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++

// 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; }

## Java

// 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

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++

// 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; }

## Java

// 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

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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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