P2260 [清华集训2012]模积和

P2260 [清华集训2012]模积和

求

$$\sum\limits_{i=1}^{n} \sum\limits_{j=1}^{m} (n \bmod i) \times (m \bmod j), i \neq j$$

mod 19940417 的值

分析

假设 $n\le m$

$\begin{aligned}\sum\limits_{i=1}^{n} \sum\limits_{j=1}^{m} (n \bmod i) \times (m \bmod j), i \neq j\end{aligned}$

$\begin{aligned} & =\sum\limits_{i=1}^{n} \sum\limits_{j=1}^{m} (n \bmod i) \times (m \bmod j)- \sum\limits_{i=1}^{n} (n \bmod i) (m\bmod i)\end{aligned}$

$\begin{aligned}=\sum\limits^n_{i=1}(n - \lfloor\dfrac{n}{i}\rfloor i)\sum\limits^m_{j=1}(m-\lfloor\dfrac{m}{j}\rfloor j)-\sum\limits^n_{i=1}(n-\lfloor\dfrac{n}{i}\rfloor i)(m-\lfloor\dfrac{m}{i}\rfloor i)\end{aligned}$

$\begin{aligned}=\sum\limits^n_{i=1}(n - \lfloor\dfrac{n}{i}\rfloor i)\sum\limits^m_{j=1}(m-\lfloor\dfrac{m}{j}\rfloor j)-\sum\limits^n_{i=1}(n-\lfloor\dfrac{n}{i}\rfloor i)(m-\lfloor\dfrac{m}{i}\rfloor i)\end{aligned}$

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

$\sum\limits^n_{i=1}i^2=\dfrac{n(n + 1) ( 2n + 1)}{6}$

加上整除分块、逆元即可