BZOJ 3994: [SDOI2015]约数个数和3994: [SDOI2015]约数个数和 莫比乌斯反演

https://www.lydsy.com/JudgeOnline/problem.php?id=3994

https://blog.csdn.net/qq_36808030/article/details/77056706

莫比乌斯反演，我现在莫比乌斯反演都不会写不会推了。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 using namespace std;
 #define LL long long
 const int maxn=;
 int mu[maxn]={},pri[maxn]={},tot=;
 bool v[maxn]={};LL f[maxn]={};
 void get_mu(int n){
     mu[]=;
     for(int i=;i<=n;++i){
         if(!v[i]){pri[++tot]=i;mu[i]=-;}
         for(int j=;j<=tot&&pri[j]*i<=n;++j){
             int w=pri[j]*i;v[w]=;
             if(i%pri[j]==){mu[w]=;break;}
             mu[w]=mu[i]*(-);
         }
     }
     for(int i=;i<=n;++i){
         mu[i]+=mu[i-];
         for(int j=,las;j<=i;j=las+){
             las=i/(i/j);
             f[i]+=(LL)(las-j+)*(LL)(i/j);
         }
     }
 }
 int main(){
     get_mu(maxn-);
     int T;scanf("%d\n",&T);
     while(T-->){
         int n,m; scanf("%d%d",&n,&m);
         if(n>m)swap(n,m);
         LL ans=;
         for(int i=,las;i<=n;i=las+){
             las=min(n/(n/i),m/(m/i));
             ans+=(LL)(mu[las]-mu[i-])*f[n/i]*f[m/i];
         }
         printf("%lld\n",ans);
     }
     return ;
 }