原来的
提供的其余部分polynom.sty并不完整（如下图所示）。
我的目标
当我上高中时，我的老师画了下面的图。:-)
因此，最后一步必须显示总余数并用 L 形曲线标记。
笔记
抱歉，之前的帖子我忘了放回-1/4原来的位置了，让你困惑了，现在没有错误了。:-)
最少代码
\documentclass{article}
\usepackage{polynom,array}
\usepackage[table]{xcolor}
\makeatletter
\def\pld@ArrangeResult#1{%
\ifx\pld@remainder\@empty
\@tempcnta\pld@maxcol\relax
\pld@InsertItems@do\pld@lastline
{\pld@firsttrue\pld@PLD{\pld@R{0}{1}}}%
\fi
\ifnum\pld@currstage>\z@
\pld@Extend\pld@allines{\pld@lastline\cr}%
\else
\pld@InsertFake\pld@lastline
\fi
\pld@iftopresult
\def\pld@lastline{\pld@PrintPoly\pld@divisor%
%====================================================================================
\quad\smash{{\color{red}\rule[-6pt]{\arrayrulewidth}{17pt}}}\kern-\arrayrulewidth&}%
%====================================================================================
\else
\let\pld@lastline\@empty
\ifx B\pld@style\else
\def\pld@lastline{\pld@leftdelim\strut\pld@rightxdelim&}%
\fi
\fi
\expandafter\pld@AR@col\expandafter\pld@PLD
\expandafter\pld@lastline#1+\relax+%
\pld@SplitQuotient
\pld@iftopresult
\let\pld@currentline\@empty
\expandafter\pld@AR@col\expandafter\pld@PLD
\expandafter\pld@currentline
\pld@quotient+\relax+%
\expandafter\pld@AR@col\expandafter\pld@XPLD
\expandafter\pld@currentline
\pld@shadow+\relax+%
\edef\pld@subline{%
\noexpand\cline{\tw@-\pld@maxcol}%
\noalign{\vskip\jot}}%
\pld@Extend\pld@currentline{\expandafter\cr\pld@subline}%
\else
\@tempcnta-\@tempcnta
\advance\@tempcnta\pld@maxcol\relax \advance\@tempcnta\@ne
\edef\pld@span{\the\@tempcnta}%
\ifx B\pld@style
\pld@AddTo\pld@lastline{%
&\multispan\pld@span${}=%
\pld@PrintPolyWithDelims\pld@divisor
\expandafter\pld@IfSum\expandafter{\pld@divisor}{}{\cdot}%
\expandafter\pld@IfSum\expandafter{\pld@quotient}\pld@true
\pld@false
\pld@if \pld@leftdelim
\pld@PrintPolyShadow
\pld@rightdelim
\else \pld@PrintPolyShadow \fi
\pld@firstfalse
\expandafter\pld@PrintRemain\expandafter{\pld@remainder}$}%
\else
\pld@AddTo\pld@lastline{%
&\multispan\pld@span$\pld@leftxdelim\strut\pld@rightdelim
\pld@div
\pld@PrintPolyWithDelims\pld@divisor=
\pld@PrintPolyShadow
\ifx\pld@remainder\@empty\else
+{}%
\setbox\z@=\hbox{$\displaystyle
\frac{\let\strut\@empty\pld@firsttrue \expandafter
\pld@PrintRemain\expandafter{\pld@remainder}}%
{\let\strut\@empty\pld@PrintPoly\pld@divisor}$}%
\dp\z@=\z@\box\z@
\fi
$}%
\fi
\fi
\expandafter\pld@AR@\pld@allines\relax}
\makeatother
\arrayrulecolor{red}
\arrayrulewidth=0.8pt
\def\strut{\rule[-6pt]{0pt}{12pt}}
\begin{document}
\polylongdiv[style=A]{\frac{3}{7}x^9+x^2-\frac{1}{4}}{\frac{9}{5}x^4-1}
\vspace{1cm}
\polylongdiv[style=A]{x^4-1}{x^2-1}
\end{document}
答案1
以下代码应该可以工作；我已添加了一些测试用例。为了使代码更简短，我没有添加您对输出样式的更改（红色，以及被除数和除数之间的横线而不是括号）。我对的更改和添加polynom已标记为!!!。
\documentclass{article}
\usepackage{polynom,array}
\makeatletter
\def\pld@DivPoly@l{%
\ifx\pld@remainder\@empty\else
\pld@IfNeedsDivision\pld@remainder\pld@divisor
{\pld@ExtendPoly\pld@quotient\pld@factor
\pld@NMultiplyPoly\pld@sub\pld@divisor\pld@factor
\pld@SubtractPoly\pld@remainder\pld@sub
\expandafter\pld@DivPoly@l}%
{\expandafter\pld@insert@remainder % !!!
