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我如何说明十进制到二进制的转换？

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我如何说明十进制到二进制的转换？

作为离散数学作业的一部分，我需要使用有符号的数值、1 的补码和 2 的补码将不同的十进制值（正数和负数）转换为 8 位二进制表示。有人能建议一种优雅的方式来呈现我的解决方案吗？

手写技术的示例如下

在此处输入图片描述

答案1

在你最初的问题中，我以为你正在寻找有关制表位模式的方法的建议，类似于前几个示例。但你的编辑表明你想显示工作情况。类似这样的方法可能可以解决这个问题，我只是按照你展示的简单无符号情况进行操作，其余的我留着，因为你说这是家庭作业:-)

在此处输入图片描述

\documentclass{article}

\newcommand\decbin[9]{%
\par\smallskip
\makebox[3cm][r]{$#1$\ }\fbox{#2}\,\fbox{#3}\,\fbox{#4}\,\fbox{#5}\,\fbox{#6}\,\fbox{#7}\,\fbox{#8}\,\fbox{#9}\par}


\def\unsignedbytecalc#1{%
\par\smallskip
\noindent$#1_{10}$\par
\smallskip
\gdef\result{}%
$\left.\begin{array}{r@{\quad}|c}\udbc{#1}\end{array}\right\}\result$\par}

\makeatletter

\def\udbc#1{%
\ifnum#1=\z@
\expandafter\@gobble
\else
\expandafter\@firstofone
\fi
{2)\!\underline{\,#1}&\edef\r{\ifodd#1 1\else 0\fi}\r\xdef\result{\r\result}\\
\expandafter\udbc\expandafter{\the\numexpr(\ifodd#1 #1-1\else#1\fi)/2\relax}%
}}


\begin{document}

\decbin{13} 00001101
\decbin{54} 00110110

\unsignedbytecalc{156}

\end{document}

答案2

更新：我在最后添加了一个 b 进制到十进制转换的逆过程的图示

我还没有尝试过使用向上箭头... 我很抱歉，因为这些操作@使得代码难以阅读（对我来说）。我最初使用等\n...\m但人们总是担心在 LaTeX 中覆盖某些内容。

该方法使用表格内的可扩展循环，以便获得右侧对齐。不对不应为负数的输入进行任何测试。

我对局部和全局赋值做了一些尝试，但也许使用一个计数变量会让代码更易于阅读。无论如何，重点是计算不应重复进行，尽管表格单元格限制了范围。

当基数为 $\leq 36$ 时，数字使用以 $10 = A$ 开头的字母表。

基数是可选参数。请不要使用1...

\documentclass{article}
\usepackage[left=1cm, right=2cm, vmargin=1cm]{geometry}

% expandable loop (used to avoid scope problems in tabular cells with the
% standard \loop)
\def\boucle #1\repeat {#1\b@@cle {#1}\repeat \repeat }
\def\b@@cle #1{\repeat #1\b@@cle {#1}}

\makeatletter
\newcount\@nn
\newcount\@mm
\newcount\@base
\newcount\@baseminusone

% please do not use this at home
% #1 must be a counter name, not something expanding to a number.
\def\@arabalpha #1{\ifcase #10\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 O\or
P\or Q\or R\or S\or T\or U\or V\or W\or X\or Y\or Z\fi}

\newcommand{\baseexpansion}[2][2]{% no negative numbers please!
\def\@digits{}%
\@base#1\relax \@baseminusone\@base\advance\@baseminusone-1
\@nn #2\relax  % this is the number to be written in base #1
%
\ifnum\@baseminusone<36
\def\onerow{#1\kern.1em\hbox{\vrule
   \vtop {\hbox{\ \the\@nn}\kern.3ex\hrule height.1ex }} &%
   \global\@mm\@nn \global\divide\@mm\@base 
   \multiply\@mm\@base \advance\@nn-\@mm 
   \the\@nn \xdef\@digits{\@arabalpha\@nn\@digits}}%
\else
\def\onerow{#1\kern.1em\hbox{\vrule
   \vtop {\hbox{\ \the\@nn}\kern.3ex\hrule height.1ex }} &%
   \global\@mm\@nn \global\divide\@mm\@base 
   \multiply\@mm\@base \advance\@nn-\@mm 
   \the\@nn \xdef\@digits{\the\@nn.\@digits}}%
\fi
%
\leavevmode\oalign{$#2_{10}:$\hfil\cr
      $\left.
      \begin{tabular}{r|l}
         \boucle \onerow \\ \ifnum\@nn>\@baseminusone\global\@nn\@mm \repeat
      \end{tabular}\right\rbrace=
      \mathtt{\@digits}_{#1}$}}     % \hfil removed from the macro

