Codeforces Round #228 (Div. 1) A. Fox and Box Accumulation-优快云博客

本文链接：https://blog.youkuaiyun.com/CriminalCode/article/details/50916526

本文深入探讨了如何使用贪心算法解决最小堆箱问题，即在给定每个盒子最大堆叠数量的情况下，计算出最少需要构建多少堆来存放所有盒子。通过从上到下选择最小数量的盒子构建堆，最终得出最少堆的数量，适用于计算机科学领域的算法优化。

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A. Fox and Box Accumulation

Fox Ciel has n boxes in her room. They have the same size and weight, but they might have different strength. The i-th box can hold at most x_i boxes on its top (we'll call x_i the strength of the box).

Since all the boxes have the same size, Ciel cannot put more than one box directly on the top of some box. For example, imagine Ciel has three boxes: the first has strength 2, the second has strength 1 and the third has strength 1. She cannot put the second and the third box simultaneously directly on the top of the first one. But she can put the second box directly on the top of the first one, and then the third box directly on the top of the second one. We will call such a construction of boxes a pile.

$\text{[math]}$

Fox Ciel wants to construct piles from all the boxes. Each pile will contain some boxes from top to bottom, and there cannot be more than x_i boxes on the top of i-th box. What is the minimal number of piles she needs to construct?

Input

The first line contains an integer n (1 ≤ n ≤ 100). The next line contains n integers x₁, x₂, ..., x_n (0 ≤ x_i ≤ 100).

Output

Output a single integer — the minimal possible number of piles.

Examples

Input

3
0 0 10

Output

Input

5
0 1 2 3 4

Output

Input

4
0 0 0 0

Output

Input

9
0 1 0 2 0 1 1 2 10

Output

Note

In example 1, one optimal way is to build 2 piles: the first pile contains boxes 1 and 3 (from top to bottom), the second pile contains only box 2.

$\text{[math]}$

In example 2, we can build only 1 pile that contains boxes 1, 2, 3, 4, 5 (from top to bottom).

$\text{[math]}$

题目链接：http://codeforces.com/problemset/problem/388/A

题目大意：n个盒子，每个盒子有一个数值，代表能放在这个盒子上的盒子个数不超过这个数。上面盒子的数值不能大于下面盒子的数值，求最少能放多少堆盒子。

解题思路：如果从下往上选大盒子累积，累积完一组后剩余的都是较小的盒子，累积的不高，累积组数可能会大于最佳情况，所以应该从上往下先选小盒子。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[105];
int vis[105];
int main(void)
{
	int n;
	memset(vis,0,sizeof(vis));
	scanf("%d",&n);
	for(int i=0;i<n;i++)
		scanf("%d",&a[i]);
	sort(a,a+n);
	int cnt=1,ans=0,s;
	bool p=true;
	while(p)
	{
		p=false;
		cnt=1;
		s=n;
		for(int i=0;i<n;i++)
		{
			if(!vis[i])
			{
				vis[i]=1;
				p=true;
				s=i;
				break;
			}
		}
		for(int i=s+1;i<n;i++)
		{
			if(a[i]>=cnt && !vis[i])
			{
				cnt++;
				vis[i]=1;
				p=true;
			}
		}
		if(p)
			ans++;
	}
	printf("%d\n",ans );
}