• CodeForces 450A 队列


    Description

    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 Input

    Input
    5 2
    1 3 1 4 2
    
    Output
    4
    
    Input
    6 4
    1 1 2 2 3 3
    
    Output
    6
    

    Hint

    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.


    //队列

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<queue>
    
    using namespace std;
    
    struct node
    {
        int a,b;
    } p,t;
    int main()
    {
        queue<node> q;
        int n,m;
        while(~scanf("%d%d",&n,&m))
        {
            int top=-1;
            for(int i=0; i<n; i++)
            {
                scanf("%d",&p.a);
                p.b=i+1;
                if(p.a-m>0)
                {
                    p.a-=m;
                    q.push(p);
                    top=p.b;
                }
            }
    //		t=q.top();
    //        printf("%d
    ",t.a);
            if(top==-1)
                printf("%d
    ",n);
            else
            {
                while(!q.empty())
                {
                    t=q.front();
                    q.pop();
                    if(t.a-m>0)
                    {
                        t.a-=m;
                        q.push(t);
                        top=t.b;
                    }
                }
                printf("%d
    ",top);
            }
    
        }
    }
    
    


  • 相关阅读:
    Element Plus表单resetFields重置表单无效
    thinkphp6通过中间件设置跨域
    centos7安装jdk
    面试
    行测-逻辑填空
    行测-语句表达
    行测-片段阅读
    行测-四大文明古国
    基于 OAuth2.0 协议的单点登录系统方案设计
    性能测试知识点+流程
  • 原文地址:https://www.cnblogs.com/lxjshuju/p/7251021.html
Copyright © 2020-2023  润新知