查找是否可以通过给定的操作使数组的所有元素相等

给定一个数组arr[] ，任务是使所有数组元素与给定的操作相等。
在单个操作中，数组的任何元素都可以乘以3或5任意次。如果可以使所有数组元素与给定的操作相等，则打印是，否则打印否。
例子：

Input: arr[] = {18, 30, 54, 90, 162} 
Output: Yes 
Explanation: 
We can perform following operations: 
162 X 5 = 810 
90 X 3 X 3 = 810 
54 X 5 X 3 = 810 
30 X 3 X 3 X 3 = 810 
18 X 5 X 3 X 3 = 810
Input: arr[] = {18, 36, 58, 90, 162} 
Output: No 
Explanation: 
There is no way you can make all elements equal. 
 

观察：

• 如果经过一些操作后，所有的数字都相等，那么它们将具有相同的质数因式分解，即每个数字将具有相同的 2、3、5 ……的幂，依此类推。
• 由于我们仅将数字乘以质数3和5 ，因此我们可以在某些运算后使所有数字的质因数分解中的3 和 5 的幂相等。
• 因此，要使所有数字相等，除3和5之外的素数分解中的素数的幂必须相等。
• 解决方案是取每个数字并从中删除3和5 的所有幂。如果所有数字都相等，则可以通过使用给定的操作使数组元素相等，否则不可能。

步骤：

1. 将数组arr[] 的每个元素除以 3 和 5，使得每个元素的质因数分解中3和5 的所有幂都为零。
2. 检查数组的所有元素是否相等。如果是，则打印是。
3. 否则打印编号

下面是上述方法的实现：

CPP

// C++ implementation to find if it's
// possible to make all elements of an
// array equal by using two operations.
#include 
using namespace std;
 
// Function to find if it's possible
// to make all array elements equal
bool canMakeEqual(int a[], int n)
{
    // Iterate over all numbers
    for (int i = 0; i < n; i++) {
 
        // If a number has a power of 5
        // remove it
        while (a[i] % 5 == 0) {
            a[i] /= 5;
        }
 
        // If a number has a power of 3
        // remove it
        while (a[i] % 3 == 0) {
            a[i] /= 3;
        }
    }
 
    int last = a[0];
 
    // Check if all elements are equal
    // in the final array
    for (int i = 1; i < n; i++) {
        if (a[i] != last) {
            return false;
        }
    }
 
    return true;
}
 
// Driver's Code
int main()
{
    int arr[] = { 18, 30, 54, 90, 162 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call to check if all
    // element in the array can be equal
    // or not.
    if (canMakeEqual(arr, n)) {
        cout << "YES" << endl;
    }
    else {
        cout << "NO" << endl;
    }
 
    return 0;
}

Java

// Java implementation to find if it's
// possible to make all elements of an
// array equal by using two operations.
class GFG{
  
// Function to find if it's possible
// to make all array elements equal
static boolean canMakeEqual(int a[], int n)
{
    // Iterate over all numbers
    for (int i = 0; i < n; i++) {
  
        // If a number has a power of 5
        // remove it
        while (a[i] % 5 == 0) {
            a[i] /= 5;
        }
  
        // If a number has a power of 3
        // remove it
        while (a[i] % 3 == 0) {
            a[i] /= 3;
        }
    }
  
    int last = a[0];
  
    // Check if all elements are equal
    // in the final array
    for (int i = 1; i < n; i++) {
        if (a[i] != last) {
            return false;
        }
    }
  
    return true;
}
  
// Driver's Code
public static void main(String[] args)
{
    int arr[] = { 18, 30, 54, 90, 162 };
  
    int n = arr.length;
  
    // Function call to check if all
    // element in the array can be equal
    // or not.
    if (canMakeEqual(arr, n)) {
        System.out.print("YES" +"\n");
    }
    else {
        System.out.print("NO" +"\n");
    }
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation to find if it's
# possible to make all elements of an
# array equal by using two operations.
 
# Function to find if it's possible
# to make all array elements equal
def canMakeEqual( a, n) :
 
    # Iterate over all numbers
    for i in range(n) :
 
        # If a number has a power of 5
        # remove it
        while (a[i] % 5 == 0) :
            a[i] //= 5;
         
        # If a number has a power of 3
        # remove it
        while (a[i] % 3 == 0) :
            a[i] //= 3;
 
    last = a[0];
 
    # Check if all elements are equal
    # in the final array
    for i in range(1,n) :
        if (a[i] != last) :
            return False;
 
    return True;
 
# Driver's Code
if __name__ == "__main__" :
 
    arr = [ 18, 30, 54, 90, 162 ];
 
    n = len(arr);
 
    # Function call to check if all
    # element in the array can be equal
    # or not.
    if (canMakeEqual(arr, n)) :
        print("YES");
     
    else :
        print("NO");
 
# This code is contributed by AnkitRai01

C#

// C# implementation to find if it's
// possible to make all elements of an
// array equal by using two operations.
using System;
 
class GFG{
 
// Function to find if it's possible
// to make all array elements equal
static bool canMakeEqual(int []a, int n)
{
    // Iterate over all numbers
    for (int i = 0; i < n; i++) {
  
        // If a number has a power of 5
        // remove it
        while (a[i] % 5 == 0) {
            a[i] /= 5;
        }
  
        // If a number has a power of 3
        // remove it
        while (a[i] % 3 == 0) {
            a[i] /= 3;
        }
    }
  
    int last = a[0];
  
    // Check if all elements are equal
    // in the final array
    for (int i = 1; i < n; i++) {
        if (a[i] != last) {
            return false;
        }
    }
  
    return true;
}
  
// Driver's Code
public static void Main(string[] args)
{
    int []arr = { 18, 30, 54, 90, 162 };
  
    int n = arr.Length;
  
    // Function call to check if all
    // element in the array can be equal
    // or not.
    if (canMakeEqual(arr, n)) {
        Console.WriteLine("YES");
    }
    else {
        Console.WriteLine("NO");
    }
}
}
 
// This code is contributed by AnkitRai01

Javascript

// javascript implementation to find if it's
// possible to make all elements of an
// array equal by using two operations.   
// Function to find if it's possible
    // to make all array elements equal
    function canMakeEqual(a , n) {
        // Iterate over all numbers
        for (i = 0; i < n; i++) {
 
            // If a number has a power of 5
            // remove it
            while (a[i] % 5 == 0) {
                a[i] /= 5;
            }
 
            // If a number has a power of 3
            // remove it
            while (a[i] % 3 == 0) {
                a[i] /= 3;
            }
        }
 
        var last = a[0];
 
        // Check if all elements are equal
        // in the final array
        for (i = 1; i < n; i++) {
            if (a[i] != last) {
                return false;
            }
        }
 
        return true;
    }
 
    // Driver's Code
     
        var arr = [ 18, 30, 54, 90, 162 ];
 
        var n = arr.length;
 
        // Function call to check if all
        // element in the array can be equal
        // or not.
        if (canMakeEqual(arr, n)) {
            document.write("YES" + "\n");
        } else {
            document.write("NO" + "\n");
        }
 
// This code contributed by umadevi9616

输出：

YES

时间复杂度： O(N)，其中 N 是数组的大小。

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