Q - Beautiful Paintings

CodeForces - 651B

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.

Example
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.

题意：排列使相邻两个数后数大于前数这种情况最多

解法：桶排序思想，创建数组a[1000]，初始化为0，输入i，使a[i]++，判断不重复的序列有几个，每个序列元素个数减1，再相加即为最后答案

代码：

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;

int main()
{
    int num,s=0;
    cin>>num;
    int a[1000];
    for(int i=0;i<1000;i++)
        a[i]=0;
    for(int i=0;i<num;i++){
        int b;
        cin>>b;
        a[b-1]++;
    }
    for(int i=0;i<1000;i++){
        int temp=0;
        int flag=1;
        for(int j=0;j<1000;j++){
            if(a[j]>0){
                a[j]-=1;
                temp++;
                flag=-1;
            }
        }
        s=s+temp-1;
        if(flag==1){
            s+=1;
            break;
        }
    }
    cout<<s<<endl;
    return 0;
}