HDU 6075 Questionnaire

Questionnaire

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

 

这道题表面很有迷惑性，但其实很水很水。给你六个数字，让你找到一个数字m，并将六个数字对m取模后出现次数最多的值视为k。输出这样的一组m，k。

刚开始的时候有点懵比，毕竟英语渣，是队友给我说的题意。后来队友想到既然可以取任意值，那直接对2取余就好了，相当于判断给定数字的奇偶性。记一下数字，输出数量多的就好了。

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;long long int a;int main(){    int T,n;    int res=0;    cin>>T;    while(T--)    {        cin>>n;        res=0;        for(int i=0;i<n;i++)        {            scanf("%lld",&a);            if(a&1)                res++;        }        if(res>=n-res)            cout<<2<<" "<<1<<endl;        else            cout<<2<<" "<<0<<endl;    }    return 0;}