Codeforces 651B Beautiful Paintings【贪心】水题

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. Thei-th painting has beautya_i. 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 ofa in any order. What is the maximum possible number of indicesi (1 ≤ i ≤ n - 1), such thata_i + 1 > a_i.

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, ..., a_n (1 ≤ a_i ≤ 1000), where a_i means the beauty of thei-th painting.

Output

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

Examples

Input

520 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.

题目大意：

给你N个数，你可以随意打乱序列，变成一个新的序列，使得序列中a【i】<a【i+1】的个数最多。

思路：

1、观察到n并不大，而且a【i】也不大，考虑设定vis【i】，表示数字i出现了多少次。

2、我们肯定是想要得到若干个递增序列，将重复出现的数字，规划到其他的序列中。

那么我们扫1000次从1到1000（表示最多我能够得到1000个递增序列），对应如果有一组数对：vis【i】>0&&vis【j】>0&&i<j，那么计数器加一，并且对应整个过程中，如果有vis【i/j】>0，vis【i/j】都要减一，表示这个数用于当前这个序列中了。

Ac代码：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int vis[15151];int main(){    int n;    while(~scanf("%d",&n))    {        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)        {            int x;            scanf("%d",&x);            vis[x]++;        }        int output=0;        for(int i=1;i<=1000;i++)        {            int pre;            int f=0;            for(int j=1;j<=1000;j++)            {                if(vis[j]>0&&f==0)                {                    f=1;                    pre=j;                    vis[j]--;                    continue;                }                if(vis[j]>0&&f==1)                {                    output++;                    pre=j;                    vis[j]--;                }            }        }        printf("%d\n",output);    }}