmptotal=\tmptotal
        \pgfmathloop
        \ifnum\digitcount=0
        \else%
            \tmptmptotal=\tmptotal
            \ifnum\digitcount>0
                \tmptmptotal=\expandafter\getonedigit\the\tmptmptotal;
            \fi
            \pgfmathparse{int(\tmptmptotal*\targetbase*10^(\digitcount-1))}%
            \let\remainder=\pgfmathresult
            \pgfmathsetcount\columntotal{\columntotal-\remainder}%
            \advance\digitcount by-1
            \pgfmathsetcount\tmptotal{\tmptotal-\tmptmptotal*(10^\digitcount)}%
            \ifnum\columntotal<\targetbase
                \node at (0,-\pgfmathcounter*2) [table node, red] (@) {$\the\columntotal$};%
                \xdef\convertednumber{\the\columntotal\convertednumber}%
                \digitcount=0%
            \else
                \node at (0,-\pgfmathcounter*2) [table node](@) {$\the\columntotal$};%
            \fi
            \draw (@.north west) -- (@.north east);
            \node at (0,-\pgfmathcounter*2+1) [table node] {$\remainder$};%
        \repeatpgfmathloop%
        \tikzset{shift={(1,-1)}}%
        \expandafter\dobaseconversiontable\expandafter{\the\nextcolumntotal}%
    \fi
}%

\begin{tabular}{c}
\baseconversiontable{478}{7}\\
\baseconversiontable{1366}{5}\\
\baseconversiontable{365}{3}\\
\baseconversiontable{711}{2}    
\end{tabular}

\end{document}

在此处输入图片描述

第三个版本。思考它工作正常。无论如何，它都有点笨拙。

\documentclass{standalone}
\usepackage{tikz}
\begin{document}

\newcount\columntotal
\newcount\nextcolumntotal
\newcount\tmptotal
\newcount\tmptmptotal

\newcount\targetbase
\newcount\digitcount

\def\digittoalpha#1{%
    \ifcase#1\relax0\or1\or2\or3\or4\or5\or6\or7\or8\or9%
    \or a\or b\or c\or d\or e\or f\or g\or h\or i\or j\or k\or l\or m%
    \or n\or p\or p\or q\or r\or s\or t\or u\or v\or w\or x\or y\or z\else?\fi%
}

\def\getonedigit#1#2;{#1}
\def\getndigits#1{%
    \begingroup%
        \tmptotal=#1
        \pgfmathloop%
        \ifnum\tmptotal<10%
        \else%
            \divide\tmptotal by10%
        \repeatpgfmathloop%
        \pgfmathsmuggle\pgfmathcounter%
    \endgroup%
    \edef\ndigits{\pgfmathcounter}%
}

\def\baseconversiontable#1#2{%
    \begingroup%
    \getndigits{#1}%
    \pgfmathsetlengthmacro\tablecolumnwidth{\ndigits*width("$4$")}%
    \gdef\convertednumber{}%
    \begin{tikzpicture}
    \node (base conversion table) {%
        \begin{tikzpicture}
        [
            table node/.style={
                anchor=north,
                text width=\tablecolumnwidth, 
                inner sep=0pt,
                align=right, 
                minimum height=0.5cm, 
                minimum width=\tablecolumnwidth+5pt
            }, 
            x=\tablecolumnwidth+5pt,
            y=0.5cm
        ]
            \targetbase=#2
            \dobaseconversiontable{#1}%
        \end{tikzpicture}%
    };%
    \draw [->](base conversion table.south east) -- ++(-0.5,0);
    \node [below, anchor=north west] at (base conversion table.south west) {$#1=\convertednumber_{#2}$};    
    \end{tikzpicture}
    \endgroup%
}


