div2题目题解：Divisors of Two Integers（桶记录数值的普通思维题）

题目：

我这个题目直接复制粘贴的，所以会有显示上的问题；
B. Divisors of Two Integers
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Recently you have received two positive integer numbers x and y. You forgot them, but you remembered a shuffled list containing all divisors of x (including 1 and x) and all divisors of y (including 1 and y). If d is a divisor of both numbers x and y at the same time, there are two occurrences of d in the list.

For example, if x=4 and y=6 then the given list can be any permutation of the list [1,2,4,1,2,3,6]. Some of the possible lists are: [1,1,2,4,6,3,2], [4,6,1,1,2,3,2] or [1,6,3,2,4,1,2].

Your problem is to restore suitable positive integer numbers x and y that would yield the same list of divisors (possibly in different order).

It is guaranteed that the answer exists, i.e. the given list of divisors corresponds to some positive integers x and y.

Input
The first line contains one integer n (2≤n≤128) — the number of divisors of x and y.

The second line of the input contains n integers d1,d2,…,dn (1≤di≤104), where di is either divisor of x or divisor of y. If a number is divisor of both numbers x and y then there are two copies of this number in the list.

Output
Print two positive integer numbers x and y — such numbers that merged list of their divisors is the permutation of the given list of integers. It is guaranteed that the answer exists.

Example
inputCopy
10
10 2 8 1 2 4 1 20 4 5
outputCopy
20 8

题意：

题意就是：然后给出一个由两个数a,b的所有因子组成的序列，然后让你求出这两个数，我们用桶记录这个序列里面的数就好了，然后拿出最大的那个数一定就是ab其中的一个，这个最大数取出所有因子去桶那里抵消一下就好了。

完整代码实现：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <climits>
#include <queue>
#include <stack>
#include <map>
//鬼畜头文件
using namespace std;
#define INF 0x3f3f3f3f
#define ULL unsigned long long
#define LL long long
//鬼畜define
int n;
int tong[10001];
int main()
{
 
	scanf("%d",&n);
	fill(tong,tong+n,0);
	int _MAX=-1;
	long long sum=1;
	for(int time=0;time<n;time++)
	{
		int num;
		scanf("%d",&num);
		tong[num]++;
		if(_MAX==-1||_MAX<num)_MAX=num;
	}
	//input finished;
	int time;
	for(time=1;time*time<_MAX;time++)
	{
		if(_MAX%time==0){tong[time]--;tong[_MAX/time]--;}
	}
	if(time*time==_MAX)tong[time]--;
	int ans=1;
	for(time=_MAX;time>=1;time--)
	{
		if(tong[time]==1){ans=time;break;}
	}
	printf("%d %d\n",_MAX,ans);
 
    return 0;
}