Given GCD G and LCM L, find number of possible pairs (a, b)

We need to find number of possible pairs (a, b) such that GCD(a, b) is equal to given G and LCM (a, b) such that LCM(a, b) is equal to given L.

Examples:

Input : G = 2, L = 12
Output : 4
Explanation : There are 4 possible pairs :
 (2, 12), (4, 6), (6, 4), (12, 2)
Input : G = 3, L = 6
Output : 2
Explanation : There are 2 possible pairs :
(3, 6), (6, 3)



Solution 1 (Simple):

Since a and b both will be less than or equal to lcm(a, b) L, so we try all possible pairs that have product equal to L * G. Note that product of a and b is same as product of gcd(a, b) and lcm(a, b), a*b = G*L.

Here is our algorithm

p = G*L
count = 0
for a = 1 to L
    if p%a == 0 and gcd(a, p/a) = G
        count++
    end if
end for
return count

C/C++

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// C++ program to find all pairs
// with given GCD and LCM.
#include <iostream>
using namespace std;
  
// C++ function to calculate GCD
// of two numbers
int gcd(int a, int b)
{
    if (a == 0)
        return b;
    return gcd(b%a, a);
}
  
// C++ function to count number
// of pairs with given GCD and LCM
int countPairs(int G, int L)
{
    // To store count
    int count = 0;
  
    // To store product a*b = G*L
    int p = G*L;
  
    // p/a will be b if a divides p
    for (int a=1; a<=L; a++)
        if (!(p%a) && gcd(a, p/a)==G)
            count++;
  
    return count;
}
  
// Driver code to test above functions
int main()
{
    int G = 2, L = 12;
    cout << "Total possible pair with GCD " << G;
    cout << " & LCM " << L;
    cout <<" = " << countPairs(G, L);
    return 0;
}

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Java

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// Java program to find all pairs
// with given GCD and LCM.
public class GCD
{
    // Java function to calculate GCD
    // of two numbers
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b%a , a);
    }
      
    // Java function to count number 
    // of pairs with given GCD and LCM
    static int countPairs(int G, int L)
    {
        // To store count
        int count = 0;
          
        // To store product a*b = G*L
        int p = G*L;
          
        // p/a will be b if a divides p
        for (int a = 1; a<=L; a++)
            if ((p%a == 0) && gcd(a, p/a) == G)
                count++;
                  
       return count;
    }
      
    public static void main (String[] args)
    {
        int G = 2, L = 12;
        System.out.print("Total possible pair with GCD " + G);
        System.out.print(" & LCM " + L);
        System.out.print(" = " + countPairs(G, L));
          
    }
      
}
  
// This code is contributed by Saket Kumar

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Python3

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# Python3 program to find all pairs
# with given GCD and LCM.
import math
  
# Function to calculate GCD
# of two numbers
def gcd(a, b):
  
    if (a == 0):
        return b
    return math.gcd(b % a, a)
  
# Function to count number of 
# pairs with given GCD and LCM
def countPairs(G, L):
  
    # To store count
    count = 0
  
    # To store product a*b = G*L
    p = G * L
  
    # p/a will be b if a divides p
    for a in range(1, L + 1):
        if (not (p % a) and 
            math.gcd(a, p // a) == G):
            count += 1
  
    return count
  
# Driver Code
if __name__ == "__main__":
  
    G = 2
    L = 12
    print ("Total possible pair with GCD "
                               G, end = "")
    print (" & LCM ", L, end = "")
    print (" = ", countPairs(G, L))
  
# This code is contributed by ita_c

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C#

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// C# program to find all pairs
// with given GCD and LCM.
using System;
  
class GCD
{
      
    // function to calculate 
    // GCD of two numbers
    static int gcd(int a, int b)
    {
        if (a == 0)
            return b;
        return gcd(b % a , a);
    }
      
    // Function to count number of  
    // pairs with given GCD and LCM
    static int countPairs(int G, int L)
    {
          
        // To store count
        int count = 0;
          
        // To store product a * 
        // b = G * L
        int p = G * L;
          
        // p / a will be b if 
        // a divides p
        for (int a = 1; a <= L; a++)
            if ((p % a == 0) && 
                 gcd(a, p / a) == G)
                count++;
                  
    return count;
    }
      
    // Driver Code
    public static void Main ()
    {
        int G = 2, L = 12;
        Console.Write("Total possible pair with GCD " + G);
        Console.Write(" & LCM " + L);
        Console.Write(" = " + countPairs(G, L));
          
    }
      
}
  
// This code is contributed by Nitin Mittal.

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PHP

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<?php
// PHP program to find all pairs
// with given GCD and LCM.
  
