17暑假多校联赛4.9 HDU 6075 Questionnaire

Questionnaire

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others) 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

  题目网址：http://acm.hdu.edu.cn/showproblem.php?pid=6075  

分析

题意：找出一对m和k，满足给出的数中对m取模等于k的数不少于对m取模不等于k的数 因为没规定m和k的取值，所以直接对2取模，取模等于1的少的话，就让k为0，反之k为1  

代码

#include <iostream> #include <cstdio> using namespace std; int a[100001]; int main() { int T; scanf("%d",&T); while(T--) { int n; scanf("%d",&n); int sum1=0,sum2=0; for(int i=0; i<n; i++) { scanf("%d",&a[i]); if(a[i]%2==1) sum1++; else sum2++; } if(sum1>=sum2) printf("2 1\n"); else printf("2 0\n"); } return 0; }