ZOJ 3869 -- Ace of Aces

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时间空间行政局决定通过投票选举出其精英部队中最具潜力的成员，荣获‘Ace of Aces’荣誉。该博客详细介绍了如何通过程序帮助行政局确定当选者。

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Ace of Aces

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 3869

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 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

Sample Output

2
Nobody
998

题意：求出出现次数最多的且是唯一的数，若有多个则输出 Nobody。

代码如下：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <map>
#include <functional>
#include <algorithm>
using namespace std;
#define N 1020
#define inf 0x3f3f3f3f
struct pp
{
	int num, v;
}p[N];
int Num[N];
int main()
{
#ifdef OFFLINE
	freopen("t.txt", "r", stdin);
#endif
	int t, i, j, k, n, m;
	scanf("%d", &t);
	while (t--){
		memset(p, 0, sizeof(p));
		memset(Num, 0, sizeof(Num));
		scanf("%d", &n);
		m = 0;
		for (i = 0; i < n; i++){
			scanf("%d", &k);
			Num[k]++;
		}
		if (n == 1){
			printf("%d\n", k); continue;
		}
		j = 0;
		for (i = 1; i <= 1000; i++){
			if (Num[i]){
				p[j].v = i;
				p[j++].num = Num[i];
			}
		}
		int max_1 = 0, max_v1 = 0, max_2 = 0, max_v2 = 0;
		for (i = 0; i<j; i++){
			if (p[i].num > max_1){//最大
				max_2 = max_1;
				max_v2 = max_v1;
				max_1 = p[i].num;
				max_v1 = p[i].v;
			}
			else if (p[i].num > max_2){//次大
				max_2 = p[i].num;
				max_v2 = p[i].v;
			}
		}
		if (max_1 == max_2)
			puts("Nobody");
		else
			printf("%d\n", max_v1);
	}
	return 0;
}