ZOJ3869 Ace of Aces【序列处理】

Ace of Aces

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

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Author: JIANG, Kai
Source: The 12th Zhejiang Provincial Collegiate Programming Contest

问题链接：ZOJ3869 Ace of Aces。

问题简述：参见上文。

问题分析：

这个题本质上是求出现次数最多的数，如果不存在则输出“Nobody”（没有最多次数，有一样多次数）。

好在值的范围不大（1<=ai<=1000），找1000个桶把相同的数装在一起（计数），然后找出其出现次数最多和第二多的数。

程序说明：（略）

题记：（略）

AC的C++语言程序如下：

/* ZOJ3869 Ace of Aces */

#include <iostream>
#include <stdio.h>
#include <string.h>

using namespace std;

const int N = 1000;
int count[N+1];

int main()
{
    int t, n, ai, maxa, first, second;
    scanf("%d", &t);
    while(t--) {
        memset(count, 0, sizeof(count));

        scanf("%d", &n);
        maxa = 0;
        first = second = 0;
        for(int i=1; i<=n; i++) {
            scanf("%d", &ai);
            count[ai]++;
            if(count[ai] >= first) {
                maxa = ai;
                second = first;
                first = count[ai];
            }
        }

        if(first == second)
            printf("Nobody
");
        else
            printf("%d
", maxa);
    }

    return 0;
}