\makeatother

\begin{document}\thispagestyle{empty}

\lineskip12pt

  \baseexpansion{1000}\hfil
  \baseexpansion[3]{1000}\hfil
  \baseexpansion[4]{1000}\hfil
  \baseexpansion[5]{1000}\hfil
  \baseexpansion[6]{1000}\hfil
  \baseexpansion[7]{1000}\hfil
  \baseexpansion[8]{1000}\hfil
  \baseexpansion[9]{1000}\hfil
  \baseexpansion[10]{1000}\hfil
  \baseexpansion[11]{1000}\hfil
  \baseexpansion[12]{1000}\hfil
  \baseexpansion[13]{1000}\hfil
  \baseexpansion[14]{1000}\hfil
  \baseexpansion[15]{1000}\hfil
  \baseexpansion[16]{1000}\hfil
  \baseexpansion{1024}\hfil
  \baseexpansion[16]{1024}\hfil
  \baseexpansion[3]{6561}\hfil
  \baseexpansion[111]{1000000}\hfil
  \baseexpansion[111]{1000000000}\hfil
  \baseexpansion[7]{1000000000}\par

\end{document}

例子

数字

\documentclass{article}
\usepackage[left=1cm, right=2cm, vmargin=1cm]{geometry}

%%%-----------------------------------------------------------------------------
%%% macros to display the computation of a decimal representation of a positive
%%% integer given in base b representation with b at most 36

\makeatletter

\newcount\@nn
\newcount\@mm
\newcount\@base

% globally sets the counter \@mm to the decimal number represented by #1
% which is either a digit 0-9 or a letter A-Z or a-z
% no check on input done!
%\def\@set@mmto #1{\@mm`#1 \advance\@mm-48 
%      \ifnum\@mm>48 \advance\@mm-39 \else
%      \ifnum\@mm>16 \advance\@mm-7 \fi\fi \global\@mm\@mm}
% I had initially:
\def\@set@mmto #1{\@mm`#1 
  \ifnum\@mm>96 \advance\@mm-87 \else
  \ifnum\@mm>64 \advance\@mm-55 \else
  \advance\@mm-48 \fi\fi \global\@mm\@mm}

% prepares the first three entries of a row of the future tabular
% \@nn represents the decimal number so far computed, and will
% turn into base times itself plus new digit
% global assignments to transcend cells               
\def\@mostofonerow#1{\xdef\@digitssofar{\@digitssofar#1}%
    \@set@mmto #1%
    $\mathtt{\@digitssofar_{\the@base}}\rightarrow\the\@mm\;$%
    &%
    ${}+\the@base\times\the\@nn$%
    \multiply\@nn\@base
    \advance\@nn\@mm
    \global\@nn\@nn
    &${}={}$&}

% this is for displaying the final result in a box
\def\@endfinalrow{\fbox{$\the\@nn_{10}$}\\}

% this is the non-boxed intermediate result ending a row
\def\@endnormalrow{$\the\@nn$\\}

% recursive construction of successive rows
\def\@makenextrows#1{\ifx#1.\@endfinalrow\else
          \@endnormalrow\@mostofonerow#1\expandafter\@makenextrows\fi}

% first step for initialization. Checks for empty argument.
\def\@makerows #1{\ifx#1.\else\@mostofonerow#1\expandafter\@makenextrows\fi}

%% actually constructs the tabular with the computation
%% [moved to \todecimal]
%\def\@displayrows #1{{\setlength{\tabcolsep}{0pt}%
%               \begin{tabular}[t]{r|lcl}#1\end{tabular}}}

\newcommand{\todecimal}[2][2]{%
\@base=#1\def\the@base{#1}%
\@nn=0
\def\@digitssofar{}%
\oalign{$\mathtt{#2_{#1}}:$\hfil\cr
%%%% replaced
%%%% \expandafter\@displayrows\expandafter{\@makerows #2.}}} 
%%%% with the simpler: 
        {%
         \setlength{\tabcolsep}{0pt}%
         \begin{tabular}[t]{r|lcl}
         \@makerows #2.
         \end{tabular}%
        }%
       }%
} % end ot \todecimal

