【Codeforces Round #565 (Div. 3) D Recover it!】思维乱搞

题意 如果一个数是质数 他只能生成质数序列的第他本身个质数 否则会生成他的最大因子
首先我们要知道 肯定是把两个质数分开 然后我们发现 只有大的质数会被小的质数生成 比如 3 是 2 生成的 5 是 3 生成的 那么我们从大到小 找到一个质数就把生成他的质数放在答案序列里面 如果不是质数 那么我们就需要把他放进答案序列里面 消除他最大因子的贡献
一开始忘记 n是 2n re了一发
ans[0]和ans[2]是想要的答案

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

/*namespace sgt
{
    #define mid ((l+r)>>1)

    #undef mid
}*/
int arr[400025];
const int MAX_N = 2751150;
int prime[MAX_N];
int flag[MAX_N+5];
int tot,num[MAX_N];
map<int,int > mp;

void make_prime()
{
    for(int i = 2;i<=MAX_N;++i)
    {
        if(!flag[i])
        {
            prime[++tot] = i;
            mp[i] = tot;
            for(int j = i + i;j<=MAX_N;j+=i)
                flag[j] = 1;
        }
    }
}
vector<int > ans[4];
bool cmp(int a,int b)
{
    return a > b;
}
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n;
    make_prime();
    scanf("%d",&n);
    for(int i = 1;i<=2*n;++i)
    {
        scanf("%d",&arr[i]);
        num[arr[i]]++;
    }
    int cnt = 0;
    sort(arr+1,arr+1+2*n,cmp);
    for(int i = 1;i<=2*n;++i)
    {
        if(num[arr[i]])
        {
            if(!flag[arr[i]])
            {
            if(num[mp[arr[i]]])
            {
                num[arr[i]]-- ;
                num[mp[arr[i]]]--;
                ans[0].push_back(arr[i]);
                ans[1].push_back(mp[arr[i]]);
            }
            else
            {
                cnt++;
                num[arr[i]]--;
                ans[2].push_back(arr[i]);
            }
            }
            else
            {
                cnt++;
                num[arr[i]]--;
                ans[2].push_back(arr[i]);
                for(int j = 2;j*j<=arr[i];++j)
                {
                    if(arr[i]%j==0)
                    {
                        num[arr[i]/j]--;
                        ans[3].push_back(arr[i]/j);
                        break;
                    }
                }
            }
        }
    }
    if(ans[1].size())
    {
    for(int i = 0;i<ans[1].size();++i)
        printf("%d ",ans[1][i]);
    }
//    for(int i = 0;i<cnt;++i)
//        dbg2(i,ans[2][i]);
//    dbg(cnt);
    if(cnt)
    {
    for(int i = 0;i<cnt;++i)
        printf("%d ",ans[2][i]);
    }

    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}