HDU 6075 Questionnaire

Questionnaire

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)

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 integerai to represent his opinion. When finished, the leader will choose a pair of positive intergesm(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 integersm and k, if there are multiple solutions, print any of them.

 

Sample Input

1623 3 18 8 13 9

 

Sample Output

5 3

题意：

n个人分别给出n个数a1,a2,...,an(1≤ai≤109),让你找出m和k使得符合 ai %m = k的ai的个数多于不符合此条件的个数

解题思路

完全是脑筋急转弯啊。直接令m = 2，余数要么是1要么是0 ，如果奇数多就让k = 1 如果偶数多就让k=0，奇数偶数相等则随便

代码

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>using namespace std;const int maxn = 100000+10;int a[maxn];int main(){    int t;    scanf("%d",&t);    while(t--){        int n;        int odd = 0;        int even = 0;        scanf("%d",&n);        for(int i = 0;i<n;++i){            scanf("%d",&a[i]);            if(a[i]%2) odd++;            else even++;        }        if(even>=odd) printf("2 0\n");        else printf("2 1\n");    }    return 0;}