codeforces 450A Jzzhu and Children

最新推荐文章于 2023-09-19 20:08:53 发布

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本文介绍了一个有趣的算法问题：如何通过模拟的方式确定在特定糖果分配规则下，哪个孩子会最后离开队伍。该问题涉及到队列的数据结构使用，并通过示例说明了算法的具体实现过程。

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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:

Give m candies to the first child of the line.
If this child still haven't got enough candies, then the child goes to the end of the line, else the child go home.
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.

        Examples
      
          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.

没别的，用队列（queue），感觉这样模拟就可以了。

#include<bits/stdc++.h>
#include<queue>
#include<cmath>
using namespace std;
typedef long long ll;
int main()
{
    int n,m,a[110],b[110]={0};
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    queue<int>ff;
    for(int i=1;i<=n;i++)
    {
        ff.push(i);
    }
    int en=1;
    while(!ff.empty())
    {
        int w=ff.front();
        b[w]+=m;
        if(b[w]>=a[w])
        {
        	en=w;
        	ff.pop();
		}
		else
		{
            ff.push(w);
            ff.pop();
		}
    }
    cout<<en<<endl;
    return 0;
}