Codeforces Round #345 (Div. 2) B. Beautiful Paintings

B. Beautiful Paintings

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

There are n pictures delivered for the new exhibition. The i-th
painting has beauty ai.
We know that a visitor becomes happy every time he passes from a painting to a more beautiful one.

We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of a in
any order. What is the maximum possible number of indices i (1 ≤ i ≤ n - 1),
such that ai + 1 > ai.

Input

The first line of the input contains integer n (1 ≤ n ≤ 1000) —
the number of painting.

The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000),
where ai means
the beauty of the i-th painting.

Output

Print one integer — the maximum possible number of neighbouring pairs, such that ai + 1 > ai,
after the optimal rearrangement.

Examples

input

5
20 30 10 50 40

output

4

input

4200 100 100 200

output

2

Note

In the first sample, the optimal order is: 10, 20, 30, 40, 50.

In the second sample, the optimal order is: 100, 200, 100, 200.

题意：一个画廊有n幅画，每幅画有一个幸福值，当游客从较小幸福值走到较大幸福值时，幸福次数+1，问，怎么安排每幅画的顺序，使得游客幸福次数达到最大值。

贪心算法可以解决，把N幅画从小到大排列以后，前哨兵遍历每幅画，后哨兵从前哨兵开始往后找，找到第一幅能使游客幸福次数上升的画，加上标记访问后幸福次数加一，标记后的画能让前哨兵访问，但不能再次让后哨兵访问，例如，1,1,1,2,2,2,会出现1,1,1,2,使用的情况下幸福次数达到3，这是不正确的。

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
const int maxn=1005;
int a[maxn],vis[maxn],n;
int main()
{
cin>>n;
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n);
int ans=0;
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if(a[i]<a[j]&&!vis[j]){
ans++;
vis[j]=1;
break;
}
}
}
cout<<ans<<endl;
return 0;
}