我在算法环境中编写伪代码。它很长,无法放在一页上。我尝试了许多选项,但无法分成两页。在另一个文件中(因为这是论文),我声明了以下包:
\usepackage[vlined,linesnumbered]{algorithm2e} % Algorithm
\usepackage{algorithmic}
\usepackage{algorithm}
\usepackage{algcompatible}
这是我的代码:
\begin{figure}[h]
\small
\hrule
\vspace{2mm}
\textbf{Reader Operation}
%\vspace{1mm}
\begin{algorithmic}[1]
\STATE $ws = 1$
\STATE FQwT([\textquoteleft 0'],$ws$); FQwT([\textquoteleft 1'],$ws$)
\STATE function $FQwT$(char [] $query,$ int $ws$)
\STATE $k = ID.length$
\STATE $L = query.length$
\STATE $c_g=k-L$
\STATE $s=log2(ws)$
\STATE broadcast([$query], s$)
\STATE $[winMatch,crcCheck] = $receiveResponses()
\IF{$nSuccess=0$}
\STATE{$Phase1$}
\IF{$isempty$($winMatch$)}
\STATE $nIdles ++$
\ELSE [$crcCheck = 0$]
\STATE $nCollisions ++$
\IF{$ws=1$}
\STATE $Store$ $c_g$ $with$ $query$ $into$ $LIFO$ $(c_g=k-L)$
\STATE $FQwT$([$query$, \textquoteleft 0'],$ws$); FQwT([$query$, \textquoteleft 1'],$ws$)
\ELSE
\STATE $ws = 1$
\STATE $FQwT$ ([$query$],$ws$); FQwT([$query$],$ws$)
\ENDIF
\ELSE [$crcCheck=1$]
\IF{$L + ws < k$}
\STATE $nGoons ++$
\STATE $ws=ws ++$
\STATE FQwT($[query,winMatch],ws$)
\ELSE [$L + ws = k$]
\STATE $ws=1$
\STATE $nSuccess ++$
\ENDIF
\ENDIF
\ELSE
\STATE{$Phase2$}
\IF{isempty($winMatch$)}
\STATE $nIdles ++$
\ELSE[$crcCheck = 0$]
\STATE $nCollisions ++$
\STATE FQwT([$query$, \textquoteleft 0'],$ws$); FQwT([$query$, \textquoteleft 1'],$ws$)
\ELSE[$crcCheck = 1$]
\IF{$L + ws < k$}
\STATE $nGoons ++$
\STATE $ ws=f(c_g,L)$
\STATE $s= log2(ws)$
\STATE FQwT([$query$, $winMatch$],$ws$);
\ELSE[$L + ws = k$]
\STATE $nSuccess ++$
\ENDIF
\ENDIF
\ENDIF
\end{algorithmic}
\vspace{1mm}
\hrule
\vspace{2mm}
\textbf{Tag Operation}
\vspace{1mm}
\begin{algorithmic}[1]
\STATE $receive(query, s)$
\STATE $L = query.length$
\STATE $ws=2^s$
\IF{$query = ID[0:L-1]$}
\STATE $CRC=$ crc($ID[L:L{+}ws]$)
\STATE backscatter($ID[L:L{+}ws],CRC$)
\ENDIF
\end{algorithmic}
\vspace{1mm}
\hrule
\caption{Pseudo-code of FQwT.
First, the operation of the reader is presented, then the tag operation.}
\label{Fig:dis53}
\end{figure}
答案1
要将算法拆分为两页,请使用 Werner 的便捷breakablealgorithm
环境,如下所述这里。
将此代码添加到您的序言中:
\makeatletter
\newenvironment{breakablealgorithm}
{% \begin{breakablealgorithm}
\begin{center}
\refstepcounter{algorithm}% New algorithm
\hrule height.8pt depth0pt \kern2pt% \@fs@pre for \@fs@ruled
\renewcommand{\caption}[2][\relax]{% Make a new \caption
{\raggedright\textbf{\ALG@name~\thealgorithm} ##2\par}%
\ifx\relax##1\relax % #1 is \relax
\addcontentsline{loa}{algorithm}{\protect\numberline{\thealgorithm}##2}%
\else % #1 is not \relax
\addcontentsline{loa}{algorithm}{\protect\numberline{\thealgorithm}##1}%
\fi
\kern2pt\hrule\kern2pt
}
}{% \end{breakablealgorithm}
\kern2pt\hrule\relax% \@fs@post for \@fs@ruled
\end{center}
}
\makeatother
然后algorithmic
用\begin{breakablealgorithm}
和包围你的\end{breakablealgorithm}
。因此:
\begin{breakablealgorithm}
\begin{algorithmic}[1]
...
