Codeforces Round #345(Div.2)--B. B. Beautiful Paintings(욕심 상승 서열 개수)

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

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.
제목: 가장 많은 상승 서열의 개수를 구하고, 맵으로 수조보다 훨씬 빠르다. 방법: 케이크를 자르듯이 한 층씩 자르기

#include<iostream>
#include<algorithm>
#include<vector>
#include<sstream>
#include<map>
using namespace std;
int main(void)
{
	int t,n,i,j,be,sum,ans;
	while (cin>>n)
	{
		map<int,int>list;
		for (i=0; i<n; i++)
		{
			cin>>be;
			list[be]++;
		}
		ans=0;
		while (n)//  
		{
			int num=0;
			for (map<int,int>::iterator it=list.begin(); it!=list.end(); it++)
			{
				if(it->second>0)
				{
					(it->second)--;
					n--;
					num++;
				}
			}
			ans=ans+num-1;//            
		}
		cout<<ans<<endl;
	}
	return 0;
}