Codeforces Round #257 (Div. 2) A. Jzzhu and Children

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本文探讨了一个关于分配糖果的策略及其在不同情况下最后离场者的身份。通过数学算法实现，揭示了每名儿童获取足够糖果的条件与顺序离场的关系。

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.

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.

妈妈说代码要写得短短短短短短短短短短短短短短才能AC。

题意：好像是老师派发苹果吧。有n个学生，每人每次发m个。然后接下来n个数表示依次每个学生的要有多少个苹果才会满足，满足了的就拿着小苹果，果果果果果、走人了，没满足的就到队伍的最后继续等着，问最后出队列的是第几个小盆友。这题要火火火火火。。。

题解：把每个小盆友要发几次苹果才会满足的次数向上取整存下来，然后比较即可，相等的当然是后面的出列。第一次wrong的原因是把数组定义成int类型了，ceil（）取值的时候相除直接是截取的整数了，淡淡的忧伤。

#include <bits/stdc++.h>
using namespace std;
int b[110];
double a[110];
int main()
{
	int n,m,i;
	scanf("%d%d",&n,&m);
	int max=-1,pos;
	for (i=1;i<=n;i++)
	{
		scanf("%lf",&a[i]);
		b[i]=ceil(a[i]/m);
		if (max<=b[i]) 
		{
			max=b[i];
			pos=i;
		}
	}
	printf("%d\n",pos);
	return 0;
}