Codeforce Round #227 Div2 C

Codeforces Round #228 (Div. 2)
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Submission	Time	Verdict
5913055	02/07/2014 12:41AM	Time limit exceeded on test 1
5913049	02/07/2014 12:40AM	Accepted
5911696	02/06/2014 08:35PM	Wrong answer on test 10

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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).

$\text{[math]}$

 1 #pragma comment(linker,"/STACK:102400000,102400000")
 2 #include <cstdio>
 3 #include <vector>
 4 #include <cmath>
 5 #include <queue>
 6 #include <set>
 7 #include <cstring>
 8 #include <iostream>
 9 #include <algorithm>
10 using namespace std;
11 #define INF 0x7fffffff 
12 #define mod 1000000007
13 #define ll long long
14 #define maxn 205
15 #define pi acos(-1.0)  
16 int n, m, k, c,t,b;
17 int a[maxn];
18 int main(){
19     int x;
20     scanf("%d", &n);
21     for (int i = 0; i < n; i++)scanf("%d", &x),a[x+1]++;
22     while (1){
23         k++;
24         t = 0;
25         for (int i = 0; i <= 101; i++)
26             while (t < i&&a[i])t++, a[i]--;
27         t = 0;
28         for (int i = 0; i <= 100; i++)t += a[i];
29         if (t == 0)break;
30     }
31     printf("%d\n", k);
32     return 0;
33 }

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posted @ 2014-02-07 04:43 HaibaraAi 阅读(168) 评论(0) 收藏举报

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