(hdu6075) 2017杭电多校联赛第四场-Questionnaire 思维题

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 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

1
6
23 3 18 8 13 9

 

Sample Output

5 3

题目大意：老师想要给学生进行acm训练，但是有很多学生不想训练，所以教练让每个学生选择一个正整数，然后需要我们找到一组m和k，使得学生选择的数字整除m之后等于k的数尽可能的多，这样才可以让学生继续参加训练。
解题思路：因为题目要求的m和k不是唯一确定的，只需要我们找到能满足题意的就可以了，这里我选择的是奇数和偶数，因为一个正整数要么是奇数要么就是偶数，所以我们选择的m是2，所以k值我们需要确定是0还是1就可以了，我们只需要处理每一个数，判断是奇数多还是偶数多，如果奇数多k值就为1，否则k就是0.
ac代码：

#include <cstdio>
int main()
{
int T,n,a,sum1,sum2;
scanf("%d", &T);
while(T--)
{
sum1=sum2=0;
scanf("%d", &n);
while(n--)
{
scanf("%d",&a);
if(a&1) sum2++;
else sum1++;
}
if(sum1>=sum2)
printf("%d %d\n",2,0);
else 	printf("%d %d\n",2,1);
}
return 0;
}