HDU6075-Questionnaire

Questionnaire

                                                                     Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
                                                                                               Total Submission(s): 653    Accepted Submission(s): 447
                                                                                                                             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
  

 

Source

2017 Multi-University Training Contest - Team 4

 

题意：给出n个数，让你找出一对m和k，使得n个数对m取模的答案中k是不小于n/2

解题思路：m取2就很好解决这个问题了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
const LL mod = 998244353;

int n;

int main()
{
	int t;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%d", &n);
		int sum1 = 0, sum2 = 0, a;
		for (int i = 0; i < n; i++)
		{
			scanf("%d", &a);
			if (a % 2) sum1++;
			else sum2++;
		}
		if (sum1 >= sum2) printf("2 1\n");
		else printf("2 0\n");
	}
	return 0;
}