2017 Multi-University Training Contest

题目：

Questionnaire

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1585    Accepted Submission(s): 914
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 integerT(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：求出m和k，使得给定数组的过半的元素值%m=k。签到题

CODE：

#include<bits/stdc++.h>using namespace std;int a[100002];int main(){        int t,n,i,j;        scanf("%d",&t);        while(t--){                scanf("%d",&n);                int x=0,y=0;                for(i=0;i<n;i++){                        scanf("%d",&a[i]);                        a[i]%2?x++:y++;                }                printf("2 %d\n",x>y?1:0);        }        return 0;}