
latex 多行公式居中对齐后与公式编号重叠了
试了好几种方法都是这种效果,无法解决。1图是重叠的;2图,是没有重叠了,但第二行左对齐;3图是我想要的效果,两行公式都居中。重叠:split\begin{eqnarray...
试了好几种方法都是这种效果,无法解决。1图是重叠的;2图,是没有重叠了,但第二行左对齐;3图是我想要的效果,两行公式都居中。
重叠:
split
\begin{eqnarray}\label{ifsdiseq} \begin{a} {d_{IFS}}(A,B) = \left( {\mu _A^{}(x) - \mu _B^{}(x),\nu _A^{}(x) - \nu _B^{}(x),(1 - \mu _A^{}(x) - \nu _A^{}(x)) - (1 - \mu _B^{}(x) - \nu _B^{}(x))} \right) \\ \left[ {\begin{array}{*{20}{c}}& \{Y\} & \{N\} & \{{Y,N}\} \\ \{Y\} & 1 & 0 & {1/2} \\ \{N\} & 0 & 1 & {1/2} \\ \{{Y,N}\} & {1/2} & {1/2} & 1 \\\end{array}} \right]\left( {\begin{array}{*{20}{c}} {\mu _A^{}(x) - \mu _B^{}(x)} \\ {\nu _A^{}(x) - \nu _B^{}(x)} \\ {(1 - \mu _A^{}(x) - \nu _A^{}(x)) - (1 - \mu _B^{}(x) - \nu _B^{}(x))} \\\end{array}} \right)\end{split}\end{eqnarray}
用nonumber
\begin{eqnarray}\label{ifsdiseq} {d_{IFS}}(A,B) = \left( {\mu _A^{}(x) - \mu _B^{}(x),\nu _A^{}(x) - \nu _B^{}(x),(1 - \mu _A^{}(x) - \nu _A^{}(x)) - (1 - \mu _B^{}(x) - \nu _B^{}(x))} \right) \nonumber \\ \left[ {\begin{array}{*{20}{c}}& \{Y\} & \{N\} & \{{Y,N}\} \\ ........... {(1 - \mu _A^{}(x) - \nu _A^{}(x)) - (1 - \mu _B^{}(x) - \nu _B^{}(x))} \\\end{array}} \right)\end{eqnarray}
不重叠左对齐:用&
\begin{eqnarray}\label{ifsdiseq} \begin{split} & {d_{IFS}}(A,B) = \left( {\mu _A^{}(x) - \mu _B^{}(x),\nu _A^{}(x) - \nu _B^{}(x),(1 - \mu _A^{}(x) - \nu _A^{}(x)) - (1 - \mu _B^{}(x) - \nu _B^{}(x))} \right) \\ &.........
\end{spilt}
\end{eqnarray}
求大神解答!!为什么会重叠,我很郁闷! 展开
重叠:
split
\begin{eqnarray}\label{ifsdiseq} \begin{a} {d_{IFS}}(A,B) = \left( {\mu _A^{}(x) - \mu _B^{}(x),\nu _A^{}(x) - \nu _B^{}(x),(1 - \mu _A^{}(x) - \nu _A^{}(x)) - (1 - \mu _B^{}(x) - \nu _B^{}(x))} \right) \\ \left[ {\begin{array}{*{20}{c}}& \{Y\} & \{N\} & \{{Y,N}\} \\ \{Y\} & 1 & 0 & {1/2} \\ \{N\} & 0 & 1 & {1/2} \\ \{{Y,N}\} & {1/2} & {1/2} & 1 \\\end{array}} \right]\left( {\begin{array}{*{20}{c}} {\mu _A^{}(x) - \mu _B^{}(x)} \\ {\nu _A^{}(x) - \nu _B^{}(x)} \\ {(1 - \mu _A^{}(x) - \nu _A^{}(x)) - (1 - \mu _B^{}(x) - \nu _B^{}(x))} \\\end{array}} \right)\end{split}\end{eqnarray}
用nonumber
\begin{eqnarray}\label{ifsdiseq} {d_{IFS}}(A,B) = \left( {\mu _A^{}(x) - \mu _B^{}(x),\nu _A^{}(x) - \nu _B^{}(x),(1 - \mu _A^{}(x) - \nu _A^{}(x)) - (1 - \mu _B^{}(x) - \nu _B^{}(x))} \right) \nonumber \\ \left[ {\begin{array}{*{20}{c}}& \{Y\} & \{N\} & \{{Y,N}\} \\ ........... {(1 - \mu _A^{}(x) - \nu _A^{}(x)) - (1 - \mu _B^{}(x) - \nu _B^{}(x))} \\\end{array}} \right)\end{eqnarray}
不重叠左对齐:用&
\begin{eqnarray}\label{ifsdiseq} \begin{split} & {d_{IFS}}(A,B) = \left( {\mu _A^{}(x) - \mu _B^{}(x),\nu _A^{}(x) - \nu _B^{}(x),(1 - \mu _A^{}(x) - \nu _A^{}(x)) - (1 - \mu _B^{}(x) - \nu _B^{}(x))} \right) \\ &.........
\end{spilt}
\end{eqnarray}
求大神解答!!为什么会重叠,我很郁闷! 展开
2个回答
展开全部
公式中有一些空隙,减小一些间距就不会重叠了。如可以在方括号和圆括号间插入若干个“\!”:
\end{array}} \right] \!\! \left( {\begin{array}{*{20}{c}}
以及缩小运算符号两边的距离:
{(1 \! - \! \mu _A^{}(x) \! - \! \nu _A^{}(x)) \! - \! (1 \! - \! \mu _B^{}(x) \! - \! \nu _B^{}(x))} \\
\end{array}} \right] \!\! \left( {\begin{array}{*{20}{c}}
以及缩小运算符号两边的距离:
{(1 \! - \! \mu _A^{}(x) \! - \! \nu _A^{}(x)) \! - \! (1 \! - \! \mu _B^{}(x) \! - \! \nu _B^{}(x))} \\
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