A. Jzzhu and Children （周赛4）

最新推荐文章于 2022-04-03 13:53:36 发布

原创最新推荐文章于 2022-04-03 13:53:36 发布 · 560 阅读

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本文探讨了一个关于糖果分配的问题，通过特定算法让孩子们按顺序回家。重点在于使用队列来模拟分配过程，并最终确定最后一个回家的孩子。

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 a_i 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, ..., a_n(1 ≤ a_i ≤ 100).

Output

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

Sample Input

Input

5 2
1 3 1 4 2

Output

Input

6 4
1 1 2 2 3 3

Output

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.

#include <iostream>
#include <stdio.h>

int n,m;
struct list
{
    int xu,a;
}st[100],x;

int main()
{
    while(~scanf("%d %d",&n,&m))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&st[i].a);
            st[i].xu=i+1;
        }


        while(n>1)
        {
            x=st[0];
            if(x.a-m>0)
            {
                x.a=x.a-m;
                for(int i=0;i<n-1;i++)
                    st[i]=st[i+1];
                st[n-1]=x;
            }

            else
            {
                for(int i=0;i<n-1;i++)
                    st[i]=st[i+1];
                n--;
            }
        }

        printf("%d\n",st[n-1].xu);

    }

    return 0;
}