// function to calculate GCD
// of two numbers
function gcd($a, $b)
{
    if ($a == 0)
        return $b;
    return gcd($b % $a, $a);
}
  
// function to count number
// of pairs with given GCD and LCM
function countPairs( $G, $L)
{
      
    // To store count
    $count = 0;
  
    // To store product a*b = G*L
    $p = $G * $L;
  
    // p/a will be b if a divides p
    for ($a = 1; $a <= $L; $a++)
        if (!($p % $a) and 
           gcd($a, $p / $a) == $G)
            $count++;
  
    return $count;
}
  
    // Driver Code
    $G = 2; 
    $L = 12;
    echo "Total possible pair with GCD " , $G;
    echo " & LCM " , $L;
    echo " = " , countPairs($G, $L);
  
// This code is contributed by anuj_67.
?>

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Output:

Total possibe pair with gcd 2 & lcm 12 = 4

Auxiliary Space : O(1)
Time Complexity: O( L * log(L) ). We have one for loop which iterates L times and in each iteration in the worst case, gcd will be called so O(log L) in the worst case for that call.

Solution 2 (Efficient):

We know that G * L = a * b
Since G is gcd(a, b), both a and b will have G as its factor
Let A = a/G
Let B = b/G

From above definitions of A and B, GCD of A and B must be 1.
We can write, a = G * A, b = G * B

G * L = G * A * G * B
A * B = L / G
Now, we need to find all possible pairs of (A, B)
such that gcd(A, B) = 1 and A*B = L/G
Let say p1, p2, ..., pk are prime factors of L/G.
Then if p1 is present in prime factorization of A then p1
can't be present in prime factorization of B because 
gcd(A, B) = 1.
Therefore each prime factor pi will be present in either
A or B. Hence total possible ways to divide all prime 
factors among A and B is 2^k, where L/G has k distinct 
prime factors.

Below is implementation of above steps.

C/C++

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// Efficient C++ program to count all
// pairs with GCD and LCM.
#include <iostream>
using namespace std;
  
// A function to find number of distinct
// prime factors of a given number n
int totalPrimeFactors(int n)
{
    // To keep track of count
    int count = 0;
  
    // 2s that divide n
    if (!(n%2))
    {
        count++;
        while (!(n%2))
            n /= 2;
    }
  
    // n must be odd at this point. So we can skip
    // one element (Note i = i +2)
    for (int i = 3; i*i <= n; i = i+2)
    {
        // i divides n
        if (!(n%i))
        {
            count++;
            while (!(n%i))
                n /= i;
        }
    }
  
    // This condition is to handle the case when n
    // is a prime number greater than 2
    if (n > 2)
        count++;
  
    return count;
}
  
// C++ function to count number
// of pair with given GCD and LCM
int countPairs(int G, int L)
{
    if (L % G != 0)
       return 0;
  
    int div = L/G;
  
    // answer is 2^totalPrimeFactors(L/G)
    return (1 << totalPrimeFactors(div));
}
  
// Driver code to test above functions
int main()
{
    int G = 2, L = 12;
    cout << "Total possibe pair with GCD " << G;
    cout << " & LCM " << L;
    cout << " = " << countPairs(G, L);
    return 0;
}

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Java

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// Efficient Java program to count all
// pairs with GCD and LCM.
public class GCD
{
    // A function to find number of distinct
    // prime factors of a given number n
    static int totalPrimeFactors(int n)
    {
        // To keep track of count
        int count = 0;
          
        // 2s that divide n
        if ((n%2 == 0))
        {
            count++;
            while ((n%2 == 0))
               n /= 2;
        }
        // n must be odd at this point. So we can skip
        // one element (Note i = i +2)
        for (int i = 3; i*i <= n; i = i+2)
        {
            // i divides n
            if ((n%i == 0))
                count++;
                while ((n%i == 0))
                    n /= 2;
        }
        // This condition is to handle the case when n
        // is a prime number greater than 2
        if (n > 2)
            count++;
         
       return count; 
    }
    // Java function to count number
    // of pair with given GCD and LCM
    static int countPairs(int G, int L)
    {
        if (L % G != 0)
            return 0;
          
        int div = L/G;
          
        // answer is 2^totalPrimeFactors(L/G)
        return (1 << totalPrimeFactors(div));
    }
      
    // Driver function
    public static void main (String[] args)
    {
        int G = 2, L = 12;
        System.out.print("Total possible pair with GCD " + G);
        System.out.print(" & LCM " + L);
        System.out.print(" = " + countPairs(G, L));
    }
}
  
// This code is contributed by Saket Kumar

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Python3

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# Efficient python3 program to count 
# all pairs with GCD and LCM. 
  
