CodeForces - 246B Increase and Decrease （模拟）水

CodeForces - 246B

Increase and Decrease

Time Limit:                                                        2000MS

Memory Limit:                                                        262144KB

64bit IO Format:                            %I64d & %I64u

SubmitStatus

Description

Polycarpus has an array, consisting of n integers a₁, a₂, ..., a_n. 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 a_i, a_j(i ≠ j);
• he simultaneously increases number a_i by 1 and decreases number a_j by 1, that is, executes a_i = a_i + 1 and a_j = a_j - 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, ..., a_n (|a_i| ≤ 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.

Sample Input

Input

2
2 1

Output

1

Input

3
1 4 1

Output

3

Sample Output

Hint

Source

Codeforces Round #151 (Div. 2)

//题意：

一个长为 n （1?≤?n?≤?10^5）的序列（元素(|ai|?≤?10^4) ），可以取其中的两个数，一个数 + 1，另一个数就 - 1，可以这样操作无数次，问最后最多能有多少个数是一样的。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main(){
	int n;
	while(~scanf("%d", &n)){
		int sum = 0, x;
		for(int i = 0; i < n; i++){
			scanf("%d", &x);
			sum += x;
		}
		if(sum%n!=0) n--;
		printf("%d\n", n);
	}
	return 0;
}