\makeatother

\begin{document}
\todecimal{0}\medskip

\todecimal{1}\medskip

\todecimal{10101110}\medskip

\todecimal[16]{AE}\medskip

\todecimal[16]{aE}\medskip

\todecimal[7]{351}\medskip

\todecimal[36]{zz}\medskip

\todecimal[36]{xYz}

\end{document}

十进制

答案3

不可避免的 LaTeX3 解决方案。

可选参数\decbin表示位数（默认为 8）；\bitcalc应在数学模式下使用。

\documentclass{article}
\usepackage{xparse}

\ExplSyntaxOn
% the needed variables
\tl_new:N \l__bee_bindigits_tl
\seq_new:N \l__bee_quotients_seq
\seq_new:N \l__bee_remainders_seq

% the user level macros
\NewDocumentCommand{\decbin}{ O{8} m }
 {
  \bee_decbin:nn { #1 } { #2 }
 }
\NewDocumentCommand{\bitcalc}{m}
 {
  \bee_bitcalc:n {#1}
 }

% the internal functions
\cs_new:Npn \bee_decbin:nn #1 #2
 {
  \int_compare:nTF { #2 >= 1 \prg_replicate:nn { #1 } { *2 } }
   {
    BAD! % too few bits
   }
   {
    \bee_print_decbin:nn { #1 } { #2 }
   }
 }
\cs_new_protected:Npn \bee_print_decbin:nn #1 #2
 {
  % compute the binary representation
  \tl_set:Nx \l__bee_bindigits_tl { \int_to_binary:n { #2 } }
  % pad with zeros
  \prg_replicate:nn { #1 - \tl_count:N \l__bee_bindigits_tl }
   { \tl_put_left:Nn \l__bee_bindigits_tl { 0 } }
  % print the original number
  $#2\sb{10}$
  % print the boxed binary digits
  \tl_map_inline:Nn \l__bee_bindigits_tl { \,\fbox{##1} }
 }

\cs_new_protected:Npn \bee_bitcalc:n #1
 {
  % print the number and start the recursion
  #1\sb{10}\to
  \seq_clear:N \l__bee_quotients_seq
  \seq_clear:N \l__bee_remainders_seq
  \bee_bitcalc_aux:n { #1 }
  % after the recursion print the binary representation
  \int_to_binary:n { #1 }\sb{2}
 }
\cs_new_protected:Npn \bee_bitcalc_aux:n #1
 {
  % if the quotient is nonzero store the current quotient
  % and the remainder of the division by 2, then repeat;
  % otherwise print the result
  \int_compare:nTF { #1 > 0 }
   {
    \seq_put_right:Nn \l__bee_quotients_seq { 2)\!\underline{\,\, #1 } }
    \seq_put_right:Nx \l__bee_remainders_seq { \int_mod:nn { #1 } { 2 } }
    \bee_bitcalc_aux:x { \int_div_truncate:nn { #1 } { 2 } }
   }
   {
    \bee_print_computation:
   }
 }
\cs_generate_variant:Nn \bee_bitcalc_aux:n { x }
\cs_new_protected:Npn \bee_print_computation:
 {
  \left\{\left.\kern-\nulldelimiterspace
   % print the array of quotients
   \begin{array}{r}\seq_use:Nnnn \l__bee_quotients_seq { \\ } { \\ } { \\ }\end{array}
   \right\uparrow
   % print the array of remainders
   \begin{array}{c}\seq_use:Nnnn \l__bee_remainders_seq { \\ } { \\ } { \\ }\end{array}
   \right\}
 }
\ExplSyntaxOff

\begin{document}
\decbin{1}

\decbin{255}

\decbin[16]{256}

\[
\bitcalc{156}
\]

\end{document}

在此处输入图片描述

通过直接更改几个宏，我们就可以扩展\bitcalc它以执行对（几乎）任意基数的转换：

\NewDocumentCommand{\bitcalc}{ O{2} m }
 {
  \bee_bitcalc:nn { #1 } { #2 }
 }
\cs_new_protected:Npn \bee_bitcalc:nn #1 #2
 {
  % print the number and start the recursion
  #2\sb{10}\to
  \seq_clear:N \l__bee_quotients_seq
  \seq_clear:N \l__bee_remainders_seq
  \bee_bitcalc_aux:nn { #1 } { #2 }
  % after the recursion print the binary representation
  \mathrm{ \int_to_base:nn { #2 } { #1 }\sb{ #1 } }
 }
\cs_new_protected:Npn \bee_bitcalc_aux:nn #1 #2
 {
  % if the quotient is nonzero store the current quotient
  % and the remainder of the division by #1, then repeat;
  % otherwise print the result. Here #1 is the base.
  \int_compare:nTF { #2 > 0 }
   {
    \seq_put_right:Nn \l__bee_quotients_seq { #1)\!\underline{\,\, #2 } }
    \seq_put_right:Nx \l__bee_remainders_seq { \int_mod:nn { #2 } { #1 } }
    \bee_bitcalc_aux:nx { #1 } { \int_div_truncate:nn { #2 } { #1 } }
   }
   {
    \bee_print_computation:
   }
 }
\cs_generate_variant:Nn \bee_bitcalc_aux:nn { nx }

