[堆][贪心]cf228cFox and Box Accumulation

最新推荐文章于 2018-03-21 21:29:08 发布

原创最新推荐文章于 2018-03-21 21:29:08 发布 · 875 阅读

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本文介绍了一种关于箱堆叠的算法问题，旨在利用不同强度的箱子构建尽可能少的堆叠，以解决Fox Ciel的问题。文章详细阐述了解决方案，并提供了一段使用C++实现的代码。

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.

Sample test(s)

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).

` $\text{[math]}$

假如没有相同大小的，我们依次选择v,v-1,v-2...1是最优的，这样没有浪费。
假如有相同大小的，我们仍先选择v,v-1,v-2...1。当其不连续的时候，若此序列只有n个，那我们必定可以插入v-n个箱子进去，显然我们要插入v-n个较大的。当其连续时，可以证明这样仍然是最优的，因为如果把其中v,v-1...v',v'-1...1换成v,v-1...v',v'...1的话，个数不能增加，但是剩下的选择的空间变小了。
经过一轮放置后，若没有放完，则继续按照这样的方式放置。

这道题显然用堆实现最方面，但是当时没有想到。

#include <cstdio>
#include <functional>
#include <algorithm>
using std::sort;
using std::greater;

int ptr[110];
int num[110];
int start[110];

int min(int a,int b,int c)
{
	if (a < b && a < c)
		return a;
	if (b < c)
		return b;
	return c;
}

int main()
{
	int n;
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%d",num+i);
	}
	sort(num+1,num+1+n,greater<int>());
	num[0] = 0x3f3f3f3f;
	num[n+1] = 0x3f3f3f3f;
	for (int i=1;i<=n;i++)
	{
		if (num[i]!=num[i-1])
		{
			ptr[++ptr[0]] = i;
			start[ptr[0]] = i;
		}
	}
	start[ptr[0]+1] = n+1;
	int grp = 0;
	int used= 0;
	while (1)
	{
		if (used == n)
			break;
		int left = -0x3f3f3f3f;
		grp ++;
		for (int i=1;i<=ptr[0];i++)
		{
			if (ptr[i] == -1)
				continue;
			if (left == -0x3f3f3f3f)
				left = num[ptr[i]];
			else
				left --;
			used ++;
			if (num[ptr[i]]==num[ptr[i]+1])
				ptr[i] ++;
			else
				ptr[i] = -1;
		}
		if (left <= 0)
			continue;
		for (int i=ptr[0];left>0 && i>=1;i--)
		{
			if (ptr[i] == -1)
				continue;
			int chose = min(num[ptr[i]],start[i+1]-ptr[i],left);
			left -= chose;
			used += chose;
			if (num[ptr[i]]==num[ptr[i]+chose])
				ptr[i] += chose;
			else
				ptr[i] = -1;
		}
	}
	printf("%d",grp);
	return 0;
}