hdu 多校联赛 Questionnaire

Questionnaire

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0
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

1623 3 18 8 13 9

 

Sample Output

5 3

 

本场比赛最简单的题 大多数人都做出来了 刚开始没看到any of them 郁闷了半天 后来才看到  只需要烤炉用2当作被除数就行了 然后看看每个数余几 记一下数量就好了 

虽然简单但是我还是tle了以此 用的全部是cin 可能是脸太黑把

ac代码：

#include<bits/stdc++.h>using namespace std;int main(){    int t;    cin>>t;    while(t--)    {        int n;        cin>>n;        int a;        int d=0,s=0;        for(int i=0;i<n;i++)        {            scanf("%d",&a);            if(a&1)            {                d++;            }            else            {                s++;            }        }        if(s>=d)        {            printf("2 0\n");        }        else        {            printf("2 1\n");        }    }    return 0;}

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