CF345div2B

题目连接

http://codeforces.com/contest/651/problem/B

Description

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.

Sample Input

5
20 30 10 50 40

Sample Output

4

题意

构造最优解的方法，每次都找到一个最长的递增数列，如果找完之后还有数的话，就继续寻找第二个递增数列。例如2 2 2 2 2 3 5 2 3 5 3 5，构造完了之后是2 2 2 2 3 2 3 5 2 3 5，一共有n-1个区间，有max-1个区间不满足条件（max=出现次数最多的数）。即12-1-（6-1）.

代码

#include<bits/stdc++.h>using namespace std;int a[1100];int main(int argc, char const *argv[]){    int n;    scanf("%d",&n);    for (int i = 0; i < n; ++i)    {        scanf("%d",&a[i]);    }    sort(a,a+n);    int t=1;    int max=1;    for (int i = 0; i < n; ++i)    {        if (a[i]==a[i+1])        {            t++;            if (t>max)            {                max=t;            }        }        else{            t=1;        }    }    printf("%d\n",n-1-max+1 );    return 0;}

P：下个月在看数论吧…这个月先把图论看完…不得不说数论好神奇…