Codeforces 1333 F. Kate and imperfection(埃筛)

• 对n个数，我们考虑它对整个gcd集合的贡献，对于质数，它只能贡献1,对于合数,它所有的因子都在集合中(gcd(A,A的因子)= A的因子)。要输出选i个数中的最小的gcd最大值。那就把前i个数的的贡献从小到大排序，输出。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<map>
#include<queue>
#include<vector>
#include<string>
#include<fstream>
 
using namespace std;
typedef long long ll;
const int N = 5e5 + 105;
const int mod = 998244353;
const double Pi = acos(- 1.0);
const int INF = 0x3f3f3f3f;
const int G = 3, Gi = 332748118;
 

int vis[N * 2];
int res[N * 2];
void prime(){
    vis[1] = 1;
    vis[2] = 0;
    for(int i = 2; i < N; ++ i){
        if(!vis[i]){
            res[i] = 1;
            for(int j = i + i; j < N; j += i){
                vis[j] = 1;
                res[j] = max(res[j], j / i);
            }
        }
    }
}
//
int n;
int main()
{   
    prime();
    res[1] = 1;
    scanf("%d",&n);
    sort(res + 1, res + n + 1);
    for(int i = 2; i <= n; ++ i){
        if(i == n) printf("%d
", res[i]);
        else printf("%d ",res[i]);
    }
    return 0;
}