题意

题目链接

Sol

min25筛的板子题，直接筛出\(g(N, \infty)\)即可

筛的时候有很多trick，比如只存\(\frac{N}{x}\)的值，第二维可以滚动数组滚动掉

#include<bits/stdc++.h>
#define LL long long
//#define int long long  
using namespace std;
const int MAXN = 2e6 + 10;
int Lim, vis[MAXN], prime[MAXN], tot;
LL N, g[MAXN], id[MAXN], cnt, pos1[MAXN], pos2[MAXN];
void Get(int N) {
    vis[1] = 1;
    for(int i = 2; i <= N; i++) {
        if(!vis[i]) prime[++tot] = i;
        for(int j = 1; j <= tot && i * prime[j] <= N; j++) {
            vis[i * prime[j]] = 1;
            if(!(i % prime[j])) break;
        }
    }
}
LL get(LL x) {
    return x <= Lim ? pos1[x] : pos2[N / x];
}
signed main() {
    cin >> N; Lim = sqrt(N);
    Get(Lim);
    for(LL i = 1, j; i <= N; i = N / j + 1) {
        j = N / i; id[++cnt] = j; g[cnt] = id[cnt] - 1;
        j <= Lim ? pos1[j] = cnt : pos2[N / j] = cnt;;
    }
    for(int j = 1; j <= tot; j++) 
        for(LL i = 1; 1ll * prime[j] * prime[j] <= id[i]; i++) 
            g[i] -= g[get(id[i] / prime[j])] - (j - 1);
    cout << g[1];
    return 0;
}