Questionnaire

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 326 Accepted Submission(s): 253
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

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

Source

给出的数不是奇数就是偶数，判断一下就好，当时想太多了，浪费了很多时间。

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          #include 
           
            #define INF 0x3f3f3f3f#define ms(x,y) memset(x,y,sizeof(x))using namespace std;typedef long long ll;const double pi = acos(-1.0);const double eps = 1e-8;const int maxn = 1e6 + 10;int main(){ int t; scanf("%d", &t); while (t--) { int n; scanf("%d", &n); int ans = 0; for (int i = 0; i < n; i++) { int p; scanf("%d", &p); if (p % 2 == 0) { ans++; } } if (ans < n / 2) { printf("2 1\n"); } else printf("2 0\n"); }}

转载于:https://www.cnblogs.com/Archger/p/8451615.html

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