\def\dobaseconversiontable#1{%
    \columntotal=#1%
    \getndigits{\columntotal}%
    \tmptotal=#1%
    \divide\tmptotal by\targetbase
    \nextcolumntotal=\tmptotal
    %
    \ifnum\columntotal<\targetbase
        \edef\currentdigit{\uppercase{\digittoalpha{\the\columntotal}}}%
        \ifnum\columntotal>9
            \edef\currentdigit{\noexpand\rm{\currentdigit}}%
        \fi
        \node at (0, 0) [table node] (@) {%
            \\% For some reason necessary.
            \ifnum\the\columntotal<10
                \color{red}$\the\columntotal$%
            \else
                $\the\columntotal$\\%
                \tikz\draw[->](0,0)(-0.75ex,0)--++(0,-0.5);\\%
                \color{red}\currentdigit%
            \fi
            };% 
        \expandafter\expandafter\expandafter\gdef\expandafter\expandafter\expandafter\convertednumber%
        \expandafter\expandafter\expandafter{\expandafter\currentdigit\convertednumber}%
    \else%
        \node at (1, 0) [table node] (@) {$\the\targetbase$};%
        \draw (@.north west) |- (@.south east);
        %
        \getndigits{\tmptotal}%
        \digitcount=\ndigits%
        %   
        \node at (0,0) [table node] {$\the\columntotal$};%
        \tmptmptotal=\tmptotal
        \pgfmathloop
        \ifnum\digitcount=0
        \else%
            \tmptmptotal=\tmptotal
            \ifnum\digitcount>0
                \tmptmptotal=\expandafter\getonedigit\the\tmptmptotal;
            \fi
            \pgfmathparse{int(\tmptmptotal*\targetbase*10^(\digitcount-1))}%
            % Hmm this is a kludge.
            \ifnum\pgfmathresult>\columntotal%
                \pgfmathparse{int(\pgfmathresult/10)}%
            \fi%
            \let\remainder=\pgfmathresult%
            \pgfmathsetcount\columntotal{\columntotal-\remainder}%
            \advance\digitcount by-1
            \pgfmathsetcount\tmptotal{\tmptotal-\tmptmptotal*(10^\digitcount)}%
            \ifnum\columntotal<\targetbase
                \edef\currentdigit{\uppercase{\digittoalpha{\the\columntotal}}}%
                \ifnum\columntotal>9
                    \edef\currentdigit{\noexpand\rm{\currentdigit}}%
                \fi
                \node at (0,-\pgfmathcounter*2) [table node] (@) {%
                    \\
                    \ifnum\columntotal<10
                        \color{red}$\the\columntotal$%
                    \else
                        $\the\columntotal$\\%
                        \tikz\draw[->](0,0)(-0.75ex,0)--++(0,-0.5);\\%
                        \color{red}\currentdigit%
                    \fi
                    };%                     
                \expandafter\expandafter\expandafter\gdef\expandafter\expandafter\expandafter\convertednumber%
                    \expandafter\expandafter\expandafter{\expandafter\currentdigit\convertednumber}%
                \digitcount=0%
            \else
                \node at (0,-\pgfmathcounter*2) [table node](@) {$\the\columntotal$};%
            \fi
            \draw (@.north west) -- (@.north east);
            \node at (0,-\pgfmathcounter*2+1) [table node] {$\remainder$};%
        \repeatpgfmathloop%
        \tikzset{shift={(1,-1)}}%
        \expandafter\dobaseconversiontable\expandafter{\the\nextcolumntotal}%
    \fi
}%

\begin{tabular}{ccc}
\baseconversiontable{3022}{16}&&
\baseconversiontable{1462}{12}\\
\baseconversiontable{5407}{19}&&
\baseconversiontable{3887}{36}  
\end{tabular}

\end{document}

在此处输入图片描述

Answer

编辑：这个答案有 3 个版本，第一个版本基于我对上述方法的误解，第二个版本（我希望）是更正确的解释。第三个版本允许最多 36 个基数的转换（但仅限于引擎的数字范围pgfmath，即 0-16383）。

