开博测试

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

\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}