Codeforces 389C Fox and Box Accumulation【贪心】

A. Fox and Box Accumulation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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

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 thanx_i boxes on the top ofi-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 containsn integers x₁, x₂, ..., x_n (0 ≤ x_i ≤ 100).

Output

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

Examples

Input

30 0 10

Output

2

Input

50 1 2 3 4

Output

1

Input

40 0 0 0

Output

4

Input

90 1 0 2 0 1 1 2 10

Output

3

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.

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

题目大意：

一共已知有N个盒子，每个盒子都有对应能够承受的压力值，假设一个盒子的压力值为4，那么其可以承担上边的盒子数就是4个，问最多需要摞几摞、

思路：

1、我们首先可以明确一点，让承受压力值低的放在上边，那么我们将盒子按照承受压力值从小到大排序。

2、然后我们用vis【i】表示第i个盒子是否已经用过了，那么我们从第一个盒子开始摞，向后扫压力值大于1的第一个盒子，然后再向后扫，找压力值大于2的第一个盒子..................依次类推，将已经用过的盒子对应vis【i】=1.标记用过了。

那么起点数，就是最少的摞数。

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int vis[150];int a[150];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);        }        int output=0;        sort(a,a+n);        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            if(vis[i]==1)continue;            vis[i]=1;            output++;            int cnt=1;            for(int j=i+1;j<n;j++)            {                if(vis[j]==1)continue;                if(a[j]>=cnt)                {                    vis[j]=1;                    cnt++;                }            }        }        printf("%d\n",output);    }}