\pld@last@remainder+\relax\relax} % !!!
\fi}
\def\pld@insert@remainder#1+#2\relax{% % !!!
\ifx\relax#1\relax\else\pld@InsertItems\@empty\@empty{#1}\fi % !!!
\ifx\relax#2\relax\else\pld@insert@remainder#2\relax\fi} % !!!
\def\pld@SubtractPoly@l#1+#2\@empty#3+#4\@empty{%
\ifx\relax#1\relax
\let\pld@last@remainder\@empty % !!!
\ifx\relax#3\relax \let\pld@next\@empty \else
\pld@AddToPoly\pld@tempoly{#3}%
\pld@if \pld@InsertItems{#3}{#3}{}\fi
\def\pld@next{\pld@SubtractPoly@l\relax+\@empty#4\@empty}%
\fi
\else
\ifx\relax#3\relax
\pld@SubtractPoly@r#1+#2\@empty
\let\pld@next\@empty
\else
\pld@IfMonomE{#1}{#3}%
{\def\pld@temp{#1+#3}%
\pld@CondenseMonomials\pld@true\pld@temp
\ifx\pld@temp\@empty\else
\pld@ExtendPoly\pld@tempoly\pld@temp
\fi
\pld@if \expandafter\pld@InsertItems\expandafter
{\pld@temp}{#3}{#1}\fi
\def\pld@next{\pld@SubtractPoly@l#2\@empty#4\@empty}}%
{\pld@IfMonomL{#1}{#3}%
{\pld@AddToPoly\pld@tempoly{#3}%
\pld@if \pld@InsertItems{#3}{#3}{}\fi
\def\pld@next{\pld@SubtractPoly@l#1+#2\@empty#4\@empty}}%
{\pld@AddToPoly\pld@tempoly{#1}%
\pld@if \pld@InsertItems{#1}{}{#1}\fi
\def\pld@next{\pld@SubtractPoly@l#2\@empty#3+#4\@empty}}%
}%
\fi \fi
\pld@next}
\def\pld@SubtractPoly@r#1+\relax+\@empty{%
\pld@AddToPoly\pld@tempoly{#1}%
\def\pld@last@remainder{#1}} % !!!
\makeatother
\def\strut{\rule[-6pt]{0pt}{12pt}}
\begin{document}
\polylongdiv{x^5-1}{x-1}
\polylongdiv{x^5-x^2}{x^2-1}
\polylongdiv{\frac{3}{7}x^9+x^2-\frac{1}{4}}{\frac{9}{5}x^4-1}
\polylongdiv{x^9+x^2-1}{x^4-x}
\polylongdiv{x^{15}+1}{x^5+x^3+x+1}
\polylongdiv{x^4-1}{x^2-1}
\end{document}
答案2
嗯，1/4 不是余数，而是第一个参数的最后一部分。您尝试更改的行是显示两个多项式的行。如果您使用稍微简单一点的参数，就可以看到这一点：
\documentclass{article}
\usepackage{polynom,array}
\usepackage[table]{xcolor}
\arrayrulecolor{red}
\arrayrulewidth=0.8pt
\def\strut{\rule[-6pt]{0pt}{12pt}}
\begin{document}
\polylongdiv[style=A]{x^3-1}{x^2+1}
\bigskip
\polylongdiv[style=B]{x^3-1}{x^2+1}
\bigskip
\polylongdiv[style=C]{x^3-1}{x^2+1}
\end{document}
编辑：好消息是其余部分可以访问：
\documentclass{article}
\usepackage{polynom}
\usepackage[table]{xcolor}
\arrayrulecolor{red}
\arrayrulewidth=0.8pt
\def\strut{\rule[-6pt]{0pt}{12pt}}
\makeatletter
\renewcommand*\polylongdiv[1][]{%
\begingroup
\let\pld@stage\maxdimen \polyset{#1}%
\pld@GetPoly{\pld@polya\pld@polyb}%
{\pld@LongDividePoly\pld@polya\pld@polyb
\pld@PrintLongDiv\\
$\expandafter\pld@PrintRemain\expandafter{\pld@remainder}$%new
\endgroup \ignorespaces}}
\begin{document}
\polylongdiv[style=A,stage=5]{x^3-1}{x^2+1}
\bigskip
\polylongdiv[style=A]{\frac{3}{7}x^9+x^2-\frac{1}{4}}{\frac{9}{5}x^4-1}
\end{document}
但我没有时间去寻找如何将其正确插入数组中。