【2017多校第二场】HDU 6075 Questionnaire【思维】

Questionnaire

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 714    Accepted Submission(s): 487
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

原题链接：http://acm.hdu.edu.cn/showproblem.php?pid=6075

题意：每个人选择一个正整数，队长选择两个数m,k,对于每个人选择的数，如果对m取余，结果为k，则得到一个‘yes’，否则得到‘No’，如果最后‘Yes’的数量大于等于‘No’，则队长将得到更多的训练机会

问，队长如何选择m,k

一个数对m取余，结果为[0,m-1],所以我们取m=2，判断奇偶即可。

AC代码：

/**  * 行有余力，则来刷题！  * 博客链接:http://blog.csdn.net/hurmishine  * 个人博客网站:http://wuyunfeng.cn/*/#include <iostream>#include <cstdio>#include <cmath>using namespace std;const int maxn=800000+5;char a[maxn],b[maxn];int main(){    int T;    cin>>T;    int n,x,y;    while(T--)    {        scanf("%d%d%d",&n,&x,&y);        scanf("%s",a);        scanf("%s",b);        int cnt=0;        for(int i=0;i<n;i++)        {            if(a[i]!=b[i])                cnt++;        }        if(abs(x-y)<=cnt&&x+y<=2*n-cnt)            printf("Not lying\n");        else            printf("Lying\n");    }    return 0;}