Divide the two given numbers by their common divisors

Given two numbers A and B, the task is to Divide the two numbers A and B by their common divisors. The numbers A and B is less than 10^8.


Examples:

Input: A = 10, B = 15
Output: A = 2, B = 3
The common factors are 1, 5

Input: A = 100, B = 150
Output: A = 2, B = 3

Naive Approach: Iterate from i = 1 to minimum of A and B and check whether i is a factor of both A and B. If i is a factor of A and B then Divide the two numbers A and B by i.

Below is the implementation of the above approach:

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// print the numbers after dividing
// them by their common factors
void divide(int a, int b)
{
    // iterate from 1 to minimum of a and b
    for (int i = 2; i <= min(a, b); i++) {
  
        // if i is the common factor
        // of both the numbers
        while (a % i == 0 && b % i == 0) {
            a = a / i;
            b = b / i;
        }
    }
  
    cout << "A = " << a << ", B = " << b << endl;
}
  
// Driver code
int main()
{
    int A = 10, B = 15;
  
    // divide A and B by their common factors
    divide(A, B);
  
    return 0;
}

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Java

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// Java implementation of above approach
import java.util.*;
  
class solution
{
  
// print the numbers after dividing
// them by their common factors
static void divide(int a, int b)
{
    // iterate from 1 to minimum of a and b
    for (int i = 2; i <= Math.min(a, b); i++) {
  
        // if i is the common factor
        // of both the numbers
        while (a % i == 0 && b % i == 0) {
            a = a / i;
            b = b / i;
        }
    }
  
    System.out.println("A = "+a+", B = "+b);
}
  
// Driver code
public static void main(String args[])
{
    int A = 10, B = 15;
  
    // divide A and B by their common factors
    divide(A, B);
  
}
}
  
// This code is contributed by
// Surendra_Gangwar

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Python3

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# Python3 implementation of above approach 
  
# print the numbers after dividing 
# them by their common factors 
def divide(a, b) :
      
    # iterate from 1 to minimum of a and b 
    for i in range(2, min(a, b) + 1) : 
  
        # if i is the common factor 
        # of both the numbers 
        while (a % i == 0 and b % i == 0) :
            a = a // i
            b = b // i
  
    print("A =", a, ", B =", b) 
  
# Driver code 
if __name__ == "__main__" :
  
    A, B = 10, 15
  
    # divide A and B by their 
    # common factors 
    divide(A, B) 
      
# This code is contributed by Ryuga

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C#

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// C# implementation of above approach
using System;
  
class GFG
{
      
// print the numbers after dividing
// them by their common factors
static void divide(int a, int b)
{
    // iterate from 1 to minimum of a and b
    for (int i = 2; i <= Math.Min(a, b); i++)
    {
  
        // if i is the common factor
        // of both the numbers
        while (a % i == 0 && b % i == 0)
        {
            a = a / i;
            b = b / i;
        }
    }
    Console.WriteLine("A = "+a+", B = "+b);
}
  
// Driver code
static public void Main ()
{
    int A = 10, B = 15;
  
    // divide A and B by their common factors
    divide(A, B);
}
}
  
// This code is contributed by ajit.

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PHP

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<?php
// PHP implementation of above approach
// print the numbers after dividing
// them by their common factors
  
function divide($a, $b)
{
    // iterate from 1 to minimum of a and b
    for ($i = 2; $i <= min($a, $b); $i++) 
    {
  
        // if i is the common factor
        // of both the numbers
        while ($a % $i == 0 && 
                $b % $i == 0) 
        {
            $a = $a / $i;
            $b = $b / $i;
        }
    }
    echo "A = ", $a, ", B = ", $b, "\n";
}
  
// Driver code
$A = 10;
$B = 15;
  
// divide A and B by their common factors
divide($A, $B);
  
// This code is contributed by Sach_Code
?>

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Output:

A = 2, B = 3

An efficient approch is to use the same concept used in Common divisors of two numbers. Calculate the greatest common divisor (gcd) of given two numbers, and then divide the numbers by their gcd.

C++

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// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to calculate gcd of two numbers
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b % a, a);
}
  
// Function to calculate all common divisors
// of two given numbers
// a, b --> input integer numbers
void commDiv(int a, int b)
{
    // find gcd of a, b
    int n = gcd(a, b);
  
    a = a / n;
    b = b / n;
  
    cout << "A = " << a << ", B = " << b << endl;
}
  
// Driver code
int main()
{
    int a = 10, b = 15;
    commDiv(a, b);
  
    return 0;
}

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Output:

A = 2, B = 3

Time comlplexity : O(n)



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