A - Ace of Aces ZOJ - 3869 -mystical_curve

最新推荐文章于 2020-09-06 21:49:48 发布

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, 在宇宙深处有一个名为时空管理局的神秘组织，今年决定从精英部队中选举一名成员授予Ace of Aces称号。根据有效选票上的候选人ID，通过编程找出获得最多提名的候选人。若出现平局，则无人当选。

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).

<h4< dd="">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：根据每组样列，求出某个数最多的那个，如果有多个一样的，就输出Nobody

解题思路：根据题中所给的数据范围，可以开一个数组存储每个个之出现的次数最后遍历一遍数组出现最多的就

为答案。

#include <iostream>
#include <cstring>
#include <vector>
#include <cstdio>
#include <cmath>
#include <algorithm>
#define pb push_back
#define mk make_pair
typedef long long ll;
using namespace std;
const int inf = 1100;
int s[inf];
int main()
{
    int t;cin>>t;
    while(t--)
    {
        int n;cin>>n;
        memset(s,0,sizeof(s));
        for(int i=0;i<n;i++)
        {
            int a;cin>>a;
            s[a]++;
        }
        /*for(int i=1;i<10;i++)
            cout<<s[i]<<" ";
        cout<<endl;*/
        int k=0,ans=0;
        for(int i=1;i<inf;i++)
        {
            if(s[i]>s[ans]){
                ans=i;k=0;
                //cout<<"-----"<<i<<endl;
            }
            else if(s[i]==s[ans]){
                k=1;//cout<<i<<"======"<<endl;
            }
        }
        if(!k){
            printf("%d\n",ans);
        }else
        {
            printf("Nobody\n");
        }
    }
    return 0;
}