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  • codeforces 637A A. Voting for Photos(水题)

    题目链接:

    A. Voting for Photos

    time limit per test

    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    After celebrating the midcourse the students of one of the faculties of the Berland State University decided to conduct a vote for the best photo. They published the photos in the social network and agreed on the rules to choose a winner: the photo which gets most likes wins. If multiple photoes get most likes, the winner is the photo that gets this number first.

    Help guys determine the winner photo by the records of likes.

    Input

    The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the total likes to the published photoes.

    The second line contains n positive integers a1, a2, ..., an (1 ≤ ai ≤ 1 000 000), where ai is the identifier of the photo which got the i-th like.

    Output

    Print the identifier of the photo which won the elections.

    Examples
    input
    5
    1 3 2 2 1
    output
    2
    input
    9
    100 200 300 200 100 300 300 100 200
    output
    300
    Note

    In the first test sample the photo with id 1 got two likes (first and fifth), photo with id 2 got two likes (third and fourth), and photo with id 3got one like (second).

    Thus, the winner is the photo with identifier 2, as it got:

    • more likes than the photo with id 3;
    • as many likes as the photo with id 1, but the photo with the identifier 2 got its second like earlier.

    AC代码:

    #include <bits/stdc++.h>
    using namespace std;
    const int N=1e6+1;
    int flag[N],x[1005];
    int main()
    {
        int n;
        scanf("%d",&n);
        memset(flag,0,sizeof(flag));
        for(int i=0;i<n;i++)
        {
            scanf("%d",&x[i]);
            flag[x[i]]++;
        }
        int ans=0;
        for(int i=0;i<n;i++)
        {
            if(flag[x[i]]>ans)ans=flag[x[i]];
        }
        memset(flag,0,sizeof(flag));
        for(int i=0;i<n;i++)
        {
            flag[x[i]]++;
            if(flag[x[i]]==ans)
            {
                cout<<x[i]<<"
    ";
                break;
            }
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zhangchengc919/p/5278969.html
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