【枚举】【莫比乌斯反演】【线性筛】bzoj2818 Gcd

#include<cstdio>
using namespace std;
#define N 10000000
bool notpri[N+5];
int pri[N+5],mu[N+5],sum[N+5];
typedef long long ll;
void shai_mu()
{
    notpri[1]=1; mu[1]=1;
    for(int i=2;i<=N;i++){
        if(!notpri[i]){
            pri[++pri[0]]=i;
            mu[i]=-1;
        }
        for(int j=1;j<=pri[0];j++){
            if((ll)i*(ll)pri[j]>(ll)N){
                break;
            }
            notpri[i*pri[j]]=1;
            mu[i*pri[j]]=-mu[i];
            if(i%pri[j]==0){
                mu[i*pri[j]]=0;
                break;
            }
        }
    }
    sum[1]=mu[1];
    for(int i=2;i<=N;i++){
        sum[i]=sum[i-1]+mu[i];
    }
}
int n;
int main(){
	shai_mu();
	scanf("%d",&n);
	ll ans=0;
	for(int i=2;i<=n;++i){
		if(!notpri[i]){
			int nn=n/i;
			for(int j=1;j<=n/i;){
				ans+=(ll)(sum[nn/(nn/j)]-sum[j-1])*(ll)(nn/j)*(ll)(nn/j);
				j=nn/(nn/j)+1;
			}
		}
	}
	printf("%lld\n",ans);
	return 0;
}