https://codeforces.com/contest/1176/problem/D

题解：线性筛+暴力

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=4000000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N],b[N],c[N];
int prime[N];
int pre[N];
int maxd[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    prime[0]=prime[1]=1;
    for(int i=2;i<N;i++){
        if(!prime[i]){
            pre[++cnt]=i;
            maxd[i]=i;
        }
        for(int j=1;j<=cnt&&i*pre[j]<N;j++){
            prime[i*pre[j]]=1;
            maxd[i*pre[j]]=min(maxd[i],maxd[pre[j]]);
            if(i%pre[j]==0)break;
        }
    }
    scanf("%d",&n);
    for(int i=1;i<=2*n;i++){
        scanf("%d",&a[i]),b[a[i]]++;
    }
    sort(a+1,a+2*n+1,greater<int>());
    for(int i=1;i<=2*n;i++){
        if(prime[a[i]]&&b[a[i]]&&b[a[i]/maxd[a[i]]]){
            b[a[i]/maxd[a[i]]]--;
            b[a[i]]--;
            c[++sum]=a[i];//cout<<a[i]<<endl;
        }else if(!prime[a[i]]&&a[i]<=199999&&b[a[i]]&&b[pre[a[i]]]){
            b[pre[a[i]]]--;//cout<<a[i]<<" "<<pre[a[i]]<<endl;;
            b[a[i]]--;
            c[++sum]=a[i];
        }//cout<<sum<<" "<<i<<endl;
    }
    for(int i=1;i<=n;i++){
        printf("%d%c",c[i]," \n"[i==n]);
    }
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}