第一个：

\documentclass{standalone}
\usepackage{tikz}
\begin{document}

\newcount\total
\newcount\lasttotal
\newcount\targetbase

\def\basetenconversiontable#1#2{%
    \begin{tikzpicture}[every node/.style={minimum width=1cm, minimum height=0.5cm}, x=1cm,y=0.5cm]
    %
    \total=#1%
    \targetbase=#2
    \def\newnumber{}
    %
    \pgfmathloop
    \ifnum\total<1
    \else
        %
        \ifnum\pgfmathcounter>1
            \node at (\pgfmathcounter, -\pgfmathcounter+1) (tmp) {\the\targetbase};
            \draw (tmp.north west) |- (tmp.south east);
            %
            \node at (\pgfmathcounter-1, -\pgfmathcounter) (tmp) {\pgfmathparse{int(\total*\targetbase)}\pgfmathresult};
            \draw (tmp.south west) -- (tmp.south east);
            %
            \pgfmathparse{int(\lasttotal-\total*\targetbase)}%
            \let\digit=\pgfmathresult
            \node at (\pgfmathcounter-1, -\pgfmathcounter-1) [text=red] {\digit};
            \edef\newnumber{\digit\newnumber}
        \fi
        %
        \ifnum\total<\targetbase
                \edef\newnumber{\the\total\newnumber}
            \node at (\pgfmathcounter, -\pgfmathcounter) [text=red]  {\the\total};
        \else
            \node at (\pgfmathcounter, -\pgfmathcounter) {\the\total};
        \fi
        \lasttotal=\total
        \divide\total by\targetbase
    \repeatpgfmathloop    
    \draw [->] (\pgfmathcounter-1,-\pgfmathcounter-1) -- ++(-0.5,0); 
    \node [anchor=west] at (1, -\pgfmathcounter-2) {$#1=\newnumber_{\the\targetbase}$};
    \end{tikzpicture}   
}

\begin{tabular}{c}
\basetenconversiontable{478}{7} \\ 
\basetenconversiontable{1362}{5} \\
\basetenconversiontable{365}{3} \\
\basetenconversiontable{637}{2}
\end{tabular}

\end{document}

在此处输入图片描述

第二个：

\documentclass{standalone}
\usepackage{tikz}
\begin{document}

\newcount\columntotal
\newcount\nextcolumntotal
\newcount\tmptotal
\newcount\tmptmptotal

\newcount\targetbase
\newcount\digitcount


\def\getonedigit#1#2;{#1}
\def\getndigits#1{%
    \begingroup%
        \tmptotal=#1
        \pgfmathloop%
        \ifnum\tmptotal<10%
        \else%
            \divide\tmptotal by10%
        \repeatpgfmathloop%
        \pgfmathsmuggle\pgfmathcounter%
    \endgroup%
    \edef\ndigits{\pgfmathcounter}%
}

\def\baseconversiontable#1#2{%
    \begingroup%
    \getndigits{#1}%
    \pgfmathsetlengthmacro\tablecolumnwidth{\ndigits*width("$4$")}%
    \def\convertednumber{}%
    \begin{tikzpicture}
    \node (base conversion table) {%
        \begin{tikzpicture}
        [
            table node/.style={
                text width=\tablecolumnwidth, 
                inner sep=0pt,
                align=right, 
                minimum height=0.5cm, 
                minimum width=\tablecolumnwidth+5pt
            }, 
            x=\tablecolumnwidth+5pt,
            y=0.5cm
        ]
            \targetbase=#2
            \dobaseconversiontable{#1}%
        \end{tikzpicture}%
    };%
    \draw [->](base conversion table.south east) -- ++(-0.5,0);
    \node [below, anchor=north west] at (base conversion table.south west) {$#1=\convertednumber_{#2}$};    
    \end{tikzpicture}
    \endgroup%
}


\def\dobaseconversiontable#1{%
    \columntotal=#1%
    \getndigits{\columntotal}%
    \tmptotal=#1%
    \divide\tmptotal by\targetbase
    \nextcolumntotal=\tmptotal
    %
    \ifnum\columntotal<\targetbase
        \node at (0, 0) [table node, red] (@) {$\the\columntotal$};%
        \xdef\convertednumber{\the\columntotal\convertednumber}%
    \else%
        \node at (1, 0) [table node] (@) {$\the\targetbase$};%
        \draw (@.north west) |- (@.south east);
        %
        \getndigits{\tmptotal}%
        \digitcount=\ndigits%
        %   
        \node at (0,0) [table node] {$\the\columntotal$};%
        \tmptmptotal=\tmptotal
        \pgfmathloop
        \ifnum\digitcount=0
        \else%
            \tmptmptotal=\tmptotal
            \ifnum\digitcount>0
                \tmptmptotal=\expandafter\getonedigit\the\tmptmptotal;
            \fi
            \pgfmathparse{int(\tmptmptotal*\targetbase*10^(\digitcount-1))}%
            \let\remainder=\pgfmathresult
            \pgfmathsetcount\columntotal{\columntotal-\remainder}%
            \advance\digitcount by-1
            \pgfmathsetcount\tmptotal{\tmptotal-\tmptmptotal*(10^\digitcount)}%
            \ifnum\columntotal<\targetbase
                \node at (0,-\pgfmathcounter*2) [table node, red] (@) {$\the\columntotal$};%
                \xdef\convertednumber{\the\columntotal\convertednumber}%
                \digitcount=0%
            \else
                \node at (0,-\pgfmathcounter*2) [table node](@) {$\the\columntotal$};%
            \fi
            \draw (@.north west) -- (@.north east);
            \node at (0,-\pgfmathcounter*2+1) [table node] {$\remainder$};%
        \repeatpgfmathloop%
        \tikzset{shift={(1,-1)}}%
        \expandafter\dobaseconversiontable\expandafter{\the\nextcolumntotal}%
    \fi
}%

