约数个数和

首先这个约数公式要记住，具体见这篇题解

然后剩下的就是比较简单的套路操作了，最后会化出来

$$\sum _{d=1}^{min(n,m)}\sum _{d|x}\sum _{d|y}\lfloor \frac{n}{x}\rfloor \lfloor \frac{m}{y}\rfloor u(d)=\sum _{d=1}^{min(n,m)}u(d)(\sum _{d|x}\lfloor \frac{n}{x}\rfloor )(\sum _{d|y}\lfloor \frac{m}{y}\rfloor )=\sum _{d=1}^{min(n,m)}u(d)(\sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\lfloor \frac{n}{dk}\rfloor )(\sum _{k=1}^{\lfloor \frac{m}{d}\rfloor }\lfloor \frac{m}{dk}\rfloor )=\sum _{d=1}^{min(n,m)}u(d)(\sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\lfloor \frac{\lfloor \frac{n}{d}\rfloor }{k}\rfloor )(\sum _{k=1}^{\lfloor \frac{m}{d}\rfloor }\lfloor \frac{\lfloor \frac{m}{d}\rfloor }{k}\rfloor )$$

令$f(x)=\sum _{k=1}^x\lfloor \frac{x}{k}\rfloor$，则求式等于

$$\sum _{d=1}^{min(n,m)}u(d)f(\lfloor \frac{n}{d}\rfloor )f(\lfloor \frac{m}{d}\rfloor )$$

预处理出$f$，然后对这个式子进行数论分块就好了