Recover it!(映射)

1. 先将a数组复制进b数组
2. 再从1~n判定ai
   • 如果ai是质数，那么添加第ai个质数进入b数组
   • 如果不是，那么添加ai除本身外的最大因数进入b数组

//#include<bits/stdc++.h>
#include<algorithm>
#include <iostream>
#include   <fstream>
#include  <cstdlib>
#include  <cstring>
#include  <cassert>
#include   <cstdio>
#include   <vector>
#include   <string>
#include    <cmath>
#include    <queue>
#include    <stack>
#include      <set>
#include      <map>
using namespace std;
#define P(a,b,c) make_pair(a,make_pair(b,c))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n;i>=a;i--)
#define CLR(vis) memset(vis,0,sizeof(vis))
#define MST(vis,pos) memset(vis,pos,sizeof(vis))
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef pair<int,pair<int,int> >pii;
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1000000007;
const int INF = 0x3f3f3f3f;
ll gcd(ll a, ll b) {
	return b ? gcd(b, a%b) : a;
}
template<class T>inline void gmax(T &A, T B){
    (A<B)&&(A=B);//if(B>A)A=B;
}
template<class T>inline void gmin(T &A, T B){
    (A>B)&&(A=B);//if(B<A)A=B;
}
template <class T>
inline bool scan_d(T &ret) {
	char c;
	int sgn;
	if(c=getchar(),c==EOF) return 0; //EOF
	while(c!='-'&&(c<'0'||c>'9')) c=getchar();
	sgn=(c=='-')?-1:1;
	ret=(c=='-')?0:(c-'0');
	while(c=getchar(),c>='0'&&c<='9') ret=ret*10+(c-'0');
	ret*=sgn;
	return 1;
}
inline void outt(int x) {
	if(x>9) outt(x/10);
	putchar(x%10+'0');
}
const int MAXN = 1e6+10; 
int vis[2750132];
int prim[2750132];
void prime(){
	int cnt = 0;
	for(int i=2;i<=2750131;i++){
		if(!vis[i]){
			prim[i]=++cnt;
			for(int j =i*2;j<=2750131;j+=i){
				if(!vis[j])vis[j]=1;
			}
		}
	}
}
int b[MAXN];
int a[MAXN];
int num[2750132];
bool cmp(int a,int b){
	return a>b;
}
int main(){
	prime();
	int n;
	cin>>n;
	rep(i,1,n*2){
		scanf("%d", &b[i]);
		++num[b[i]];
	}
	sort(b+1,b+n*2+1,cmp);
	rep(i,1,n*2){
		if(num[b[i]]){
			if(!vis[b[i]]){
				num[b[i]]--;
				num[prim[b[i]]]--;
				//cout<<b[i]<<" "<<prim[b[i]]<<endl;
				printf("%d ",prim[b[i]]);
			}else{
				printf("%d ", b[i]);
				--num[b[i]];
				int x;
				rep(j,2,b[i]){
					if(b[i]%j==0){
						x = b[i]/j;
						break;
					}
				}
				--num[x];
			}
		}
	}
	return 0;
}