codeforces389 C. Fox and Box Accumulation【简单贪心】

C. Fox and Box Accumulation

time limit per test1 second

memory limit per test256 megabytes

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 xi boxes on its top (we’ll call xi 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.

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 xi 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, …, xn (0 ≤ xi ≤ 100).

Output

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

Examples

input

3

0 0 10

output

2

input

5

0 1 2 3 4

output

1

input

4

0 0 0 0

output

4

input

9

0 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).

题意： 给你一些大小重量相同的盒子，已知每个盒子都有一个力量，如果该盒子的力量为2就说明该盒子上面可以放两个盒子，现在给你一些盒子，问你分成的最小堆为多少

分析： 我们先排序，然后贪心从小到大来排即可，每次取一堆，因为范围不大，直接暴力即可，然后用一个数组标记选没选过

总结： 总感觉这个题可以用优先队列来优化，暂时没思路，有思路的可以留言交流哈~

参考代码

#include <bits/stdc++.h>

using namespace std;
const int maxn = 1e3 + 10;

int a[maxn];
bool h[maxn];
int main() {
int n;cin>>n;
for (int i = 0; i < n; i++) {
cin>>a[i];
}
sort(a,a+n);
int res = 0;
while (true) {
int len = 0;
for (int i = 0; i < n; i++) {
if(h[i]) continue;
if(a[i] >= len) {
h[i] = true;
len++;
}
}
if(len > 0)
res++;
else break;
}
cout<<res<<endl;
return 0;
}