...
...
\end{algorithmic}
\end{breakablealgorithm}
编辑: 如果我把你的长算法塞进Werner的环境中,那么它就能完美地运行。
\documentclass{article}
\usepackage{algorithm,algpseudocode,float}
\usepackage{lipsum}
\makeatletter
\newenvironment{breakablealgorithm}
{% \begin{breakablealgorithm}
\begin{center}
\refstepcounter{algorithm}% New algorithm
\hrule height.8pt depth0pt \kern2pt% \@fs@pre for \@fs@ruled
\renewcommand{\caption}[2][\relax]{% Make a new \caption
{\raggedright\textbf{\ALG@name~\thealgorithm} ##2\par}%
\ifx\relax##1\relax % #1 is \relax
\addcontentsline{loa}{algorithm}{\protect\numberline{\thealgorithm}##2}%
\else % #1 is not \relax
\addcontentsline{loa}{algorithm}{\protect\numberline{\thealgorithm}##1}%
\fi
\kern2pt\hrule\kern2pt
}
}{% \end{breakablealgorithm}
\kern2pt\hrule\relax% \@fs@post for \@fs@ruled
\end{center}
}
\makeatother
\begin{document}
\listofalgorithms
\section{Some section}
\lipsum[1]
\begin{breakablealgorithm}
\caption{Euclid’s algorithm}
\begin{algorithmic}[1]
\State $ws = 1$
\State FQwT([\textquoteleft 0'],$ws$); FQwT([\textquoteleft 1'],$ws$)
\State function $FQwT$(char [] $query,$ int $ws$)
\State $k = ID.length$
\State $L = query.length$
\State $c_g=k-L$
\State $s=log2(ws)$
\State broadcast([$query], s$)
\State $[winMatch,crcCheck] = $receiveResponses()
\If{$nSuccess=0$}
\State{$Phase1$}
\If{$isempty$($winMatch$)}
\State $nIdles ++$
\Else [$crcCheck = 0$]
\State $nCollisions ++$
\If{$ws=1$}
\State $Store$ $c_g$ $with$ $query$ $into$ $LIFO$ $(c_g=k-L)$
\State $FQwT$([$query$, \textquoteleft 0'],$ws$); FQwT([$query$, \textquoteleft 1'],$ws$)
\Else
\State $ws = 1$
\State $FQwT$ ([$query$],$ws$); FQwT([$query$],$ws$)
\EndIf
\Else [$crcCheck=1$]
\If{$L + ws < k$}
\State $nGoons ++$
\State $ws=ws ++$
\State FQwT($[query,winMatch],ws$)
\Else [$L + ws = k$]
\State $ws=1$
\State $nSuccess ++$
\EndIf
\EndIf
\Else
\State{$Phase2$}
\If{isempty($winMatch$)}
\State $nIdles ++$
\Else[$crcCheck = 0$]
\State $nCollisions ++$
\State FQwT([$query$, \textquoteleft 0'],$ws$); FQwT([$query$, \textquoteleft 1'],$ws$)
\Else[$crcCheck = 1$]
\If{$L + ws < k$}
\State $nGoons ++$
\State $ ws=f(c_g,L)$
\State $s= log2(ws)$
\State FQwT([$query$, $winMatch$],$ws$);
\Else[$L + ws = k$]
\State $nSuccess ++$
\EndIf
\EndIf
\EndIf
\end{algorithmic}
\end{breakablealgorithm}
\end{document}
如果它对你有用,那么所有的功劳都归功于 Werner。