hdu 6075 Questionnaire【贪心】

原创于 2018-11-14 21:23:01 发布 · 269 阅读

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本文介绍了一种竞赛编程策略，通过分析团队成员的选择，找到一组数中满足特定条件的m和k，使得a[i]%m==k的数量大于等于a[i]%m!=k的数量。提供了一种简单有效的解决方案，即比较奇数和偶数个数，选择更优的一方。

题目链接

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

23 3 18 8 13 9

Sample Output

5 3

Source

2017 Multi-University Training Contest - Team 4

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题意：

给你n个数，找出一对数m,k，满足这a[i]%m==k的个数大于等于a[i]%m!=k，输出任意一种答案

分析：

判断这n个数偶数和奇数的个数，选择%2之后余为1还是0，若偶数多k为0，否则为1

代码：

#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);
    }
}