我在一篇论文中有三个主要例子，我一直在回顾它们。因此，我想按以下方式构建内容。一种反转编号/字母的解决方案（例如示例 A1、A2、B1、B2 等）在这个答案，但我更喜欢下面的样式。需要澄清的是，该解决方案通常需要与 hyperref 和 cross-ref 兼容。
我也愿意接受针对该问题的其他创造性解决方案。
结构
示例 1A。这是第一个例子的定义以及一些基本属性。
示例 2A。这是第二个示例的定义及其一些基本属性。
示例 3A。以下是第三个示例的定义及其一些基本属性。
（在引入定理 1 等之后）
示例 1B。定理 1 在示例 1 中的应用示例。
示例 3B。定理1到例3的一个例子应用。
（在引入定理 2 等之后）
示例 1C。定理 2 在示例 1 中的应用示例。
（本文稍后刊登）
示例 2B。对示例 2 的其他内容的示例应用。
（编辑：修改示例以强调编号/字母不是结构化的。）
答案1
我相信你需要提前规划一下你心中有多少种例子。
\documentclass{article}
\usepackage{amsthm}
\usepackage{hyperref}
\newtheorem{theorem}{Theorem}
\newcommand{\newexample}[1]{%
\theoremstyle{definition} % upright type for examples
\newtheorem{example#1}{Example}%
\expandafter\renewcommand\csname theexample#1\endcsname{#1\Alph{example#1}}%
}
\newexample{1}
\newexample{2}
\newexample{3}
\begin{document}
\begin{example1}\label{ex1-first}
Here is the definition of the first example with some elementary properties.
\end{example1}
\begin{example2}
Here is the definition of the second example with some elementary properties.
\end{example2}
\begin{example3}
Here is the definition of the third example with some elementary properties.
\end{example3}
\begin{theorem}
This is the first theorem
\end{theorem}
\begin{example1}
An example application of Theorem 1 to Example 1.
\end{example1}
\begin{example2}\label{ex2-second}
An example application of Theorem 1 to Example 2.
\end{example2}
\begin{example3}
An example application of Theorem 1 to Example 3.
\end{example3}
\begin{theorem}
This is the second theorem
\end{theorem}
\begin{example1}
An example application of Theorem 2 to Example 1.
\end{example1}
\begin{example2}
An example application of Theorem 2 to Example 2.
\end{example2}
\begin{example3}
An example application of Theorem 2 to Example 3.
\end{example3}
\ref{ex1-first} and \ref{ex2-second}
\end{document}
如果您始终具有相同的结构，即三个示例品种始终组合在一起，您可能会像下面这样操作。
\documentclass{article}
\usepackage{amsthm}
\usepackage{hyperref}
\newtheorem{theorem}{Theorem}
\theoremstyle{definition} % upright type for examples
\newtheorem{exampleinner}{Example}
\newcounter{examples}
\newcounter{example}
\newenvironment{examples}
{\setcounter{examples}{0}\stepcounter{example}}
{}
\newenvironment{example}
{%
\renewcommand{\theexampleinner}{\theexamples\Alph{example}}%
\stepcounter{examples}%
\exampleinner
}
{\endexampleinner}
\begin{document}
\begin{examples}
\begin{example}\label{ex1-first}
Here is the definition of the first example with some elementary properties.
\end{example}
\begin{example}
Here is the definition of the second example with some elementary properties.
\end{example}
\begin{example}
Here is the definition of the third example with some elementary properties.
\end{example}
\end{examples}
\begin{theorem}
This is the first theorem
\end{theorem}
\begin{examples}
\begin{example}
An example application of Theorem 1 to Example 1.
\end{example}
\begin{example}\label{ex2-second}
An example application of Theorem 1 to Example 2.
\end{example}
\begin{example}
An example application of Theorem 1 to Example 3.
\end{example}
\end{examples}
\begin{theorem}
This is the second theorem
\end{theorem}
\begin{examples}
\begin{example}
An example application of Theorem 2 to Example 1.
\end{example}
\begin{example}
An example application of Theorem 2 to Example 2.
\end{example}
\begin{example}
An example application of Theorem 2 to Example 3.
