不妨设 \( n \leq m \)。 那原式：

$= \sum_{i = 1} ^ n (n \bmod i) \times \sum_{j = 1} ^ m (m \bmod j) - \sum_{i = 1} ^ n (n \bmod i) \times (m \bmod i)$

$= \sum_{i = 1} ^ n \left( n - \left\lfloor \frac{n}{i} \right\rfloor \times i \right) \times \sum_{j = 1} ^ m \left( m - \left\lfloor \frac{m}{j} \right\rfloor \times j \right) - \sum_{i = 1} ^ n \left( n - \left\lfloor \frac{n}{i} \right\rfloor \times i \right) \left( m - \left\lfloor \frac{m}{i} \right\rfloor \times i \right)$

$= \sum_{i = 1} ^ n \left( n - \left\lfloor \frac{n}{i} \right\rfloor \times i \right) \times \sum_{j = 1} ^ m \left( m - \left\lfloor \frac{m}{j} \right\rfloor \times j \right) - \sum_{i = 1} ^ n \left( nm - n \times i \times \left\lfloor \frac{m}{i} \right\rfloor - m \times i \times \left\lfloor \frac{n}{i} \right\rfloor + i ^ 2 \times \left\lfloor \frac{n}{i} \right\rfloor \times \left\lfloor \frac{m}{i} \right\rfloor \right)$

显然对这 3 坨数论分块就可以了