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.
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.
Print one integer — the maximum possible number of neighbouring pairs, such that ai + 1 > ai, after the optimal rearrangement.
5
20 30 10 50 40
4
4
200 100 100 200
2
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：在一个序列中，当遇到递增的数时，会觉得开心，现在给一组数，问可以开心几次
分析：求出这个序列中总共有多少个上升的序列，并且没个上升序列中有多少个元素
题解：先将序列从小到大排序，记录所有重复出现的数的次数，去掉出现次数最多的求剩下的数的和即可
#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD double
#define MAX 3000
#define mod 100
#define dian 1.000000011
#define INF 0x3f3f3f
using namespace std;
int s[MAX];
int vis[MAX];
int main()
{
int n,m,j,i,t,k;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
scanf("%d",&s[i]);
sort(s,s+n);
for(i=0;i<=n;i++)
vis[i]=1;
k=0;
for(i=0;i<n;i++)
{
if(s[i]==s[i+1])
vis[k]++;
else
k++;
}
int sum=0;
sort(vis,vis+k);
for(i=0;i<k-1;i++)
sum+=vis[i];
printf("%d
",sum);
}
return 0;
}