ZOJ - 3869 Ace of Aces (水)

本文介绍了一个编程竞赛题目——ZOJ-3869 AceofAces的问题背景、输入输出要求及示例,并提供了一段C++代码实现,用于找出票选中获得最多提名的候选人。
ZOJ - 3869
Time Limit: 2000MS Memory Limit: 65536KB 64bit IO Format: %lld & %llu

Status

Description

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 Ai 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".

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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 Ai (1 <= Ai <= 1000).

<h4< body="">

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.

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Sample Input

3
5
2 2 2 1 1
5
1 1 2 2 3
1
998
<h4< body="">

Sample Output

2
Nobody
998

Hint

Source

The 12th Zhejiang Provincial Collegiate Programming Contest
//题意:
找出出现次数最多的数,如果有多个输出 Nobody
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN = 1010;
int num[MAXN];
int p[MAXN];
int main(){
	int T, N;
	scanf("%d", &T);
	while(T--){
		memset(num, 0, sizeof(num));
		memset(p, 0, sizeof(p));
		scanf("%d", &N);
		int x;
		for(int i = 0; i < N; i++){
			scanf("%d", &x);
			num[x]++;
		}
		int ans = 0, temp = 0;
		for(int i = 0; i < MAXN; i++){
			if(num[i] >= temp){
				temp = num[i];
				ans = i;
				p[temp]++;
			}
		}
		if(p[num[ans]] > 1){
			puts("Nobody");
		}
		else
			printf("%d\n", ans);
	}
	return 0;
}


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