CodeForces 651B Beautiful Paintings-优快云博客

本文介绍了一道关于重新排列画作以最大化观众愉悦次数的问题。通过输入画作数量及每幅画的美观值，利用算法求解最优排列方案，使得相邻两画之间的美观值递增次数最多。

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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

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>

using namespace std;
int a[1005];
int b[1005];
int tag[1005];
int tag2[1005];
int num[1005];
int n;
int main()
{
    scanf("%d",&n);
    memset(tag2,0,sizeof(tag2));
    int cnt=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        if(!tag2[a[i]])
        {
            tag2[a[i]]=1;
            b[++cnt]=a[i];
        }
    }
    sort(b+1,b+cnt+1);
    for(int i=1;i<=cnt;i++)
        tag[b[i]]=i;
    memset(num,0,sizeof(num));
    for(int i=1;i<=n;i++)
        num[tag[a[i]]]++;
    int ans=0;
    for(int i=1;i<=cnt;i++)
    {
        for(int j=i-1;j>=1;j--)
        {
                if(num[j]<=num[i])
                {
                    ans+=num[j];
                    num[j]=0;
                    num[i]-=num[j];
                }
                else
                {
                    ans+=num[i];
                    num[i]=0;
                    num[j]-=num[i];

                }
        }

    }
    printf("%d\n",ans);
    return 0;

}