• (CF#257)A. Jzzhu and Children


    There are n children in Jzzhu's school. Jzzhu is going to give some candies to them. Let's number all the children from 1 to n. The i-th child wants to get at least ai candies.

    Jzzhu asks children to line up. Initially, the i-th child stands at the i-th place of the line. Then Jzzhu start distribution of the candies. He follows the algorithm:

    1. Give m candies to the first child of the line.
    2. If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home.
    3. Repeat the first two steps while the line is not empty.

    Consider all the children in the order they go home. Jzzhu wants to know, which child will be the last in this order?

    Input

    The first line contains two integers n, m (1 ≤ n ≤ 100; 1 ≤ m ≤ 100). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100).

    Output

    Output a single integer, representing the number of the last child.

    Sample test(s)
    input
    5 2
    1 3 1 4 2
    
    output
    4
    
    input
    6 4
    1 1 2 2 3 3
    
    output
    6
    
    Note

    Let's consider the first sample.

    Firstly child 1 gets 2 candies and go home. Then child 2 gets 2 candies and go to the end of the line. Currently the line looks like [3, 4, 5, 2] (indices of the children in order of the line). Then child 3 gets 2 candies and go home, and then child 4 gets 2 candies and goes to the end of the line. Currently the line looks like [5, 2, 4]. Then child 5 gets 2 candies and goes home. Then child 2 gets two candies and goes home, and finally child 4 gets 2 candies and goes home.

    Child 4 is the last one who goes home.


    水题,就不多说了,贴代码吧0.0
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    int a[110],num[110];
    int n,m;
    int main()
    {
        while(cin>>n>>m)
        {
            memset(num,0,sizeof(num));
            for(int i=1;i<=n;i++)
            {
                cin>>a[i];
                while(a[i]>0)
                {
                    a[i]-=m;
                    num[i]++;
                }
            }
            int k=1,maxn=0;
            for(int i=1;i<=n;i++)
            {
                if(num[i]>=maxn)//此处注意是>=
                {
                    k=i;
                    maxn=num[i];
                }
            }
            cout<<k<<endl;
        }
        return 0;
    }


  • 相关阅读:
    May LeetCoding Challenge3 之 HashMap的使用
    May LeetCoding Challenge2 之 HashSet的使用
    May LeetCoding Challenge1 之 标准二分查找法
    30-Day Leetcoding Challenge Day21
    30-Day Leetcoding Challenge Day19
    30-Day Leetcoding Challenge Day18
    ALV 单元格编辑回车响应
    ALV模版1
    web dynpro 开发配置
    convert fm
  • 原文地址:https://www.cnblogs.com/yjbjingcha/p/6844382.html
Copyright © 2020-2023  润新知