求小于N的素数个数（N>1e9）

#include<bits/stdc++.h>
#define il inline
#define pb push_back
#define ms(_data,v) memset(_data,v,sizeof(_data))
#define SZ(a) int((a).size())
using namespace std;
typedef long long ll;
const ll inf=0x3f3f3f3f;
const int N=320005;
//il int Add(int &x,ll y) {return x=x+y>=mod?x+y-mod:x+y;}
//il int Mul(int &x,ll y) {return x=x*y>=mod?x*y%mod:x*y;}
ll phi[10005][105], p2[N], ans[N];
int len, vis[N];
void init() {
	len = 0;
	for(int i=2; i<N; i++) {
		if(!vis[i]) {
			for(int j=i+i; j<N; j+=i) vis[j]=1;
			p2[len++] = i;
			ans[i] = ans[i-1]+1;
			continue;
		}
		ans[i] = ans[i-1];
	}
	for(int i=0; i<=10000; i++) {
		phi[i][0] = (ll)i;
		for(int j=1; j<=100; j++) {
			phi[i][j] = phi[i][j-1] - phi[i/p2[j-1]][j-1];
		}
	}
}
ll solve_phi(ll m, ll n) {
	if(!n) return m;
	if(p2[n - 1] >= m) return 1;
	if(m<=10000 && n<=100) return phi[m][n];
	return solve_phi(m, n-1) - solve_phi(m/p2[n-1], n-1);
}
ll solve_p2(ll m) {
	if(m < (ll)N) return ans[m];
	ll y = (int)cbrt(m*1.0);
	ll n = ans[y];
	ll sum = solve_phi(m, n) + n -1;
	for(ll i=n; p2[i]*p2[i]<=m; i++)
		sum = sum - solve_p2(m/p2[i])+solve_p2(p2[i])-1;
	return sum;
}
int main(){
    init();
    scanf("%lld", &n);
    printf("%lld\n",solve_p2(n));
    return 0;
}