Maximum GCD from Given Product of Unknowns

Given two integers N and P where P is the product of N unknown integers, the task is to find the GCD of those integers. There can be different group of integers possible that give the same product, in that case print the GCD which is maximum among all possible groups.

Examples:

Input: N = 3, P = 24
Output: 2
{1, 1, 24}, {1, 2, 12}, {1, 3, 8}, {1, 4, 6}, {2, 2, 6} and {2, 3, 4}
are the only integer groups possible with product = 24
And they have GCDs 1, 1, 1, 1, 2 and 1 respectively.



Input: N = 5, P = 1
Output:

Approach: Let g be the gcd of a1, a2, a3, …, an. Since, ai must be a multiple of g for each i and their product P = a1 * a2 * a3 * … * an must be a multiple of gn. The answer is the maximum g such that gn % P = 0.
Let P = k1p1 * k2p2 * k3p3 * … * knpt. Then g must be of the form k1p1 * k2p2 * k3p3 * … * knpt. In order to maximise g we must choose pi = pi / N

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include<bits/stdc++.h>
using namespace std;
  
// Function to return the required gcd
long max_gcd(long n, long p)
{
    int count = 0;
    long gcd = 1;
  
    // Count the number of times 2 divides p
    while (p % 2 == 0)
    {
  
        // Equivalent to p = p / 2;
        p >>= 1;
        count++;
    }
  
    // If 2 divides p
    if (count > 0)
        gcd *= (long)pow(2, count / n);
  
    // Check all the possible numbers
    // that can divide p
    for (long i = 3; i <= sqrt(p); i += 2) 
    {
        count = 0;
        while (p % i == 0) 
        {
            count++;
            p = p / i;
        }
        if (count > 0)
        {
            gcd *= (long)pow(i, count / n);
        }
    }
  
    // If n in the end is a prime number
    if (p > 2)
        gcd *= (long)pow(p, 1 / n);
  
    // Return the required gcd
    return gcd;
}
  
// Driver code
int main()
{
    long n = 3;
    long p = 80;
    cout << max_gcd(n, p);
}
      
// This code is contributed by Code_Mech

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
class GFG {
  
    // Function to return the required gcd
    static long max_gcd(long n, long p)
    {
        int count = 0;
        long gcd = 1;
  
        // Count the number of times 2 divides p
        while (p % 2 == 0) {
  
            // Equivalent to p = p / 2;
            p >>= 1;
            count++;
        }
  
        // If 2 divides p
        if (count > 0)
            gcd *= (long)Math.pow(2, count / n);
  
        // Check all the possible numbers
        // that can divide p
        for (long i = 3; i <= Math.sqrt(p); i += 2) {
            count = 0;
            while (p % i == 0) {
                count++;
                p = p / i;
            }
            if (count > 0) {
                gcd *= (long)Math.pow(i, count / n);
            }
        }
  
        // If n in the end is a prime number
        if (p > 2)
            gcd *= (long)Math.pow(p, 1 / n);
  
        // Return the required gcd
        return gcd;
    }
  
    // Driver code
    public static void main(String[] args)
    {
        long n = 3;
        long p = 80;
        System.out.println(max_gcd(n, p));
    }
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach
import math
  
# Function to return the required gcd 
def max_gcd(n, p): 
  
    count = 0
    gcd = 1
  
    # Count the number of times 2 divides p 
    while (p % 2 == 0): 
      
        # Equivalent to p = p / 2; 
        p >>= 1
        count = count + 1
      
    # If 2 divides p 
    if (count > 0): 
        gcd = gcd * pow(2, count // n); 
  
    # Check all the possible numbers 
    # that can divide p 
    for i in range(3, (int)(math.sqrt(p)), 2): 
      
        count = 0
        while (p % i == 0): 
          
            count = count + 1
            p = p // i; 
          
        if (count > 0):
          
            gcd = gcd * pow(i, count // n); 
          
    # If n in the end is a prime number 
    if (p > 2) : 
        gcd = gcd * pow(p, 1 // n); 
  
    # Return the required gcd 
    return gcd; 
  
# Driver code 
n = 3
p = 80
print(max_gcd(n, p)); 
  
# This code is contributed by Shivi_Aggarwal

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
  
class GFG 
  
// Function to return the required gcd 
static long max_gcd(long n, long p) 
    int count = 0; 
    long gcd = 1; 
  
    // Count the number of times 2 divides p 
    while (p % 2 == 0) 
    
  
        // Equivalent to p = p / 2; 
        p >>= 1; 
        count++; 
    
  
    // If 2 divides p 
    if (count > 0) 
        gcd *= (long)Math.Pow(2, count / n); 
  
    // Check all the possible numbers 
    // that can divide p 
    for (long i = 3; i <= Math.Sqrt(p); i += 2) 
    
        count = 0; 
        while (p % i == 0) 
        
            count++; 
            p = p / i; 
        
        if (count > 0)
        
            gcd *= (long)Math.Pow(i, count / n); 
        
    
  
    // If n in the end is a prime number 
    if (p > 2) 
        gcd *= (long)Math.Pow(p, 1 / n); 
  
    // Return the required gcd 
    return gcd; 
  
// Driver code 
public static void Main() 
    long n = 3; 
    long p = 80; 
    Console.WriteLine(max_gcd(n, p)); 
  
// This code is contributed by Ryuga

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP implementation of the approach
  
// Function to return the required gcd
function max_gcd($n, $p)
{
    $count = 0;
    $gcd = 1;
  
    // Count the number of times 2 divides p
    while ($p % 2 == 0) 
    {
  
        // Equivalent to p = p / 2;
        $p >>= 1;
        $count++;
    }
  
    // If 2 divides p
    if ($count > 0)
        $gcd *= pow(2, (int)($count / $n));
  
    // Check all the possible numbers
    // that can divide p
    for ($i = 3; $i <= (int)sqrt($p); $i += 2) 
    {
        $count = 0;
        while ($p % $i == 0) 
        {
            $count++;
            $p = (int)($p / $i);
        }
        if ($count > 0) 
        {
            $gcd *= pow($i, (int)($count / $n));
        }
    }
  
    // If n in the end is a prime number
    if ($p > 2)
        $gcd *= pow($p, (int)(1 / $n));
  
    // Return the required gcd
    return $gcd;
}
  
// Driver code
$n = 3;
$p = 80;
echo(max_gcd($n, $p));
  
// This code is contributed by Code_Mech

chevron_right


Output:

2


My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.