杭电Questionnaire_杭电距离向量算法(d-v)-优快云博客

本文链接：https://blog.youkuaiyun.com/ESESZB/article/details/76616884

本文主要探讨了杭州电子科技大学的一项问卷调查，详细分析了收集到的数据，揭示了学生在学术、生活和校园活动方面的观点和反馈。

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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

题意：给你n个数，让你找一对数，m,k，使得这n个数 模上m等于k的个数 大于等于 模上m不等于k的个数   m>1  0<=k<m
思路：所有的数模上2之后为0或者1

AC代码如下：
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=1e5+10;
int a[maxn];

int main()
{
    int t;
    int n;
    cin>>t;
    while(t--)
    {
        cin>>n;
        int num=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            if(a[i]%2==1) num++;
        }
        if(num>=n-num) cout<<2<<" "<<1<<endl;
        else
            cout<<2<<" "<<0<<endl;
    }
    return 0;
}