\begin{tabular}{c}
\baseconversiontable{478}{7}\\
\baseconversiontable{1366}{5}\\
\baseconversiontable{365}{3}\\
\baseconversiontable{711}{2}    
\end{tabular}

\end{document}

在此处输入图片描述

第三个版本。思考它工作正常。无论如何，它都有点笨拙。

\documentclass{standalone}
\usepackage{tikz}
\begin{document}

\newcount\columntotal
\newcount\nextcolumntotal
\newcount\tmptotal
\newcount\tmptmptotal

\newcount\targetbase
\newcount\digitcount

\def\digittoalpha#1{%
    \ifcase#1\relax0\or1\or2\or3\or4\or5\or6\or7\or8\or9%
    \or a\or b\or c\or d\or e\or f\or g\or h\or i\or j\or k\or l\or m%
    \or n\or p\or p\or q\or r\or s\or t\or u\or v\or w\or x\or y\or z\else?\fi%
}

\def\getonedigit#1#2;{#1}
\def\getndigits#1{%
    \begingroup%
        \tmptotal=#1
        \pgfmathloop%
        \ifnum\tmptotal<10%
        \else%
            \divide\tmptotal by10%
        \repeatpgfmathloop%
        \pgfmathsmuggle\pgfmathcounter%
    \endgroup%
    \edef\ndigits{\pgfmathcounter}%
}

\def\baseconversiontable#1#2{%
    \begingroup%
    \getndigits{#1}%
    \pgfmathsetlengthmacro\tablecolumnwidth{\ndigits*width("$4$")}%
    \gdef\convertednumber{}%
    \begin{tikzpicture}
    \node (base conversion table) {%
        \begin{tikzpicture}
        [
            table node/.style={
                anchor=north,
                text width=\tablecolumnwidth, 
                inner sep=0pt,
                align=right, 
                minimum height=0.5cm, 
                minimum width=\tablecolumnwidth+5pt
            }, 
            x=\tablecolumnwidth+5pt,
            y=0.5cm
        ]
            \targetbase=#2
            \dobaseconversiontable{#1}%
        \end{tikzpicture}%
    };%
    \draw [->](base conversion table.south east) -- ++(-0.5,0);
    \node [below, anchor=north west] at (base conversion table.south west) {$#1=\convertednumber_{#2}$};    
    \end{tikzpicture}
    \endgroup%
}


