ZOJ 3869-Ace of Aces【模拟众数】

Ace of Aces Time Limit: 2 Seconds       Memory Limit: 65536 KB

There is a mysterious organization called Time-Space Administrative Bureau (TSAB) in the deep universe that we humans have not discovered yet. This year, the TSAB decided to elect an outstanding member from its elite troops. The elected guy will be honored with the title of "Ace of Aces".

After voting, the TSAB received N valid tickets. On each ticket, there is a number A_i denoting the ID of a candidate. The candidate with the most tickets nominated will be elected as the "Ace of Aces". If there are two or more candidates have the same number of nominations, no one will win.

Please write program to help TSAB determine who will be the "Ace of Aces".

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (1 <= N <= 1000). The next line contains N integers A_i (1 <= A_i <= 1000).

Output

For each test case, output the ID of the candidate who will be honored with "Ace of Aces". If no one win the election, output "Nobody" (without quotes) instead.

Sample Input

3
5
2 2 2 1 1
5
1 1 2 2 3
1
998

Sample Output

2
Nobody
998

Author:  JIANG, Kai

Source: The 12th Zhejiang Provincial Collegiate Programming Contest

解题思路：

找到出现次数最多的数，如果有多个就输出Nobody。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int map[50000];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		memset(map,0,sizeof(map));
		for(int i=0;i<n;i++)
		{
			int aa;
			scanf("%d",&aa);
			map[aa]++;	
		}
		int max=-1;
		int ans;
		int ge=0; 
		for(int i=1;i<=1200;i++)
		{
			if(map[i]>max)
			{
				max=map[i];
				ge=0;
				ans=i;
			}
			if(map[i]==max)
			{
				ge++;
			}
		}
		if(ge>1)
		{
			printf("Nobody\n");
		}
		else
		{
			printf("%d\n",ans);
		}
	}
	return 0;	
}