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)
```

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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;

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;

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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