\def\dobaseconversiontable#1{%
    \columntotal=#1%
    \getndigits{\columntotal}%
    \tmptotal=#1%
    \divide\tmptotal by\targetbase
    \nextcolumntotal=\tmptotal
    %
    \ifnum\columntotal<\targetbase
        \edef\currentdigit{\uppercase{\digittoalpha{\the\columntotal}}}%
        \ifnum\columntotal>9
            \edef\currentdigit{\noexpand\rm{\currentdigit}}%
        \fi
        \node at (0, 0) [table node] (@) {%
            \\% For some reason necessary.
            \ifnum\the\columntotal<10
                \color{red}$\the\columntotal$%
            \else
                $\the\columntotal$\\%
                \tikz\draw[->](0,0)(-0.75ex,0)--++(0,-0.5);\\%
                \color{red}\currentdigit%
            \fi
            };% 
        \expandafter\expandafter\expandafter\gdef\expandafter\expandafter\expandafter\convertednumber%
        \expandafter\expandafter\expandafter{\expandafter\currentdigit\convertednumber}%
    \else%
        \node at (1, 0) [table node] (@) {$\the\targetbase$};%
        \draw (@.north west) |- (@.south east);
        %
        \getndigits{\tmptotal}%
        \digitcount=\ndigits%
        %   
        \node at (0,0) [table node] {$\the\columntotal$};%
        \tmptmptotal=\tmptotal
        \pgfmathloop
        \ifnum\digitcount=0
        \else%
            \tmptmptotal=\tmptotal
            \ifnum\digitcount>0
                \tmptmptotal=\expandafter\getonedigit\the\tmptmptotal;
            \fi
            \pgfmathparse{int(\tmptmptotal*\targetbase*10^(\digitcount-1))}%
            % Hmm this is a kludge.
            \ifnum\pgfmathresult>\columntotal%
                \pgfmathparse{int(\pgfmathresult/10)}%
            \fi%
            \let\remainder=\pgfmathresult%
            \pgfmathsetcount\columntotal{\columntotal-\remainder}%
            \advance\digitcount by-1
            \pgfmathsetcount\tmptotal{\tmptotal-\tmptmptotal*(10^\digitcount)}%
            \ifnum\columntotal<\targetbase
                \edef\currentdigit{\uppercase{\digittoalpha{\the\columntotal}}}%
                \ifnum\columntotal>9
                    \edef\currentdigit{\noexpand\rm{\currentdigit}}%
                \fi
                \node at (0,-\pgfmathcounter*2) [table node] (@) {%
                    \\
                    \ifnum\columntotal<10
                        \color{red}$\the\columntotal$%
                    \else
                        $\the\columntotal$\\%
                        \tikz\draw[->](0,0)(-0.75ex,0)--++(0,-0.5);\\%
                        \color{red}\currentdigit%
                    \fi
                    };%                     
                \expandafter\expandafter\expandafter\gdef\expandafter\expandafter\expandafter\convertednumber%
                    \expandafter\expandafter\expandafter{\expandafter\currentdigit\convertednumber}%
                \digitcount=0%
            \else
                \node at (0,-\pgfmathcounter*2) [table node](@) {$\the\columntotal$};%
            \fi
            \draw (@.north west) -- (@.north east);
            \node at (0,-\pgfmathcounter*2+1) [table node] {$\remainder$};%
        \repeatpgfmathloop%
        \tikzset{shift={(1,-1)}}%
        \expandafter\dobaseconversiontable\expandafter{\the\nextcolumntotal}%
    \fi
}%

\begin{tabular}{ccc}
\baseconversiontable{3022}{16}&&
\baseconversiontable{1462}{12}\\
\baseconversiontable{5407}{19}&&
\baseconversiontable{3887}{36}  
\end{tabular}

\end{document}

在此处输入图片描述

Question 2

即使没有 TikZ，使用标准环境也可以做得很好array：

\documentclass[10pt,letterpaper]{amsart}
\usepackage{xcolor}

\newcommand\myrule[1]{\multicolumn{1}{c|}{#1}}
\newcommand\myred[1]{\textcolor{red}{#1}}

\begin{document}

\begin{gather*}
\begin{array}{*{7}c}
4 & 7 & \myrule{8} & 7 \\
\cline{4-5}
4 & 2 & & 6 & \myrule{8} & 7 \\
\cline{6-6}
& 5 & 8 & 6 & 3 & \myrule{9} & 7 \\
\cline{4-5}\cline{7-7}
& 5 & 6 & & \myred{5} & 7 & \myred{1} \\
\cline{2-3}\cline{6-6}
& & \myred{2} & & & \myred{2}
\end{array} \\
\longleftarrow \\
478 = 1252_{7}
\end{gather*}

