// Program by Akash Tripathi (@proakash256)
#include <stdio.h>
int LCM1(int a , int b)
{
int l;
for(int n = 1; ; n = n + 1)
{
if(n % a == 0 && n % b == 0)
{
l = n;
return l;
}
}
}
int HCF1(int a , int b)
{
int num , den , h = 0 , r;
if(a > b)
{
num = a;
den = b;
}
else
{
num = b;
den = a;
}
while(den > 1)
{
r = num % den;
if(r == 0)
{
h = den;
break;
}
else
{
num = den;
den = r;
}
}
if(den == 1)
h = 1;
return h;
}
int HCF2(int a , int b)
{
int x = a , y = b , h = 0;
while(x != y)
{
if(x > y)
{
x = x - y;
h = x;
}
else
{
y = y - x;
h = y;
}
}
return h;
}
int LCM2(int a , int b , int h)
{
int l = (a * b) / h;
return l;
}
int main()
{
int a , b , c;
printf("Enter two numbers of which you want
to find HCF and LCM :\n");
scanf("%d" , &a);
scanf("%d" , &b);
printf("\n");
printf("LCM and HCF by Multiplication and Division method :");
printf("\n");
c = LCM1(a , b);
printf("The LCM of %d and %d is %d.\n" , a , b , c);
c = HCF1(a , b);
printf("The HCF of %d and %d is %d.\n" , a , b , c);
printf("\n");
printf("LCM and HCF by Subtraction method :");
printf("\n");
c = HCF2(a , b);
printf("The HCF of %d and %d is %d.\n" , a , b , c);
c = LCM2(a , b , c);
printf("The LCM of %d and %d is %d.\n" , a , b , c);
return 0;
}
Comments
Post a Comment