\end{example}
\end{examples}
\ref{ex1-first} and \ref{ex2-second}
\end{document}
输出与以前相同。
答案2
也许我误解了要求，但以下内容不需要对示例数量进行硬编码。结构也不是固定的，即，您可以在每个块中使用任意数量的项目，跳过数字或字母等。这确实需要将\refstepcounter（或可能\setcounter）放在相关位置，无论是数字还是字母。
它的工作原理是定义两个计数器，一个用于字母，一个用于数字，每次字母计数器增加时，数字计数器都会重置\counterwithin。这会重置为 0，以便打印1A等等，而不是0A打印计数器值+1。
然后myexample定义一个环境，打印格式化的示例数字，设置命令的文本，并为with\label创建锚点。在环境结束时，数字会自动增加。hyperref\phantomsection
每当您想要转到下一个“块”时，您都需要手动调用\refstepcounter{exletter}，即，它不与任何定理定义相关联。
\documentclass{article}
\usepackage{hyperref}
\usepackage{chngcntr}
\newcounter{exletter}
\newcounter{exnumber}
\setcounter{exletter}{1}
\setcounter{exnumber}{0}
\counterwithin{exnumber}{exletter}
\makeatletter
\newenvironment{myexample}{%
\textbf{Example \the\numexpr\value{exnumber}+1\relax\Alph{exletter}}%
\def\@currentlabel{\the\numexpr\value{exnumber}+1\relax\Alph{exletter}}%
\phantomsection%
}{\refstepcounter{exnumber}\par}
\makeatother
\begin{document}
\begin{myexample}
Here is the definition of the first example with some elementary properties.
\end{myexample}
\begin{myexample}
Here is the definition of the second example with some elementary properties.
\end{myexample}
\begin{myexample}
\label{thirdex}
Here is the definition of the third example with some elementary properties.
\end{myexample}
\refstepcounter{exletter}
\begin{myexample}
An example application of Theorem 1 to Example 1
\end{myexample}
\begin{myexample}
An example application of Theorem 1 to Example 2
\end{myexample}
\begin{myexample}
An example application of Theorem 1 to Example 3
\end{myexample}
\refstepcounter{exletter}
\begin{myexample}
An example application of Theorem 2 to Example 1
\end{myexample}
\refstepcounter{exnumber}
\begin{myexample}
\label{exappl}
An example application of Theorem 2 to Example 3
\end{myexample}
In Example \ref{exappl} a theorem is applied to Example \ref{thirdex}.
\end{document}
结果：
答案3
如果您的定理/示例设置足够一致，则以下将给出您想要的输出：
\documentclass{article}
\usepackage{amsthm}
\usepackage{hyperref}
\newtheorem{theorem}{Theorem}
\theoremstyle{definition}
\newtheorem{example}{Example}[theorem]
% option 1
\renewcommand{\theexample}{%
\addtocounter{theorem}{1}%
\arabic{example}\Alph{theorem}%
\addtocounter{theorem}{-1}%
}
% option 2
%\renewcommand{\thetheorem}{\Alph{theorem}}
%\renewcommand{\theexample}{\arabic{example}\thetheorem}
\newcommand{\exampleautorefname}{Example}
\begin{document}
\begin{example}\label{egone}
Here is the definition of the first example with some elementary properties.
\end{example}
\begin{example}\label{egtwo}
Here is the definition of the second example with some elementary properties.
\end{example}
\begin{example}\label{egthree}
Here is the definition of the third example with some elementary properties.
\end{example}
\begin{theorem}\label{thmone}
This is the first theorem
\end{theorem}
\begin{example}
An example application of \autoref{thmone} to \autoref{egone}.
\end{example}
\begin{example}\label{thmoneEgtwo}
An example application of \autoref{thmone} to \autoref{egtwo}.
\end{example}
\begin{example}
An example application of \autoref{thmone} to \autoref{egthree}.
\end{example}
\begin{theorem}\label{thmtwo}
This is the second theorem
\end{theorem}
\begin{example}
An example application of \autoref{thmtwo} to \autoref{egone}.
\end{example}
\begin{example}
An example application of \autoref{thmtwo} to \autoref{egtwo}.
\end{example}
\begin{example}
An example application of \autoref{thmtwo} to \autoref{egthree}.
\end{example}
\autoref{egone} and \autoref{thmoneEgtwo}
\end{document}
但我发现示例 #C 引用了定理 2，这令人困惑。使用选项 2，可得到以下输出：
