HDU 6075 Questionnaire

Questionnaire

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 274    Accepted Submission(s): 212
Special Judge

Problem Description

In order to get better results in official ACM/ICPC contests, the team leader comes up with a questionnaire. He asked everyone in the team whether to have more training.



Picture from Wikimedia Commons


Obviously many people don't want more training, so the clever leader didn't write down their words such as ''Yes'' or ''No''. Instead, he let everyone choose a positive integer  ai  to represent his opinion. When finished, the leader will choose a pair of positive interges  m(m>1)  and  k(0≤k<m) , and regard those people whose number is exactly  k  modulo  m  as ''Yes'', while others as ''No''. If the number of ''Yes'' is not less than ''No'', the leader can have chance to offer more training.

Please help the team leader to find such pair of  m  and  k .

 

Input

The first line of the input contains an integer  T(1≤T≤15) , denoting the number of test cases.

In each test case, there is an integer  n(3≤n≤100000)  in the first line, denoting the number of people in the ACM/ICPC team.

In the next line, there are  n  distinct integers  a1,a2,...,an(1≤ai≤109) , denoting the number that each person chosen.

 

Output

For each test case, print a single line containing two integers  m  and  k , if there are multiple solutions, print any of them.

 

Sample Input

  

   1
6
23 3 18 8 13 9
  

 

Sample Output

  

   5 3
  

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6075

题意：输入n个数a[i](1<=i<n)，输出满足a[i]%m=k的m和k（任意一组即可）。

解题思路：最开始看到题目有点懵，但是转念一想，任意一组，那么对2取余就好了。因为对2取余只有1和0两种，而一个数不是奇数就是偶数。

AC代码：

#include<cstdio>
#include<algorithm>
using namespace std;
int main(void)
{
	int T;
	int n;
	int odd,even;
	scanf("%d",&T);
	while(T--)
	{
		odd=0;//统计奇数个数 
		even=0;//统计偶数个数 
		scanf("%d",&n);
		for(int i=0;i<n;i++)
		{
			int num;
			scanf("%d",&num);
			if(num%2==1) odd++;
			else even++;
		}
		if(odd>even) printf("2 1\n");
		else printf("2 0\n");
	}
	return 0;
}