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

Difficulty Level :
Medium
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++
Java
Python3
C#
PHP
Javascript
// C++ program to find all pairs// with given GCD and LCM.#include <iostream>using namespace std; // C++ function to calculate GCD// of two numbersint 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 LCMint 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 functionsint main(){    int G = 2, L = 12;    cout << "Total possible pair with GCD " << G;    cout << " & LCM " << L;    cout <<" = " << countPairs(G, L);    return 0;}
// 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
# Python3 program to find all pairs# with given GCD and LCM.import math # Function to calculate GCD# of two numbersdef gcd(a, b):     if (a == 0):        return b    return math.gcd(b % a, a) # Function to count number of# pairs with given GCD and LCMdef 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 Codeif __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
// 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.
<?php// PHP program to find all pairs// with given GCD and LCM. // function to calculate GCD// of two numbersfunction gcd($a, $b){    if ($a == 0)        return $b;    return gcd($b % $a, $a);} // function to count number// of pairs with given GCD and LCMfunction 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.?>
<script>// javascript program to find all pairs// with given GCD and LCM. // javascript function to calculate GCD    // of two numbers    function gcd(a , b) {        if (a == 0)            return b;        return gcd(b % a, a);    } // javascript function to count number    // of pairs with given GCD and LCM    function countPairs(G , L) {        // To store count        var count = 0;         // To store product a*b = G*L        var p = G * L;         // p/a will be b if a divides p        for (let a = 1; a <= L; a++)            if ((p % a == 0) && gcd(a, p / a) == G)                count++;         return count;    }         var G = 2, L = 12;        document.write("Total possible pair with GCD " + G);        document.write(" & LCM " + L);        document.write(" = " + countPairs(G, L)); // This code is contributed by todaysgaurav</script>
Total possible 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++
Java
Python3
C#
PHP
Javascript
// 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 possible pair with GCD" << G;          cout << " &LCM " << L;          cout << " = " << countPairs(G, L);          return 0;         }
// 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
# Efficient python3 program to count# all pairs with GCD and LCM. # A function to find number of distinct# prime factors of a given number ndef 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 LCMdef countPairs(G, L):     if (L % G != 0):        return 0;     div = int(L / G);     # answer is 2^totalPrimeFactors(L/G)    return (1 << totalPrimeFactors(div)); # Driver CodeG = 2;L = 12;print("Total possible pair with GCD",           G, "& LCM", L, end = "");print(" =", countPairs(G, L)); # This code is contributed by mits
// 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.
<?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 nfunction 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 $count;} // function to count number// of pair with given GCD and LCMfunction countPairs($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 possible pair with GCD " , $G;    echo " & LCM " , $L;    echo " = " ,countPairs($G, $L);    return 0; // This code is contributed by nitin mittal.?>
<script> // Efficient javascript 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        var count = 0;         // 2s that divide n        if ((n % 2 == 0)) {            count++;            while ((n % 2 == 0))                n = parseInt(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 == 0))                count++;            while ((n % i == 0))                n = parseInt(n/2);        }        // This condition is to handle the case when n        // is a prime number greater than 2        if (n > 2)            count++;         return count;    } // javascript function to count number    // of pair with given GCD and LCM    function countPairs(G , L) {        if (L % G != 0)            return 0;         var div = L / G;         // answer is 2^totalPrimeFactors(L/G)        return (1 << totalPrimeFactors(div));    }     // Driver function             var G = 2, L = 12;        document.write("Total possible pair with GCD " + G);        document.write(" & LCM " + L);        document.write(" = " + countPairs(G, L)); // This code is contributed by aashish1995 </script>
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.This article is contributed by Pratik Chhajer. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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