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

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：

给你n个数，ai《ai+1的数有几对，感觉这题比A题更水，题目简单，数据量小，果断暴力。

AC代码：

#include<iostream>
#include<functional>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>
#include<cstdio>
#include<cmath>
#include<map>
using namespace std;
#define CRL(a) memset(a,0,sizeof(a))
#define QWQ ios::sync_with_stdio(0)
typedef unsigned __int64 LL;
typedef __int64 ll;
#define CMP bool cmp(const node& a,const node& b){ return a.R<b.R||(a.R==b.R&&a.L<b.L); }
const int T = 1555;
const int mod = 1000000007;

int a[T],b[T];

int main()
{
#ifdef zsc
freopen("input.txt","r",stdin);
#endif

int i,j,k,n,m;

while(~scanf("%d",&n))
{
fill(b,b+T,0);
for(i=0,k=0;i<n;++i){
scanf("%d",&a[i]);
b[a[i]]++;
k ++;
}
int c,cnt;
c = cnt = 0;
while(k>1)
{
cnt = 0;
for(i=1;i<=1000;i++){
if(b[i])b[i]--,k--,cnt++;
}
if(cnt>1)
c+= cnt-1;
}
printf("%d\n",c);
}

return 0;
}