P2260

\[\begin{align}\nonumber &\sum_{i=1}^{n}\sum_{j=1}^{m}(n\ mod\ i)\times(m\ mod\ j)(i\ne j)\\\nonumber &=\sum_{i=1}^{n}(n\ mod\ i)\times\left(\sum_{j=1}^{m}(m\ mod\ j)-(m\ mod\ i)\right)\\\nonumber &=\sum_{i=1}^{n}(n\ mod\ i)\times\sum_{j=1}^{m}(m\ mod\ j)-\sum_{i=1}^{min(n,m)}(n\ mod\ i)\times(m\ mod\ i)\\\nonumber &=\sum_{i=1}^{n}(n-\left\lfloor\frac{n}{i}\right\rfloor\times i)\times\sum_{j=1}^{m}(m-\left\lfloor\frac{m}{j}\right\rfloor\times j)-\sum_{i=1}^{min(n,m)}(n-\left\lfloor\frac{n}{i}\right\rfloor\times i)\times(m-\left\lfloor\frac{m}{i}\right\rfloor\times i)\\\nonumber &=\left(n^2-\sum_{i=1}^{n}(\left\lfloor\frac{n}{i}\right\rfloor\times i)\right)\times\left(m^2-\sum_{j=1}^{m}(\left\lfloor\frac{m}{j}\right\rfloor\times j)\right)-\sum_{i=1}^{min(n,m)}(n\times m-\left\lfloor\frac{n}{i}\right\rfloor\times i\times m-n\times \left\lfloor\frac{m}{i}\right\rfloor\times i+\left\lfloor\frac{n}{i}\right\rfloor\times i\times \left\lfloor\frac{m}{i}\right\rfloor\times i)\\ \end{align} \]

我们来考虑低于线性复杂度地计算\(\sum_{i=1}^{n}\left\lfloor\frac{n}{i}\right\rfloor\times i\)

首先,有许多的i,\(\left\lfloor\frac{n}{i}\right\rfloor\)是一样的

这些中的最大值为\(\frac{n}{\left\lfloor\frac{n}{i}\right\rfloor}\)

所以我们每次计算一个块内的和

\[i_{new}=\frac{n}{\left\lfloor\frac{n}{i}\right\rfloor}\\ \text 该区间的贡献为 \frac{\left\lfloor\frac{n}{i}\right\rfloor \times (i_{new}+i)\times(i_{new}-i+1)}{2} \]

接着是后半部分

\[\begin{align} &\sum_{i=1}^{min(n,m)}(n\times m-\left\lfloor\frac{n}{i}\right\rfloor\times m\times i-n\times \left\lfloor\frac{m}{i}\right\rfloor\times i+\left\lfloor\frac{n}{i}\right\rfloor\times \left\lfloor\frac{m}{i}\right\rfloor\times i^2)\\ &=min(n,m)\times n\times m-m\times\sum_{i=1}^{min(n,m)}\left\lfloor\frac{n}{i}\right\rfloor\times i-n\times\sum_{i=1}^{min(n,m)}\left\lfloor\frac{m}{i}\right\rfloor\times i+\sum_{i=1}^{min(n,m)}\left\lfloor\frac{n}{i}\right\rfloor\times \left\lfloor\frac{m}{i}\right\rfloor\times i^2 \end{align} \]

第一个式子o1,第二、第三同上,第四个式子我们有

\[1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6} \]