数学之皮

$$\sum_{i=1}^n\sum_{j=1}^m[gcd(x,y)==1]$$ $$=\sum_{i=1}^n\sum_{j=1}^m\sum_{d|gcd(i,j)}\mu(d)$$ $$=\sum_{p}\sum_{d=1}^{\lfloor \frac{min(n,m)}{p}\rfloor}\mu(d)\lfloor\frac{n}{pd}\rfloor\lfloor\frac{m}{pd}\rfloor$$ $$2^n=\sum_{i=0}^{n}\binom{n}{i}$$ [积性函数](http://blog.csdn.net/skywalkert/article/details/50500009) [某大佬のblog](https://blog.csdn.net/hhaannyyii/article/details/79335883)