codeforces 1008C Reorder the Array

C. Reorder the Array

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an array of integers. Vasya can permute (change order) its integers. He wants to do it so that as many as possible integers will become on a place where a smaller integer used to stand. Help Vasya find the maximal number of such integers.

For instance, if we are given an array $[10,20,30,40][10,20,30,40],\; we\; can\; permute\; it\; so\; that\; it\; becomes [20,40,10,30][20,40,10,30].\; Then\; on\; the\; first\; and\; the\; second\; positions\; the\; integers\; became\; larger\; (20>1020>10, 40>2040>20)\; and\; did\; not\; on\; the\; third\; and\; the\; fourth,\; so\; for\; this\; permutation,\; the\; number\; that\; Vasya\; wants\; to\; maximize\; equals 22.\; Read\; the\; note\; for\; the\; first\; example,\; there\; is\; one\; more\; demonstrative\; test\; case.$

Help Vasya to permute integers in such way that the number of positions in a new array, where integers are greater than in the original one, is maximal.

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1051\le n\le 105) —\; the\; length\; of\; the\; array.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1091\le ai\le 109) —\; the\; elements\; of\; the\; array.}$

Output

Print a single integer — the maximal number of the array's elements which after a permutation will stand on the position where a smaller element stood in the initial array.

Examples

input

Copy

7
10 1 1 1 5 5 3

output

4

input

5
1 1 1 1 1

output

Copy

0

Note

In the first sample, one of the best permutations is $[1,5,5,3,10,1,1][1,5,5,3,10,1,1].\; On\; the\; positions\; from\; second\; to\; fifth\; the\; elements\; became\; larger,\; so\; the\; answer\; for\; this\; permutation\; is\; 4.$

In the second sample, there is no way to increase any element with a permutation, so the answer is 0.

 

题意：有n个数，重新排列组合看看最多有多少个比原来大的，sort一遍暴力即可（其实一开始没想到暴力能过的。。。瞎写了两发WA）

#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define fio ios::sync_with_stdio(false);cin.tie(0);
#define pii pair<int,int>
#define vi vector<int>
#define vc vector<char>
#define pi 3.1415926
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1

const int INF=0x3f3f3f3f;
const int N=1e5+5;

typedef long long ll;
typedef double db;
typedef unsigned long long ull;
using namespace std;
int a[N];
int c[N];
int vis[N];
bool cmp(int a,int b)
{
    return a>b;
}
int main()
{
    int n;
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
    }
    sort(a+1,a+n+1,cmp);
    if(a[1]==a[n])
    {
        printf("0\n");
        return 0;
    }
    int cnt=0;
    for (int i=1;i<=n;i++)
    {
        bool flag=false;
        for (int j=i+1;j<=n;j++)
        {
            if(vis[j]) continue;
            if(a[i]>a[j]&&vis[j]==0)
            {
                cnt++;
                vis[j]=1;
                flag=true;
                break;
            }
        }
        if(!flag) break;
    }
    printf("%d\n",cnt);

}