无需更改\bee_print_computation:。输入

\[
\bitcalc[4]{156}\quad\bitcalc[16]{156}
\]

将产生

在此处输入图片描述

该函数\int_to_base:nn支持从 2 到 36 的基数。

几年后的一些修复

此函数\int_to_binary:n已被弃用。我添加了一个 * 版本，\bitcalc将其添加到&箭头之前，用于对齐。

\documentclass{article}
\usepackage{amsmath}
\usepackage{xparse}

\ExplSyntaxOn
% the needed variables
\tl_new:N \l__bee_bindigits_tl
\seq_new:N \l__bee_quotients_seq
\seq_new:N \l__bee_remainders_seq

% the user level macros
\NewDocumentCommand{\decbin}{ O{8} m }
 {
  \bee_decbin:nn { #1 } { #2 }
 }
\NewDocumentCommand{\bitcalc}{ s O{2} m }
 {
  \bee_bitcalc:nnn { #2 } { #3 } { \IfBooleanT{#1}{&} }
 }

% the internal functions
\cs_new:Npn \bee_decbin:nn #1 #2
 {
  \int_compare:nTF { #2 >= 1 \prg_replicate:nn { #1 } { *2 } }
   {
    BAD! % too few bits
   }
   {
    \bee_print_decbin:nn { #1 } { #2 }
   }
 }
\cs_new_protected:Npn \bee_print_decbin:nn #1 #2
 {
  % compute the binary representation
  \tl_set:Nx \l__bee_bindigits_tl { \int_to_bin:n { #2 } }
  % pad with zeros
  \prg_replicate:nn { #1 - \tl_count:N \l__bee_bindigits_tl }
   { \tl_put_left:Nn \l__bee_bindigits_tl { 0 } }
  % print the original number
  $#2\sb{10}$
  % print the boxed binary digits
  \tl_map_inline:Nn \l__bee_bindigits_tl { \,\fbox{##1} }
 }

\cs_new_protected:Npn \bee_bitcalc:nnn #1 #2 #3
 {
  % print the number and start the recursion
  #2\sb{10}#3\to
  \seq_clear:N \l__bee_quotients_seq
  \seq_clear:N \l__bee_remainders_seq
  \bee_bitcalc_aux:nn { #1 } { #2 }
  % after the recursion print the binary representation
  \mathrm{ \str_upper_case:f { \int_to_base:nn { #2 } { #1 } }\sb{ #1 } }
 }
\cs_new_protected:Npn \bee_bitcalc_aux:nn #1 #2
 {
  % if the quotient is nonzero store the current quotient
  % and the remainder of the division by #1, then repeat;
  % otherwise print the result. Here #1 is the base.
  \int_compare:nTF { #2 > 0 }
   {
    \seq_put_right:Nn \l__bee_quotients_seq { #1)\!\underline{\,\, #2 } }
    \seq_put_right:Nx \l__bee_remainders_seq { \int_mod:nn { #2 } { #1 } }
    \bee_bitcalc_aux:nx { #1 } { \int_div_truncate:nn { #2 } { #1 } }
   }
   {
    \bee_print_computation:
   }
 }
\cs_generate_variant:Nn \bee_bitcalc_aux:nn { nx }
\cs_new_protected:Npn \bee_print_computation:
 {
  \left\{
   % print the array of quotients
   \begin{array}{r}\seq_use:Nn \l__bee_quotients_seq { \\ } \end{array}
   \middle\uparrow
   % print the array of remainders
   \begin{array}{c}\seq_use:Nn \l__bee_remainders_seq { \\ } \end{array}
   \right\}
 }
\ExplSyntaxOff

\begin{document}
\decbin{1}

\decbin{255}

\decbin[16]{256}

\[
\bitcalc{4096}
\]

\begin{align*}
\bitcalc*{156}      & \bitcalc*[3]{156} \\
\bitcalc*[8]{5656} & \bitcalc*[16]{5656}
\end{align*}

\end{document}

在此处输入图片描述

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