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  • 第十二届浙江省大学生程序设计大赛-Ace of Aces 分类: 比赛 2015-06-26 14:25 12人阅读 评论(0) 收藏

    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 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”.
    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).
    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

    #include <bits/stdc++.h>
    
    using namespace std;
    
    struct node
    {
        int sum;
        int pos;
    }po[1100];
    
    bool cmp(node a,node b)
    {
        return a.sum>b.sum;
    }
    
    int main()
    {
        int t,n,i,a;
        bool flag;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&n);
            flag=true;
            memset(po,0,sizeof(po));
            for(i=0; i<n; i++)
            {
                scanf("%d",&a);
                po[a].sum++;
                po[a].pos=a;
    
            }
            sort(po,po+1001,cmp);
            for(i=1 ;i<=1000 ;i++)
            {
                if(po[i].sum==po[0].sum)
                {
                    flag=false;
                    break;
                }
    
            }
            //cout<<po[0].pos<<endl;
            if(flag)
            {
                printf("%d
    ",po[0].pos);
            }
            else
            {
                printf("Nobody
    ");
            }
        }
        return 0;
    }

    版权声明:本文为博主原创文章,未经博主允许不得转载。

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  • 原文地址:https://www.cnblogs.com/juechen/p/4721979.html
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