我试图在 Overleaf 中将冗长的方程式横跨两列页面。当我使用 \begin{strip} 时，数字会以黑白形式显示两个方程式

Question

    \documentclass[twocolumn]{article}
\usepackage{graphicx} 
\usepackage{cuted}
\usepackage{amsmath}



\begin{document}
\maketitle

\section{Introduction}
\subsection{Equations}
In this section, we will explore how to compose lengthy equations intended to be formatted within a single column on a two-column paper.\\
\newline

\begin{strip}
    \begin{equation}
        A = Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \left(1 - \frac{r}{2}\right)- \frac{-Q^{u} \left(1 - r \cdot c\right)}{r \cdot c} \left(1 - \frac{r}{2}\right) - Q_{\text{rand}}^{u} \left(\frac{r}{2} - 1\right)
    \end{equation}
\end{strip}

\begin{strip}
   \begin{equation}\label{eq13}
        A = Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \left(1 - \frac{r}{2}\right)- \frac{-Q^{u} \left(1 - r \cdot c\right)}{r \cdot c} \left(1 - \frac{r}{2}\right) - Q_{\text{rand}}^{u} \left(\frac{r}{2} - 1\right)
    \end{equation}
\end{strip}

\begin{strip}
   \begin{equation}\label{eq13}
        A = Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \left(1 - \frac{r}{2}\right)- \frac{-Q^{u} \left(1 - r \cdot c\right)}{r \cdot c} \left(1 - \frac{r}{2}\right) - Q_{\text{rand}}^{u} \left(\frac{r}{2} - 1\right) 
    \end{equation}
\end{strip}

\end{document}

Answer 1

正如@Zarko 在评论中已经建议的那样，请考虑将三个独立的方程式放置在适当选择的align环境中gather。

\documentclass[twocolumn]{article}
\usepackage{cuted,amsmath,lipsum}

\begin{document}

\section{Introduction}
\subsection{Equations}
\lipsum[1][1-9] % filler text

\begin{strip}
\begin{align}
A &= Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \bigl(1 - \tfrac{r}{2}\bigr)
     - \frac{-Q^{u} (1 - r \cdot c)}{r \cdot c} \bigl(1 - \tfrac{r}{2}\bigr) 
     - Q_{\textrm{rand}}^{u} \bigl(\tfrac{r}{2} - 1\bigr)
     \\
A &= Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \bigl(1 - \tfrac{r}{2}\bigr)
    - \frac{-Q^{u} (1 - r \cdot c)}{r \cdot c} \bigl(1 - \tfrac{r}{2}\bigr) 
    - Q_{\textrm{rand}}^{u} \bigl(\tfrac{r}{2} - 1\bigr) 
    \label{eq23} \\
A &= Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \bigl(1 - \tfrac{r}{2}\bigr)
    - \frac{-Q^{u} (1 - r \cdot c)}{r \cdot c} \bigl(1 - \tfrac{r}{2}\bigr) 
    - Q_{\textrm{rand}}^{u} \bigl(\tfrac{r}{2} - 1\bigr) 
    \label{eq33}
\end{align}
\end{strip}

\null % insert an invisible marker

\end{document}

Answer 2

好吧，我找到答案了。在上面的例子中，我们在一个 \strip 命令内添加了所有三个方程。

\begin{strip}

\begin{equation}
        A = Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \left(1 - \frac{r}{2}\right)- \frac{-Q^{u} \left(1 - r \cdot c\right)}{r \cdot c} \left(1 - \frac{r}{2}\right) - Q_{\text{rand}}^{u} \left(\frac{r}{2} - 1\right)

\begin{equation}\label{eq13}
 A = Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \left(1 - \frac{r}{2}\right)- \frac{-Q^{u} \left(1 - r \cdot c\right)}{r \cdot c} \left(1 - \frac{r}{2}\right) - Q_{\text{rand}}^{u} \left(\frac{r}{2} - 1\right)
\end{equation}

\begin{equation}\label{eq13}
        A = Q^{u+1} - \frac{Q^{u+1}}{r \cdot c} \left(1 - \frac{r}{2}\right)- \frac{-Q^{u} \left(1 - r \cdot c\right)}{r \cdot c} \left(1 - \frac{r}{2}\right) - Q_{\text{rand}}^{u} \left(\frac{r}{2} - 1\right) 
    \end{equation}

\end{strip}