$\sum \frac{1}{n^2}$
\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}
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\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}
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开博测试
原文：http://www.cnblogs.com/towan/p/4340423.html
