# C Program to find GCD Of Two Numbers

## GCD of Two numbers in C

Here, in this section, we will discuss the GCD of two numbers in C. The Highest Common Multiple or the Greatest Common Divisor is the greatest number that exactly divides both numbers.

Example:-

```The G.C.D of 24 and 40 is 8.

The G.C.D of 18 and 12 is 6```

## 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)

### C Code (Method 1)

Run
```#include<stdio.h>

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;
}

printf("The GCD of %d and %d is: %d", n1, n2, gcd);

return 0;
}
// Time complexity : O(N)
// Space complexity : O(1)
```

### Output

```The GCD of 18 and 45 is: 9
```

## 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<stdio.h>

// 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;
}

printf("The GCD: %d", n1, n2, gcd);

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<stdio.h>

// 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);

printf("The GCD of %d and %d is: %d", n1, n2, gcd);

return 0;
}
```

### Output

```The GCD of 45 and 18 is: 9
```

## 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<stdio.h>

// 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);

printf("The GCD of %d and %d is: %d", n1, n2, gcd);

return 0;
}```

### Output

```The GCD of 45 and 18 is: 9
```

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