执行给定操作后数组中最大数目的相等数

给定一个整数数组。任务是在多次应用给定操作后，在数组中找到最大数目的相等计数。
在操作中：

1. Choose two elements of the array a[i], a[j] (such that i is not equals to j) and, 
2. Increase number a[i] by 1 and decrease number a[j] by 1 i.e., a[i] = a[i] + 1 and a[j] = a[j] – 1. 
 

例子：

Input: a = { 1, 4, 1 }
Output: 3
after first step { 1, 3, 2}
after second step { 2, 2, 2}

Input: a = { 1, 2 }
Output: 1

方法 ：

1. 计算数组元素的总和。
2. 如果总和可被n整除，其中n是数组中元素的数量，则答案也将为n。
3. 否则，答案将为n-1。

下面是上述方法的实现：

C++

// CPP program to find the maximum
// number of equal numbers in an array
 
#include 
using namespace std;
 
// Function to find the maximum number of
// equal numbers in an array
int EqualNumbers(int a[], int n)
{
    // to store sum of elements
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += a[i];
 
    // if sum of numbers is not divisible
    // by n
    if (sum % n)
        return n - 1;
 
    return n;
}
 
// Driver Code
int main()
{
    int a[] = { 1, 4, 1 };
 
    // size of an array
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << EqualNumbers(a, n);
 
    return 0;
}

Java

// Java program to find the maximum
// number of equal numbers in an array
 
public class GFG{
     
    // Function to find the maximum number of
    // equal numbers in an array
    static int EqualNumbers(int a[], int n)
    {
        // to store sum of elements
        int sum = 0;
        for (int i = 0; i < n; i++)
            sum += a[i];
 
        // if sum of numbers is not divisible
        // by n
        if (sum % n != 0)
            return n - 1;
 
        return n;
    }
 
     
    // Driver code
    public static void main(String args[])
    {
            int a[] = { 1, 4, 1 };
 
            // size of an array
            int n = a.length;
 
            System.out.println(EqualNumbers(a, n));
    }
    // This code is contributed by ANKITRAI1
}

Python3

# Python3 program to find the maximum
# number of equal numbers in an array
 
# Function to find the maximum
# number of equal numbers in an array
def EqualNumbers(a, n):
 
    # to store sum of elements
    sum = 0;
    for i in range(n):
        sum += a[i];
 
    # if sum of numbers is not
    # divisible by n
    if (sum % n):
        return n - 1;
 
    return n;
 
# Driver Code
a = [1, 4, 1 ];
 
# size of an array
n = len(a);
 
print(EqualNumbers(a, n));
 
# This code is contributed by mits

C#

// C# program to find the maximum
// number of equal numbers in an array
using System;
 
class GFG
{
// Function to find the maximum
// number of equal numbers in an array
static int EqualNumbers(int []a, int n)
{
    // to store sum of elements
    int sum = 0;
    for (int i = 0; i < n; i++)
        sum += a[i];
 
    // if sum of numbers is not
    // divisible by n
    if (sum % n != 0)
        return n - 1;
 
    return n;
}
 
// Driver code
static public void Main ()
{
    int []a = { 1, 4, 1 };
 
    // size of an array
    int n = a.Length;
 
    Console.WriteLine(EqualNumbers(a, n));
}
}
 
// This code is contributed by jit_t

---

Javascript

 
// JavaScript program to find the maximum
// number of equal numbers in an array
 
// Function to find the maximum number of
// equal numbers in an array
function EqualNumbers(a, n)
{
    // to store sum of elements
    let sum = 0;
    for (let i = 0; i < n; i++)
        sum += a[i];
 
    // if sum of numbers is not divisible
    // by n
    if (sum % n)
        return n - 1;
 
    return n;
}
 
// Driver Code
 
    let a = [ 1, 4, 1 ];
 
    // size of an array
    let n = a.length;
 
    document.write(EqualNumbers(a, n));
 
// This code is contributed by Surbhi Tyagi.
 

输出：

3