A - Divisors of Two Integers CodeForces - 1108B

最新推荐文章于 2020-01-17 19:28:09 发布

秦曼巴

最新推荐文章于 2020-01-17 19:28:09 发布

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本文介绍了一种算法，用于从一组数字中找出两个数，使这组数字成为这两个数的所有因数的组合。通过确定最大数和另一个不能被最大数整除的最大数，解决了这一问题。

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特别水的题，可惜我还是不会做

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
Input
10
10 2 8 1 2 4 1 20 4 5
Output
20 8

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大致意思是这样的，给出一连串的数字，让你找出两个数，使得其他的数都是这两个数的因数！！！

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思路就更简单了：
首先最大的那个数肯定是一个，所以只需要找出另一个就可以了，另一个数就是不能被第一个数整除的最大的那个数，并且第二个数不和与它相邻的数相等。

再找第二个数的时候，可以再定义一个数组，遍历给出的数，如果是第一个数的因数且不和相邻的数相等，就把第二个数组对应的值设为1；其他的设为0；那么第二个就找到了。

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AC 代码

#include<stdio.h>
#include<iostream>
#include<stack>
#include<vector>
#include<map>
#include<string.h>
#include<set>
#include<algorithm>
#include<list>
#include<queue>
#include<string>

using namespace std;

int main()
{
	int n, a, b, x[100010], d[100010];
	while (scanf("%d", &n) != EOF)
	{
		for (int i = 0; i < n; i++)
		{
			cin >> x[i];
		}
		sort(x, x + n);
		a = x[n - 1];
		b = 0;
		for (int i = 0; i < n; i++)
		{
			if (a%x[i] == 0 && x[i] != x[i + 1])
			{
				d[i] = 1;
			}
			else
				d[i] = 0;
		}
		for (int i = n - 2; i >= 0; i--)
		{
			if (!d[i])
			{
				b = x[i];
				break;
			}
		}
		cout << a << " " << b << endl;
	}
	return 0;
}