CodeForces 246B B. Increase and Decrease【思维】

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本文介绍了一种通过操作数组元素使数组中具有尽可能多的相同元素的方法。具体操作为选择两个不同的元素，其中一个加1而另一个减1。文章提供了一个算法实现的例子，展示了如何计算在无限次操作后能得到的最大相同元素的数量。

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Polycarpus has an array, consisting of n integers a1, a2, ..., an. Polycarpus likes it when numbers in an array match. That's why he wants the array to have as many equal numbers as possible. For that Polycarpus performs the following operation multiple times:

he chooses two elements of the array ai, aj (i ≠ j);
he simultaneously increases number ai by 1 and decreases number aj by 1, that is, executes ai = ai + 1and aj = aj - 1.

The given operation changes exactly two distinct array elements. Polycarpus can apply the described operation an infinite number of times.

Now he wants to know what maximum number of equal array elements he can get if he performs an arbitrary number of such operation. Help Polycarpus.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the array size. The second line contains space-separated integers a1, a2, ..., an (|ai| ≤ 104) — the original array.

Output

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

Examples

input
2
2 1

output
1

input
3
1 4 1

output
3

只要能想明白，那么这道题就是大水......

按图中给出的操作方式，只要能整除，肯定是可以平均分开的.....

否则就把多于的放在一个数上...强势背锅....

/*
http://blog.youkuaiyun.com/liuke19950717
*/
#include<cstdio>
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		int sum=0;
		for(int i=0;i<n;++i)
		{
			int tp;
			scanf("%d",&tp);
			sum+=tp;
		}
		int ans=n;
		if(sum%n)
		{
			ans=n-1;
		}
		printf("%d\n",ans);
	}
	return 0;
}