# A function to find number of distinct 
# prime factors of a given number n 
def totalPrimeFactors(n): 
      
    # To keep track of count 
    count = 0
  
    # 2s that divide n 
    if ((n % 2) == 0): 
        count += 1
        while ((n % 2) == 0): 
            n //= 2
  
    # n must be odd at this point. 
    # So we can skip one element 
    # (Note i = i +2)
    i = 3;
    while (i * i <= n): 
          
        # i divides n 
        if ((n % i) == 0): 
            count += 1
            while ((n % i) == 0): 
                n //= i; 
        i += 2;
  
    # This condition is to handle the 
    # case when n is a prime number 
    # greater than 2 
    if (n > 2): 
        count += 1
  
    return count; 
  
# function to count number 
# of pair with given GCD and LCM 
def countPairs(G, L): 
  
    if (L % G != 0): 
        return 0
  
    div = int(L / G); 
  
    # answer is 2^totalPrimeFactors(L/G) 
    return (1 << totalPrimeFactors(div)); 
  
# Driver Code 
G = 2
L = 12
print("Total possibe pair with GCD"
           G, "& LCM", L, end = ""); 
print(" =", countPairs(G, L)); 
  
# This code is contributed by mits

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C#

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// Efficient C# program to count all
// pairs with GCD and LCM.
using System;
  
class GFG {
      
    // A function to find number
    // of distinct prime factors
    // of a given number n
    static int totalPrimeFactors(int n)
    {
          
        // To keep track of count
        int count = 0;
          
        // 2s that divide n
        if ((n % 2 == 0))
        {
            count++;
            while ((n % 2 == 0))
            n /= 2;
        }
          
        // n must be odd at this 
        // point. So we can skip
        // one element (Note i = i+2)
        for (int i = 3; i * i <= n; 
                        i = i + 2)
        {
              
            // i divides n
            if ((n % i == 0))
                count++;
                while ((n % i == 0))
                    n /= 2;
        }
          
        // This condition is to 
        // handle the case when n
        // is a prime number 
        // greater than 2
        if (n > 2)
            count++;
          
    return count; 
    }
      
    // function to count number
    // of pair with given GCD and LCM
    static int countPairs(int G, int L)
    {
        if (L % G != 0)
            return 0;
          
        int div = L/G;
          
        // answer is 2^totalPrimeFactors(L/G)
        return (1 << totalPrimeFactors(div));
    }
      
    // Driver Code
    public static void Main (String[] args)
    {
        int G = 2, L = 12;
        Console.Write("Total possible pair with GCD " + G);
        Console.Write(" & LCM " + L);
        Console.Write(" = " + countPairs(G, L));
    }
}
  
// This code is contributed by Anshul Aggarwal.

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PHP

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<?php
// Efficient php program to count all
// pairs with GCD and LCM.
  
// A function to find number of distinct
// prime factors of a given number n
function totalPrimeFactors($n)
{
      
    // To keep track of count
    $count = 0;
  
    // 2s that divide n
    if (!($n % 2))
    {
        $count++;
        while (!($n % 2))
            $n /= 2;
    }
  
    // n must be odd at this point.
    // So we can skip one element 
    // (Note i = i +2)
    for ($i = 3; $i * $i <= $n; $i = $i + 2)
    {
          
        // i divides n
        if (!($n % $i))
        {
            $count++;
            while (!($n % $i))
                $n /= $i;
        }
    }
  
    // This condition is to
    // handle the case when n
    // is a prime number greater
    // than 2
    if ($n > 2)
        $count++;
  
    return $count;
}
  
// function to count number
// of pair with given GCD and LCM
function countPairs($G, $L)
{
    if ($L % $G != 0)
    return 0;
  
    $div = $L/$G;
  
    // answer is 2^totalPrimeFactors(L/G)
    return (1 << totalPrimeFactors($div));
}
  
    // Driver Code
    $G = 2; 
    $L = 12;
    echo "Total possibe pair with GCD " , $G;
    echo " & LCM " , $L;
    echo " = " ,countPairs($G, $L);
    return 0;
  
// This code is contributed by nitin mittal.
?>

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Output:

Total possible pair with GCD 2 & LCM 12 = 4

Analysis of above algorithm
Auxiliary Space : O(1)
Time Complexity : O(sqrt(L/G) * log(L/G)). For time complexity to find number of distinct prime factors we need O(sqrt(L/G) * log (L/G)) time, Here sqrt(L) iterations are there in the worst case and in each iteration O(log L/G) iterations again.

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