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开博测试

时间:2015-03-15 22:49:23      阅读:107      评论:0      收藏:0      [点我收藏+]

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$\sum \frac{1}{n^2}$

\alpha


$\sum \frac{1}{n^2}$

\alpha


\begin{equation}\label{equation 1} st_i = \{     \begin{array}{ll}         d_i, \quad 2(d_i - d_{i - 1})  > tx_{i - 1} + g; &   \\         d_i + tx_{i - 1} + g - 2(d_i - d_{i - 1}), & otherwise.     \end{array} \end{equation}
\begin{equation}\label{equation 1} st_i = \{     \begin{array}{ll}         d_i, \quad 2(d_i - d_{i - 1})  > tx_{i - 1} + g; &   \\         d_i + tx_{i - 1} + g - 2(d_i - d_{i - 1}), & otherwise.     \end{array} \end{equation}

\begin{equation}\label{equation 1} st_i = \{     \begin{array}{ll}         d_i, \quad 2(d_i - d_{i - 1})  > tx_{i - 1} + g; &   \\         d_i + tx_{i - 1} + g - 2(d_i - d_{i - 1}), & otherwise.     \end{array} \end{equation}


开博测试

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原文地址:http://www.cnblogs.com/towan/p/4340423.html

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