\end{document}

在此处输入图片描述

后 奎伯比尔贝尔的建议来改变数字之间的间距，可以得到：

\documentclass[10pt,letterpaper]{amsart}
\usepackage{xcolor}

\newcommand\myrule[1]{\multicolumn{1}{@{}c|}{#1}}
\newcommand\myred[1]{\textcolor{red}{#1}}

\begin{document}

\begin{gather*}
\begin{array}{c@{\,}c@{\,}cc@{\,}ccc}
4 & 7 & \myrule{8} & 7 \\
\cline{4-5}
4 & 2 & & 6 & \myrule{8} & 7 \\
\cline{6-6}
& 5 & 8 & 6 & 3 & \multicolumn{1}{c|}{9} & 7 \\
\cline{4-5}\cline{7-7}
& 5 & 6 & & \myred{5} & 7 & \myred{1} \\
\cline{2-3}\cline{6-6}
& & \myred{2} & & & \myred{2}
\end{array} \\
\longleftarrow \\
478 = 1252_{7}
\end{gather*}

\end{document}

在此处输入图片描述

TikZ 解决方案：

\documentclass[10pt,letterpaper]{amsart}
\usepackage{tikz}
\usetikzlibrary{matrix}

\begin{document}

\begin{tikzpicture}
\matrix[matrix of nodes,column sep=-5.5pt,nodes in empty cells] (mat)
{
4 & 7 & 8 & [7pt]7 & & [7pt] \\
4 & 2 & & 6 & 8 & 7 & [7pt]{} \\
& 5 & 8 & 6 & 3 & 9 & 7 \\
& 5 & 6 & & |[red]|5 & 7 & |[red]|1 \\
& & |[red]|2 & & & |[red]|2 \\
};
\draw ([xshift=-1.5pt,yshift=-2pt]mat-1-4.north west) |- ([xshift=-3.5pt]mat-1-6.south west) |- ([xshift=-3.5pt]mat-2-7.south west) |- (mat-3-7.south east);
\draw (mat-4-2.south west) -- (mat-4-3.south east);
\draw (mat-3-4.south west) -- (mat-3-5.south east);
\draw (mat-4-6.south west) -- (mat-4-6.south east);

\node at ([yshift=-10pt]current bounding box.south)
  {$\longleftarrow$};
\node at ([yshift=-10pt]current bounding box.south)
  {$478=1252_{7}$};
\end{tikzpicture}

\end{document}

在此处输入图片描述

在一条评论中，要求对前七个进行特殊对齐：

\documentclass[10pt,letterpaper]{amsart}
\usepackage{xcolor}

\newcommand\myrule[1]{\multicolumn{1}{@{}c|}{#1}}
\newcommand\myred[1]{\textcolor{red}{#1}}

\begin{document}

\begin{gather*}
\begin{array}{c@{\,}c@{\,}cc@{\,}ccc}
4 & 7 & \myrule{8} & \multicolumn{2}{c}{7} \\
\cline{4-5}
4 & 2 & & 6 & \myrule{8} & 7 \\
\cline{6-6}
& 5 & 8 & 6 & 3 & \multicolumn{1}{c|}{9} & 7 \\
\cline{4-5}\cline{7-7}
& 5 & 6 & & \myred{5} & 7 & \myred{1} \\
\cline{2-3}\cline{6-6}
& & \myred{2} & & & \myred{2}
\end{array} \\
\longleftarrow \\
478 = 1252_{7}
\end{gather*}

\end{document}

在此处输入图片描述

Answer

即使没有 TikZ，使用标准环境也可以做得很好array：

\documentclass[10pt,letterpaper]{amsart}
\usepackage{xcolor}

\newcommand\myrule[1]{\multicolumn{1}{c|}{#1}}
\newcommand\myred[1]{\textcolor{red}{#1}}

\begin{document}

\begin{gather*}
\begin{array}{*{7}c}
4 & 7 & \myrule{8} & 7 \\
\cline{4-5}
4 & 2 & & 6 & \myrule{8} & 7 \\
\cline{6-6}
& 5 & 8 & 6 & 3 & \myrule{9} & 7 \\
\cline{4-5}\cline{7-7}
& 5 & 6 & & \myred{5} & 7 & \myred{1} \\
\cline{2-3}\cline{6-6}
& & \myred{2} & & & \myred{2}
\end{array} \\
\longleftarrow \\
478 = 1252_{7}
\end{gather*}

