题面
\[\begin{align*}
&\sum_{i = 1}^n \sum_{j = 1}^m a_i \cdot b_j \cdot (i \bmod j) \\
=&\sum_{i = 1}^n \sum_{j = 1}^m a_i \cdot b_j \cdot (i - j\left\lfloor\frac{i}{j}\right\rfloor) \\
=&\sum_{i = 1}^n \sum_{j = 1}^m a_i \cdot i \cdot b_j - a_i \cdot b_j \cdot j\left\lfloor\frac{i}{j}\right\rfloor \\
=&\left(\sum_{j = 1}^m b_j\right)\left(\sum_{i = 1}^n a_i \cdot i\right) - \sum_{j = 1}^m b_j \cdot j \sum_{i = 1}^n a_i \left\lfloor\frac{i}{j}\right\rfloor
\end{align*}
\]