Min25筛求素数和

复杂度n^0.75/logn，实际上比n^2/3的杜教筛要快一些。

//#include <bits/stdc++.h>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>

using namespace std;

const int N = 1000010;

typedef long long LL;
LL TT,nn,k;
namespace Min25 {

    int prime[N], id1[N], id2[N], flag[N], ncnt, m;

    LL g[N], sum[N], a[N], T, n;
    inline void fff()
    {
        for(int i=0;i<=N;i++){
            prime[i]=0;
            id1[i]=0;
            id2[i]=0;
            flag[i]=0;
            g[i]=0;
            sum[i]=0;
            a[i]=0;
        }
        ncnt=0;
        m=0;
        T=0;
        n=0;
    }
    inline int ID(LL x) {
        return x <= T ? id1[x] : id2[n / x];
    }

    inline LL calc(LL x) {
        return x * (x + 1) / 2 - 1;
    }

    inline LL f(LL x) {
        return x;
    }

    inline void init() {
        T = sqrt(n + 0.5);
        for (int i = 2; i <= T; i++) {
            if (!flag[i]) prime[++ncnt] = i, sum[ncnt] = sum[ncnt - 1] + i;
            for (int j = 1; j <= ncnt && i * prime[j] <= T; j++) {
                flag[i * prime[j]] = 1;
                if (i % prime[j] == 0) break;
            }
        }
        for (LL l = 1; l <= n; l = n / (n / l) + 1) {
            a[++m] = n / l;
            if (a[m] <= T) id1[a[m]] = m; else id2[n / a[m]] = m;
            g[m] = calc(a[m]);
        }
        for (int i = 1; i <= ncnt; i++)
            for (int j = 1; j <= m && (LL)prime[i] * prime[i] <= a[j]; j++)
                g[j] = g[j] - (LL)prime[i] * (g[ID(a[j] / prime[i])] - sum[i - 1]);
    }

    inline LL solve(LL x) {
        if (x <= 1) return x;
        return n = x, init(), g[ID(n)];
    }

}

void extend_gcd(LL a,LL b,LL &x,LL &y)
{
    if(b==0) {
        x=1,y=0;
        return;
    }
    extend_gcd(b,a%b,x,y);
    LL tmp=x;
    x=y;
    y=tmp-(a/b)*y;
}
LL mod_inverse(LL a,LL m)
{
    LL x,y;
    extend_gcd(a,m,x,y);
    return (m+x%m)%m;
}
/*int main()
{
    LL sum;
    cin>>TT;
    while(TT--){
        sum=0;

        scanf("%lld%lld",&nn,&k);
        if(nn==1){
            printf("0
");
        }else if(nn==2){
            printf("6
");
        }else{
        LL a=mod_inverse(2,k);
        //cout<<a<<endl;
        sum=(((((nn%k)*(nn%k))%k+(nn*3)%k)%k)*a)%k;
        //cout<<sum<<endl;
        Min25::fff();
        sum=(sum-4+Min25::solve(nn+1))%k;
        printf("%lld
",sum%k);
        }
    }
    return 0;
}*/
int main()
{
    int n;
    scanf("%d",&n);
    printf("%d
",Min25::solve(n));
}