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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;

}