ICPC 国内予選2011A チェビシェフの定理 AOJ 1172

解法：エラトステネス、累積和
やるだけ。難易度-5

#include <bits/stdc++.h>
#define chmin(a, b) ((a)=min((a), (b)))
#define chmax(a, b) ((a)=max((a), (b)))
#define fs first
#define sc second
#define eb emplace_back
using namespace std;
 
typedef long long ll;
typedef pair<int, int> P;
typedef tuple<int, int, int> T;
 
const ll MOD=1e9+7;
const ll INF=1e18;
 
int dx[]={1, -1, 0, 0};
int dy[]={0, 0, 1, -1};

int sum[250005];
bool prime[250005];

int main(){
    prime[1] = true;
    int a = 0;
    for(int i = 2;i < 250000;i++){
        if(prime[i] == false){
            a++;
            for(int j = 1;i*j < 250000;j++){
                prime[i*j] = true;
            }
        }
        sum[i] = a;
    }
    while(true){
        int n;
        cin >> n;
        if(n == 0) break;
        cout << sum[n*2] - sum[n] << endl;
    }
}