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gcd (int a, int b) {
   return  b == 0 ? a : gcd(b, a%b);
}

lcm (int a, int b) {
    return a * b / gcd (a, b);
}

First of all you need to clarify what you need to do if 0 is one of the arguments. I will just skip this case.

The gcd function below uses Euclidean algorithm. Do a search if you don't know it.

int gcd(int a, int b)
{
  while(b)
  {
    int tmp = b;
    b = a mod b;
    a = tmp;
   }
  return tmp;
}

int lcm(int a, int b)
{
  int gcd = gcd(a,b);
  return a*b/gcd;
}

LCM(m, n) X GCD(m, n) = m X n

Find GCD first, then...

LCM(m, n) = m X (n / GCD(m, n))

divide both numbers , update numbers with quotients(if remainder is 0) from number 2 to square root of the larger quotient remaining.

at each step multiply if any number divides any quotient with reminder 0, at last multiply the remaining quotients with the existing

int getLCM(int a, int b) {
int max = Math.max(a,b);
int counter = 2;
int x = max;
while(true) {
if(x%a ==0 and x%b == 0) {
return x;
} else { x = max * counter; counter++;}
}
}

#include<stdio.h>

int main()
{
	int a=12,b=27,r;
	int t,m;
	int r1;

	t=(a>b)?a:b;
	m=a+b-t;
	
	r1=r=a*b;

	while(r>t)
	{
		r=r-t;
		if(r%m==0)
			r1=r;
	}

	printf("%d\n",r1);
	
	return 0;
}

This approach uses only minus operator to calculate gcd.

int gcd(int a,int b)
{
        int r;
        r=a>b?(a-b):(b-a);
        while(r)
        {
                a=b;
                b=r;
                r=a>b?(a-b):(b-a);
        }
        return a;
}
 
int main()
{
        int a=12,b=27,r;
        
        printf("%d ",a*b/gcd(a,b));
        return 0;
}

what about negative integers as inputs ? should we consider it ?