思维 CodeForces - 651B Beautiful Paintings

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), wherea_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.

题目意思：对一列数排列使其递增数最多。

解法：思维题。用a[x]来记录数字x的出现次数，再用一个for循环每次从1到1000记录a[1]~a[1000]中递增数。这道题挺考验思维，值得好好体会。

#include <iostream>#include <cstdio>#include <cstring>using namespace std;int a[1000+20];int main(){    int n, x, ans = 0;    scanf("%d", &n);    for(int i = 0; i < n; i++) {        scanf("%d", &x);        a[x]++;    }    while(1) {        int cnt = 0;        if(n == 0) break;        for(int i = 1; i <= 1000 ;i++) {            if(a[i]) {                a[i]--;                cnt++;                n--;            }        }//        printf("cnt = %d\n", cnt);        ans += cnt-1;    }    printf("%d\n", ans);}