hdu 多校联赛 Questionnaire_when finished, the leader will choose a pair of po-优快云博客

本文介绍了一道竞赛编程题目，旨在寻找一种策略，通过选取特定的参数来确定团队成员是否支持增加训练强度。输入包括多组测试数据，每组数据包含一组整数，目标是找到合适的参数m和k，使得符合特定条件的人数不少于不符合该条件的人数。

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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

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

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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