zoukankan      html  css  js  c++  java
  • Ace of Aces(含map做法)

    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).

    <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.

    <h4< dd="">Sample Input

    3
    5
    2 2 2 1 1
    5
    1 1 2 2 3
    1
    998
    

    <h4< dd="">Sample Output

    2
    Nobody
    998
    #include <iostream>
    #include <string.h>
    using namespace std;
    #include <stdio.h>
    int main()
    {
        int t,n,a[10000],i,flag[3333]={0};
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&n);
            memset(a,0,sizeof(a));
            memset(flag,0,sizeof(flag));
            for(i=0;i<n;i++)
            {
                scanf("%d",&a[i]);
                flag[a[i]]++;
            }
            int maxx=0,maxx2=0,c=0,d=0;
            for(i=0;i<3333;i++)
            {
                if(flag[i]>maxx)
                {
                    maxx=flag[i];
                    c=i;
                }
            }
            for(i=0;i<3333;i++)
            {
                if(flag[i]>maxx2&&i!=c)
                {
                    maxx2=flag[i];
                    d=i;
                }
            }
            if(maxx==maxx2)
                printf("Nobody
    ");
            else
                printf("%d
    ",c);
        }
        return 0;
    }

    下面是map做法

    #include <stdio.h>
    #include <algorithm>
    #include <iostream>
    #include <map>
    using namespace std;
    map<int,int> mp;
    int main()
    {
        int t,n,i,a;
        map<int,int>::iterator it;
        scanf("%d",&t);
        while(t--)
        {
            int k=0,ans=0,m=0;
            scanf("%d",&n);
            mp.clear();
            for(i=0;i<n;i++)
            {
                scanf("%d",&a);
                mp[a]++;
                k=max(mp[a],k);
            }
            for(it=mp.begin();it!=mp.end();it++)
            {
                if(it->second==k)
                {
                    ans=it->first;
                    m++;
                }
            }
            if(m==1)
                printf("%d
    ",ans);
            else
                printf("Nobody
    ");
        }
        return 0;
    }
  • 相关阅读:
    51NOD 1773:A国的贸易——题解
    BZOJ4553:[HEOI2016/TJOI2016]序列——题解
    BZOJ4597:[SHOI2016]随机序列——题解
    BZOJ1858:[SCOI2010]序列操作——题解
    BZOJ5157 & 洛谷3970:[TJOI2014]上升子序列——题解
    BZOJ3173:[TJOI2013]最长上升子序列 & HDU3564:Another LIS——题解
    BZOJ4755: [JSOI2016]扭动的回文串——题解
    洛谷2000:拯救世界——题解
    PPP中的PAP和CHAP的区别
    Linux C 实现一个简单的线程池
  • 原文地址:https://www.cnblogs.com/zcy19990813/p/9702809.html
Copyright © 2011-2022 走看看