如何在嵌套对齐中对齐方程式?
\documentclass{article}
\usepackage{amsmath,amssymb,amsfonts}
\usepackage{flexisym}
\begin{document}
\begin{equation*}
CT_i = \left\{\begin {aligned}
&CT_{i1} = \tau \\
&CT_{i2} = H(M).e(g, g)^{s\alpha} \\
&CT_{i3} = h^s \\
&CT_{i4} = H(M)^{t.H_1(ID_{Ui})} \\
&CT_{i5} = H(M)^{t.H_1(ID_{ESP_i})} \\
&\left.\begin{aligned}
&CT_{i6} = g^{q_y(0)} \\
&CT_{i7} = H(attr(y))^{q_y(0)}
\end{aligned}\right\}\forall j \in \mathbb{A}_{Ui}
\end{aligned}\right.
\end{equation*}
\end{document}
在这种情况下,最后两行没有与其余行对齐(在 C 处对齐,= g/h)。
请帮忙。提前致谢。
答案1
使用包中的 cases
和环境。没有任何与符号:rcases
mathtools
\documentclass{article}
\usepackage{mathtools,% for rcases, also load amsmath
amssymb} % amsfonts is loaded by amssymb
%\usepackage{flexisym}% not needed in this MWE
\begin{document}
\[
CT_i = \begin{cases}
CT_{i1} = \tau \\
CT_{i2} = H(M).e(g, g)^{s\alpha} \\
CT_{i3} = h^s \\
CT_{i4} = H(M)^{t.H_1(ID_{Ui})} \\
CT_{i5} = H(M)^{t.H_1(ID_{ESP_i})} \\
\kern-\nulldelimiterspace\begin{rcases}
CT_{i6} = g^{q_y(0)} \\
CT_{i7} = H(attr(y))^{q_y(0)}
\end{rcases}\forall j \in \mathbb{A}_{Ui}
\end{cases}
\]
\end{document}
您的方程式中的点的含义不清楚,因此在上面的 MWE 中,我保留了它们的原样。您可能更喜欢使用 `\cdot˙:
\documentclass{article}
\usepackage{mathtools,% for rcases, also load amsmath
amssymb} % amsfonts is loaded by amssymb
%\usepackage{flexisym}% not needed in this MWE
\begin{document}
\[
CT_i = \begin{cases}
CT_{i1} = \tau \\
CT_{i2} = H(M)\cdot e(g, g)^{s\alpha} \\
CT_{i3} = h^s \\
CT_{i4} = H(M)^{t\cdot H_1(ID_{U_i})} \\
CT_{i5} = H(M)^{t\cdot H_1(ID_{\mathrm{ESP}_i})} \\
\kern-\nulldelimiterspace
\begin{rcases}
CT_{i6} = g^{q_y(0)} \\
CT_{i7} = H\bigl(\mathrm{attr}(y)\bigr)^{q_y(0)}
\end{rcases}\forall j \in \mathbb{A}_{Ui}
\end{cases}
\]
\end{document}