HDU6075 Questionnaire(思路，2017 HDU多校联赛 第4场)

题目：

Questionnaire

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 26    Accepted Submission(s): 20
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

1623 3 18 8 13 9

 

Sample Output

5 3

 

Source

2017 Multi-University Training Contest - Team 4

 

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思路：

这题问的是给你n个数，你要找到一个m和k，使得x%m==k,要在这些数里面有至少n/2个满足条件。

刚开始我们做的时候很捉鸡，无处下手，最后发现任何数模2不是0就是1，所以统计一下奇数和偶数的个数。。然后直接输出就好

代码：

#include <cstdio>#include <cstring>#include <cctype>#include <string>#include <set>#include <iostream>#include <stack>#include <cmath>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define N 300#define ll long longusing namespace std;int main(){int t,x,n;scanf("%d",&t);while(t--){int sum1=0,sum2=0;scanf("%d",&n);while(n--){scanf("%d",&x);if(x&1)sum1++;else sum2++;}if(sum1>sum2) puts("2 1");else puts("2 0");}return 0;}