GCD of Two Numbers in C++

June 29, 2023

C++ Program to find the GCD of two numbers

Here, in this section we will discuss GCD of two numbers in C++. GCD (Greatest Common Divisor) of two numbers is the largest number that divides both numbers.

Example : The GCD of 45 and 30 will be :

                   Factors of 45 are 3 X 3 X 5

                   Factors of 30 are 2 X 3 X 5

                   Common factors of 45 and 30 are : 3 X 5 , So the required GCD will be 15

GCD of two number in C++

Various Methods to calculate the GCD

  • Using Prime Factorization,
  • Euclid’s Algorithm,
  • Lehmer’s GCD algorithm

Method 1

For a user input n1 & n2.

  • Use a for loop linearly traversing such that (i <= num1 || i <= num2)
  • Find max i value such that both n1 and n2 are divisible by i
    • (num1 % i == 0 && num2 % i == 0)
While loop in C

C++ Code (Method 1)

Run 
#include<iostream>
using namespace std;

int main()
{
    int n1 = 18, n2 = 45, gcd = 1;
    
    for(int i = 1; i <= n1 || i <= n2; i++) {
        if(n1 % i == 0 && n2 % i == 0)
            gcd = i;
    }
    
    cout<<"The GCD is "<< gcd;
    
    return 0;
}
// Time complexity : O(N)
// Space complexity : O(1)

Output

The GCD of 18 and 45 is: 9

Lowest Common Multiple (LCM)

 

Binary to Decimal to conversion

Octal to Decimal conversion

Hexadecimal to Decimal conversion

Method 2 

Euclidean Algorithm (also known as repeated Subtraction) is used for this method 

  • Euclidean Algorithm: GCD of two numbers remains constant even if we subtract the smaller number from the higher number.

Below is an example of a manual calculation that you can understand before moving ahead with the code.

C++ Code (Method 2)

Run 
#include<iostream>
using namespace std;

// Using Euclidean Algorithm (Repeated Subtraction)
int main()
{
    int n1 = 104, n2 = 24, gcd = 1;
    
    while (n1 != n2)
    {
        if (n1 > n2)
            n1 -= n2;
        else
            n2 -= n1;
    }
    
    cout<<"The GCD : "<<n1;
    
    return 0;
}

Output

The GCD: 8

Method 3

Euclidean Algorithm (also known as repeated Subtraction) is again used for this method 

This method has two changes –

  • Uses recursive approach
  • Also, handles if one of the numbers is 0

C++ Code (Method 3)

Run 
#include<iostream>
using namespace std;

// Using Recursive Euclidean Algorithm (Repeated Subtraction)
int calcGCD(int n1, int n2)
{
    // Takes care of the case n1 is 0
    // We have given explanation above
    if (n1 == 0)
       return n2;
    
    // Takes care of the case n2 is 0
    // We have given explanation above
    if (n2 == 0)
       return n1;
  
    // base case
    if (n1 == n2)
        return n1;
  
    // n1 is greater
    if (n1 > n2)
        return calcGCD(n1 - n2, n2);

    // n2 is greater
    return calcGCD(n1, n2 - n1);
}

int main()
{
    int n1 = 45, n2 = 18;

    int gcd = calcGCD(n1, n2);
    
    cout<<"The GCD is : "<<gcd;
    
    return 0;
}

Output

The GCD is : 9
While loop in C

Method 4

Euclidean Algorithm (also known as repeated Subtraction) is again used for this method 

This method has two changes –

  • Reduced number of subtraction
  • Modulo operation helps us achieve these reduce subtraction

C++ Code (Method 4)

Run 
#include<iostream>
using namespace std;

// Improved the time complexity in comparision to repeated subtraction
// Done by efficient usage of modulo operator in euclidean algorithm
int calcGCD(int a, int b)
{
    return b == 0 ? a : calcGCD(b, a % b);   
}

int main()
{
    int n1 = 45, n2 = 18;

    int gcd = calcGCD(n1, n2);
    
    cout<<"The GCD is: "<<gcd;
    
    return 0;
}

Output

The GCD is : 9

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While loop in C

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