C++ Program To Find The GCD Of Two Numbers

In this tutorial, we will write a c++ program to find the GCD of two numbers.

GCD (Greatest common divisor) is also known as HCF (Highest Common Factor).

For example

36 = 2 2 3 * 3

60 = 2 2 3 * 5

The highest common factor of the two numbers is 2, 2, and 3.

So, the HCF of two numbers is 2 2 3 = 12

20 = 2 2 5

28 = 2 2 7

The highest common factor of the two numbers is 2 and 2.

So, the HCF of two numbers is 2 * 2 = 4.

Approach 1

The problem can be solved by finding all the prime factors of the two numbers and then find common factors and return their product.

✅ Find the factors of the first number
✅ Find the factors of the second number
✅ Find common factors of both numbers
✅ Multiply them to get GCD

C++ code

Example


#include <bits/stdc++.h>
#define MAXFACTORS 1024 // Let us consider max factors can be 1024
using namespace std;

typedef struct 
{

	int size;
	int factor[MAXFACTORS + 1];
	int exponent[MAXFACTORS + 1];

} FACTORIZATION;

// Function to find the factorization of M and N
void FindFactorization(int x, FACTORIZATION* factorization)
{
	int i, j = 1;
	int n = x, c = 0;
	int k = 1;
	factorization->factor[0] = 1;
	factorization->exponent[0] = 1;

	for (i = 2; i <= n; i++) {
		c = 0;

		while (n % i == 0) {
			c++;

			// factorization->factor[j]=i;
			n = n / i;
			// j++;
		}

		if (c > 0) {
			factorization->exponent[k] = c;
			factorization->factor[k] = i;
			k++;
		}
	}

	factorization->size = k - 1;
}

// function to find the gcd using Middle School procedure
int gcdMiddleSchoolProcedure(int m, int n)
{

	FACTORIZATION mFactorization, nFactorization;

	int r, mi, ni, i, k, x = 1, j;

	// Step 1.
	FindFactorization(m, &mFactorization);

	// Step 2.
	FindFactorization(n, &nFactorization);

	// Steps 3 and 4.
	// Procedure algorithm for computing the
	// greatest common divisor.
	int min;
	i = 1;
	j = 1;
	while (i <= mFactorization.size && j <= nFactorization.size) {
		if (mFactorization.factor[i] < nFactorization.factor[j])
			i++;

		else if (nFactorization.factor[j] < mFactorization.factor[i])
			j++;

		else /* if arr1[i] == arr2[j] */
		{
			min = mFactorization.exponent[i] > nFactorization.exponent[j]
					? nFactorization.exponent[j]
					: mFactorization.exponent[i];

			x = x * mFactorization.factor[i] * min;
			i++;
			j++;
		}
	}

	return x;
}

int main()

{

	int m = 36, n = 60;
	cout << "GCD of " << m << " and " << n << " is "
		<< gcdMiddleSchoolProcedure(m, n);

	return (0);
}

Output


GCD of 36 and 60 is 12

Approach 2

Approach 1 can be solved in an efficient way using the Euclidean algorithm. This algorithm follows the idea that GCD does not change if the smaller number gets subtracted from the larger one.

C++ code

Example


#include <iostream>
using namespace std;

int gcd(int a, int b) // The function runs recursive in nature to return GCD
{
 
    if (a == 0) // If a becomes zero
       return b; // b is the GCD 
    if (b == 0)// If b becomes zero
       return a;// a is the GCD 
  
   
    if (a == b) // The case of equal numbers 
        return a; // return any one of them 
  
   // Apply case of substraction 
    if (a > b) // if a is greater subtract b 
        return gcd(a-b, b);
    return gcd(a, b-a); //otherwise subtract a 
}
  

int main()
{
    int a = 36, b = 60;
    cout<<"GCD of "<<a<<" and "<<b<<" is "<<gcd(a, b);
    return 0;
}

Output


GCD of 36 and 60 is 12

Approach 3

Approach two can be optimized further to use a modulo operator instead of subtraction.

C++ code

Example


#include <iostream>
using namespace std;

int gcd(int a, int b) // The function runs recursive in nature to return GCD
{
    if (b == 0) // if b becomes 0 return a 
        return a;
    return gcd(b, a % b); // divide to a by b to make smaller number 
     
}
  

int main()
{
    int a = 60, b = 36;
    cout<<"GCD of "<<a<<" and "<<b<<" is "<<gcd(a, b);
    return 0;
}

Output


GCD of 60 and 36 is 12

Input Required