\end{document}

在此处输入图片描述

后 奎伯比尔贝尔的建议来改变数字之间的间距，可以得到：

\documentclass[10pt,letterpaper]{amsart}
\usepackage{xcolor}

\newcommand\myrule[1]{\multicolumn{1}{@{}c|}{#1}}
\newcommand\myred[1]{\textcolor{red}{#1}}

\begin{document}

\begin{gather*}
\begin{array}{c@{\,}c@{\,}cc@{\,}ccc}
4 & 7 & \myrule{8} & 7 \\
\cline{4-5}
4 & 2 & & 6 & \myrule{8} & 7 \\
\cline{6-6}
& 5 & 8 & 6 & 3 & \multicolumn{1}{c|}{9} & 7 \\
\cline{4-5}\cline{7-7}
& 5 & 6 & & \myred{5} & 7 & \myred{1} \\
\cline{2-3}\cline{6-6}
& & \myred{2} & & & \myred{2}
\end{array} \\
\longleftarrow \\
478 = 1252_{7}
\end{gather*}

\end{document}

在此处输入图片描述

TikZ 解决方案：

\documentclass[10pt,letterpaper]{amsart}
\usepackage{tikz}
\usetikzlibrary{matrix}

\begin{document}

\begin{tikzpicture}
\matrix[matrix of nodes,column sep=-5.5pt,nodes in empty cells] (mat)
{
4 & 7 & 8 & [7pt]7 & & [7pt] \\
4 & 2 & & 6 & 8 & 7 & [7pt]{} \\
& 5 & 8 & 6 & 3 & 9 & 7 \\
& 5 & 6 & & |[red]|5 & 7 & |[red]|1 \\
& & |[red]|2 & & & |[red]|2 \\
};
\draw ([xshift=-1.5pt,yshift=-2pt]mat-1-4.north west) |- ([xshift=-3.5pt]mat-1-6.south west) |- ([xshift=-3.5pt]mat-2-7.south west) |- (mat-3-7.south east);
\draw (mat-4-2.south west) -- (mat-4-3.south east);
\draw (mat-3-4.south west) -- (mat-3-5.south east);
\draw (mat-4-6.south west) -- (mat-4-6.south east);

\node at ([yshift=-10pt]current bounding box.south)
  {$\longleftarrow$};
\node at ([yshift=-10pt]current bounding box.south)
  {$478=1252_{7}$};
\end{tikzpicture}

\end{document}

在此处输入图片描述

在一条评论中，要求对前七个进行特殊对齐：

\documentclass[10pt,letterpaper]{amsart}
\usepackage{xcolor}

\newcommand\myrule[1]{\multicolumn{1}{@{}c|}{#1}}
\newcommand\myred[1]{\textcolor{red}{#1}}

\begin{document}

\begin{gather*}
\begin{array}{c@{\,}c@{\,}cc@{\,}ccc}
4 & 7 & \myrule{8} & \multicolumn{2}{c}{7} \\
\cline{4-5}
4 & 2 & & 6 & \myrule{8} & 7 \\
\cline{6-6}
& 5 & 8 & 6 & 3 & \multicolumn{1}{c|}{9} & 7 \\
\cline{4-5}\cline{7-7}
& 5 & 6 & & \myred{5} & 7 & \myred{1} \\
\cline{2-3}\cline{6-6}
& & \myred{2} & & & \myred{2}
\end{array} \\
\longleftarrow \\
478 = 1252_{7}
\end{gather*}

\end{document}

在此处输入图片描述

Question 3

另一个TiKZ解决方案。代码由杰利迪亚兹在塞万提斯，（西班牙 TeX 组）邮件列表。

每个连续的分区都绘制了matrix of nodes与前一个分区的相对位置。它还标记了每个提醒，以帮助在它们之间划线或放置其他信息。

它不显示完整的除法，只显示被除数、除数和余数。商是下一个除法被除数。

\documentclass[tikz]{standalone}

\usetikzlibrary{matrix,positioning}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Macros for ``successive divisions'' 
%
\def\Division#1#2#3{ % Dividend, divisor, remainder
 \matrix (D) [matrix of nodes,
              below=0pt of D-1-2.south east,
              row sep=1pt, column sep=1pt,
              every node/.append style={minimum width=12mm}] {
   #1 \pgfmatrixnextcell #2 \\
   |[marcar] (R#1)| #3      \\
 };
 \draw[shorten >=2pt, shorten <=2pt]
   (D-1-2.north west) |- (D-1-2.south east);
}
\def\FinDivision#1{
\node[marcar, below=2pt of D-1-2.south] (C)(C)  {#1};
}
\tikzset{marcar/.style={circle,draw,inner sep=2pt,minimum width=0pt,
fill=yellow!10}}

\begin{document}
\begin{tikzpicture}
  \coordinate (D-1-2) at (0,0) {}; % We must start with this command.
  \Division{25}{2}{1} % First dividend, divisor, remainder
  \Division{12}{2}{0} % Dividend (previous quotient), divisor, remainder
  \Division{6}{2}{0}  
  \Division{3}{2}{1}  
  \FinDivision{1}     % Last remainder.

