BZOJ 4805: 欧拉函数求和 杜教筛

https://www.lydsy.com/JudgeOnline/problem.php?id=4805

给出一个数字N，求sigma(phi(i)),1<=i<=N

https://blog.csdn.net/popoqqq/article/details/45023331 ←杜教筛的一些讲解

杜教筛用来求积性函数前缀和，本题同bzoj 3944，bzoj 3944多了一个求sigma( μ ( i ) )

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<queue>
 using namespace std;
 #define LL long long
 const int maxn=;
 LL n;
 LL phi[maxn]={};
 LL pri[maxn/]={},tot=;
 bool v[maxn]={};
 void get_phi(int m){
     phi[]=;
     for(int i=;i<=m;++i){
         if(!v[i]){pri[++tot]=i;phi[i]=i-;}
         for(int j=;j<=tot&&pri[j]*i<=m;++j){
             int w=pri[j]*i;v[w]=;
             if(i%pri[j]==){phi[w]=phi[i]*pri[j];break;}
             phi[w]=phi[i]*(pri[j]-);
         }
     }
     //for(int i=1;i<=10;++i)cout<<phi[i]<<endl;
     for(int i=;i<=m;++i){
         phi[i]+=phi[i-];
     }
 }
 LL solve(LL m){
     if(m<=maxn-)return phi[m];
     LL ans=m*(m+)/;
     for(LL i=,las;i<=m;i=las+){
         las=m/(m/i);
         ans-=(LL)(las-i+)*solve(m/i);
     }
     return ans;
 }
 int main(){
     get_phi(maxn-);
     scanf("%lld",&n);
     printf("%lld\n",solve(n));
     return ;
 }