使数组的所有数字相等

给定一个数组arr [] ，任务是使所有数组元素与给定操作相等。在单个操作中，数组的任何元素都可以乘以2或3 。如果有可能使所有数组元素与给定操作相等，则打印“是”，否则打印“否” 。

例子：

Input: arr[] = {50, 75, 100} 
Output: Yes 
{50 * 2 * 3, 75 * 2 * 2, 100 * 3} = {300, 300, 300}

Input: arr[] = {10, 14} 
Output: No 

方法：任何正整数都可以分解并写为2 ^a * 3 ^b * 5 ^c * 7 ^d *…..
我们可以将给定数字乘以2和3，以便为它们增加a和b 。因此，我们可以通过将a和b增加到相同的大值(例如100)来使它们相等。但是我们无法更改其他质数的幂，因此它们从一开始就必须相等。我们可以通过将输入中的所有数字尽可能多地除以2和3来检查它。那么所有的人都必须平等。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that returns true if all
// the array elements can be made equal
// with the given operation
bool EqualNumbers(int a[], int n)
{
    for (int i = 0; i < n; i++) {
 
        // Divide number by 2
        while (a[i] % 2 == 0)
            a[i] /= 2;
 
        // Divide number by 3
        while (a[i] % 3 == 0)
            a[i] /= 3;
 
        if (a[i] != a[0]) {
            return false;
        }
    }
 
    return true;
}
 
// Driver code
int main()
{
    int a[] = { 50, 75, 150 };
 
    int n = sizeof(a) / sizeof(a[0]);
 
    if (EqualNumbers(a, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of above approach
class GFG
{
 
    // Function that returns true if all
    // the array elements can be made equal
    // with the given operation
    static boolean EqualNumbers(int a[], int n)
    {
        for (int i = 0; i < n; i++)
        {
 
            // Divide number by 2
            while (a[i] % 2 == 0)
            {
                a[i] /= 2;
            }
 
            // Divide number by 3
            while (a[i] % 3 == 0)
            {
                a[i] /= 3;
            }
 
            if (a[i] != a[0])
            {
                return false;
            }
        }
 
        return true;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int a[] = {50, 75, 150};
 
        int n = a.length;
 
        if (EqualNumbers(a, n))
        {
            System.out.println("Yes");
        }
        else
        {
            System.out.println("No");
        }
    }
}
 
// This code is contributed by Rajput-JI

Python3

# Python3 implementation of the approach
 
# Function that returns true if all
# the array elements can be made equal
# with the given operation
def EqualNumbers(a, n):
 
    for i in range(0, n):
 
        # Divide number by 2
        while a[i] % 2 == 0:
            a[i] //= 2
 
        # Divide number by 3
        while a[i] % 3 == 0:
            a[i] //= 3
 
        if a[i] != a[0]:
            return False
 
    return True
 
# Driver code
if __name__ == "__main__":
 
    a = [50, 75, 150]
    n = len(a)
 
    if EqualNumbers(a, n):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by Rituraj Jain

C#

// C# implementation of above approach
using System;
 
class GFG
{
 
    // Function that returns true if all
    // the array elements can be made equal
    // with the given operation
    static bool EqualNumbers(int []a, int n)
    {
        for (int i = 0; i < n; i++)
        {
 
            // Divide number by 2
            while (a[i] % 2 == 0)
            {
                a[i] /= 2;
            }
 
            // Divide number by 3
            while (a[i] % 3 == 0)
            {
                a[i] /= 3;
            }
 
            if (a[i] != a[0])
            {
                return false;
            }
 
        }
 
        return true;
    }
 
    // Driver code
    public static void Main()
    {
        int []a = {50, 75, 150};
 
        int n = a.Length;
 
        if (EqualNumbers(a, n))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}
 
// This code is contributed by Ryuga

Javascript

 
// Javascript implementation of above approach
 
// Function that returns true if all
// the array elements can be made equal
// with the given operation
function EqualNumbers(a, n)
{
    for(let i = 0; i < n; i++)
    {
         
        // Divide number by 2
        while (a[i] % 2 == 0)
        {
            a[i] = parseInt(a[i] / 2, 10);
        }
 
        // Divide number by 3
        while (a[i] % 3 == 0)
        {
            a[i] = parseInt(a[i] / 3, 10); 
        }
 
        if (a[i] != a[0])
        {
            return false;
        }
    }
    return true;
}
 
// Driver code
let a = [ 50, 75, 150 ];
let n = a.length;
 
if (EqualNumbers(a, n))
{
    document.write("Yes");
}
else
{
    document.write("No");
}
 
// This code is contributed by divyeshrabadiya07  
 

输出：

Yes

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