% We can draw an arrow jumping from one remainder 
% to the next one. Every reminder is a node called
% Rdividend. Last remainder is node C.
  \draw[shorten <=1mm, ->, dashed] (C) to[out=-150,in=-65] (R3);
  \draw[shorten <=1mm, ->, dashed] (R3) to[out=-150,in=-65] (R6);
  \draw[shorten <=1mm, ->, dashed] (R6) to[out=-150,in=-65] (R12);
  \draw[shorten <=1mm, ->, dashed] (R12) to[out=-150,in=-65] (R25);

% Some more information:
  \node (MSB) at ([yshift=-1.3cm]R6.south) {Most significant bit (MSB)};    
  \node (LSB) at ([yshift=-2mm]MSB.south) {Less significant bit (LSB)}; 
\draw[ ->] (MSB.east) to[out=30,in=-55] (C);
\draw[ ->] (LSB.west) to[out=150,in=-95] (R25);
\end{tikzpicture}
\end{document}

在此处输入图片描述

Answer

另一个TiKZ解决方案。代码由杰利迪亚兹在塞万提斯，（西班牙 TeX 组）邮件列表。

每个连续的分区都绘制了matrix of nodes与前一个分区的相对位置。它还标记了每个提醒，以帮助在它们之间划线或放置其他信息。

它不显示完整的除法，只显示被除数、除数和余数。商是下一个除法被除数。

\documentclass[tikz]{standalone}

\usetikzlibrary{matrix,positioning}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Macros for ``successive divisions'' 
%
\def\Division#1#2#3{ % Dividend, divisor, remainder
 \matrix (D) [matrix of nodes,
              below=0pt of D-1-2.south east,
              row sep=1pt, column sep=1pt,
              every node/.append style={minimum width=12mm}] {
   #1 \pgfmatrixnextcell #2 \\
   |[marcar] (R#1)| #3      \\
 };
 \draw[shorten >=2pt, shorten <=2pt]
   (D-1-2.north west) |- (D-1-2.south east);
}
\def\FinDivision#1{
\node[marcar, below=2pt of D-1-2.south] (C)(C)  {#1};
}
\tikzset{marcar/.style={circle,draw,inner sep=2pt,minimum width=0pt,
fill=yellow!10}}

\begin{document}
\begin{tikzpicture}
  \coordinate (D-1-2) at (0,0) {}; % We must start with this command.
  \Division{25}{2}{1} % First dividend, divisor, remainder
  \Division{12}{2}{0} % Dividend (previous quotient), divisor, remainder
  \Division{6}{2}{0}  
  \Division{3}{2}{1}  
  \FinDivision{1}     % Last remainder.

% We can draw an arrow jumping from one remainder 
% to the next one. Every reminder is a node called
% Rdividend. Last remainder is node C.
  \draw[shorten <=1mm, ->, dashed] (C) to[out=-150,in=-65] (R3);
  \draw[shorten <=1mm, ->, dashed] (R3) to[out=-150,in=-65] (R6);
  \draw[shorten <=1mm, ->, dashed] (R6) to[out=-150,in=-65] (R12);
  \draw[shorten <=1mm, ->, dashed] (R12) to[out=-150,in=-65] (R25);

% Some more information:
  \node (MSB) at ([yshift=-1.3cm]R6.south) {Most significant bit (MSB)};    
  \node (LSB) at ([yshift=-2mm]MSB.south) {Less significant bit (LSB)}; 
\draw[ ->] (MSB.east) to[out=30,in=-55] (C);
\draw[ ->] (LSB.west) to[out=150,in=-95] (R25);
\end{tikzpicture}
\end{document}

在此处输入图片描述

重复除法 - 从十进制转换为其他进制

答案1

答案2

答案3

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