CodeForces - 651B Beautiful Paintings

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
4
200 100 100 200
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.
我那又臭又长的代码，过不了，有BUG，无奈借鉴大佬写的
思路：暴力跑一下，对进行过比较的大数标记，只有没被标记的才能继续比较，很nice的思路

#include<bits/stdc++.h>
using namespace std;
int f[1001];
int main()
{
	int n,flag;
	int count=0;
	int a[1001];
	cin>>n;
	for(int i=0;i<n;i++)
	cin>>a[i];
	sort(a,a+n);
	for(int i=0;i<n;i++)
	    for(int j=i+1;j<n;j++)
	    {
	    	if(a[i]<a[j]&&!f[j])
	    	{
	    		count++;
	    		f[j]=1;
	    		break;
			}
		}
	cout<<count;
}