Warm up 15 Cinderella

Cinderella

Time Limit: 10000ms, Special Time Limit:25000ms, Memory Limit:65536KB

Total submit users: 56, Accepted users: 50

Problem 12768 : No special judgement

Problem description

Cinderella is given a task by her Stepmother before she is allowed to go to the Ball. There are N (1 ≤ N ≤ 1000) bottles with water in the kitchen. Each bottle contains Li (0 ≤ Li ≤ 106) ounces of water and the maximum capacity of each is 10^9 ounces. To complete the task Cinderella has to pour the water between the bottles to fill them at equal measure. Cinderella asks Fairy godmother to help her. At each turn Cinderella points out one of the bottles. This is the source bottle. Then she selects any number of other bottles and for each bottle specifies the amount of water to be poured from the source bottle to it. Then Fairy godmother performs the transfusion instantly. Please calculate how many turns Cinderella needs to complete the Stepmother’s task.

Input

The first line of input contains an integer number N (1 ≤ N ≤ 1000) — the total number of bottles. On the next line integer numbers Li are contained (0 ≤ Li ≤ 106) — the initial amount of water contained in ith bottle.

Output

Output a single line with an integer S — the minimal number of turns Cinderella needs to complete her task.

Sample Input

Sample Input 1
3
5 7 7

Sample Input 2
3
21 10 2012

Sample Input 3
1
100

Sample Output

Sample Output 1
2

Sample Output 2
1

Sample Output 3
0

Problem Source

HNU Contest 

Submit   Discuss   Judge Status  Problems  Ranklist 

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <map>
#include <queue>
#include <vector>
#include <string>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
#define maxn 1305
#define ll long long
#define mod 1000000007
#define INF 0x7fffffff
#define eps 1e-8
int n,m;
int a[maxn];
int main(){
    /*int t;
    scanf("%d", &t);
    while (t--){
        scanf("%I64d", &n);
        
    }*/
    while (~scanf("%d", &n)){
        ll s = 0;
        for (int i = 0; i < n; i++){ scanf("%d", &a[i]); s += a[i]; }
        s = s / n;
        int x = 0;
        for(int i=0;i<n;i++)if (a[i]>s)x++;
        printf("%d\n", x);
    }
    return 0;
}