LCM of two numbers
Algorithm
Input Two integer a and b Output GCD of a and b. complexity O(d) where d is the number of digits in the smaller number gcd(a, b) 1 if (b = 0) then 2 Return a 3 else 4 Return gcd(b, a mod b)
Algorithm Description
LCM () function takes two parameter m and n as an integers. If b=0 then gcd (Greatest common divisor) is a. a and b are temporary variable.
C Code
//To Find LCM of two numbers
//Input : Any non zero integer number(s)
#include< stdio.h>
#include< conio.h>
#include< stdlib.h>
int main()
{
int a,b,m,n,lcm=1,
printf("Enter two numbers ");
scanf("%d%d",&a,&b);
printf("LCM of %d and %d is ",a,b);
m=a;
n=b;
while(a!=b)
{ if (a< b)
a=a+m;
else
b=b+n;
}
printf("%d",a);
}
