題目連結

Questionnaire

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1204    Accepted Submission(s): 800
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

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題意：

給你n個數，找出一對數m,k，滿足這a[i]%m==k的個數大於等於a[i]%m!=k，輸出任意一種答案

分析：

判斷這n個數偶數和奇數的個數，選擇%2之後餘為1還是0，若偶數多k為0，否則為1

程式碼：

#include<bits/stdc++.h>
#define ll long long
#define mod 1000000007
using namespace std;
int a[100005];
int main()
{
    int t,n,i,j,aa,bb,x;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        aa=bb=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&x);
            if(x%2)
                aa++;
            else
                bb++;
        }
        if(aa>=bb)
            i=1;
        else
            i=0;
        printf("2 %d\n",i);
    }
}