Radical and array

Radical and array

时间限制: 1 Sec  内存限制: 128 MB
提交: 46  解决: 27
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题目描述

Radical has an array , he wants the array to have as many equal numbers as possible.

He can performs the following operation as many times as he wants:

1.he chooses two elements of the array a[i] , a[j] (i!=j).

2.he simultaneously increases number a[i] by 1 and decreases number a[j] by 1

Now he want to know what maximum number of equal array elements he can get if he performs an arbitary number of such operation.

输入

The first line contains integer n (1 ≤ n ≤ 10⁵) — the array size. The second line contains space-separated integers a1, a2, ..., an (ai ≤ 100000) — the original array.

输出

Print a single integer — the maximum number of equal array elements Radical can get if he performs an arbitrary number of the given operation.

样例输入

2
2 3
3
2 4 3

样例输出

1
3

经过任意次，在数组里任意取数，一个数相加，一个数相减后，最多可以得到几个相等的数
先求出平均值想必是极好哒。我们当然想尽量把所有数都凑成平均值，如果一定会有的数凑不成，那这些数就可以让我们随意拿来减或者加，以便让其他数减或加凑成平均值。而这些数之和与平均值的余数

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

#define N 100005

int main()
{
    int n, a[N], sum;

    while(cin >> n)
    {
        memset(a, 0, sizeof(a));
        sum = 0;

        for(int i = 0; i < n; i++)
        {
            cin >> a[i];
            sum += a[i];
        }
        int x = sum / n;
        int y = sum % n;

        if(y == 0)
            printf("%d
", n);
        else{
            printf("%d
", n-1-y/(x+1));
        }
    }
    return 0;
}