mcfx's blog - 莫比乌斯反演 /category/mob/ BZOJ 2694 & 4659 Lcm /archives/181/ 2017-02-28T00:16:00+08:00 ###Description ![fa(1).jpg][1] ###Solution $$\sum_i^n\sum_j^m[\mu(gcd(i,j))\neq 0]lcm(i,j)$$ $$=\sum_k^n|\mu(k)|\sum_i^{\frac{n}{k}}\sum_j^{\frac{n}{k}}[gcd(i,j)=1]i\cdot j\cdot k$$ $$=\sum_k^n|\mu(k)|\cdot k\sum_d^{\frac{n}{k}}\mu(d)\cdot d^2\sum_i^{\frac{n}{k\cdot d}}\sum_j^{\frac{m}{k\cdot d}}i\cdot j$$ $$=\sum_k^n\sum_d[d|k]\mu(d)\cdot d^2\cdot|\mu(\frac{k}{d})|\cdot\frac{p}{d}\cdot\frac{\frac{n}{k}\cdot(\frac{n}{k}+1)}{2}\cdot\frac{\frac{m}{k}\cdot(\frac{m}{k}+1)}{2}$$ 最后的式子,前面是积性函数,后面根号分块 ###Code ```c++ #define _GLIBCXX_IOSTREAM #include #define N 4000001 int pri[N],pm,f[N],ps[N],pc[N]; bool np[N]; int main() { for(int i=2;i