Question

Fidn the gcd of two numbers using recursion

Answer 1

Public class GCD
{
public static void main (string args [])
{
int n1 =366, n2 =60 :
int hcf =hcf (n1, n2)
system. out. print f (GCD of percentage d and percentage d)
{
public static int hcf (n1, n2)
{
if (n2 is unequal to 0)
return hcf (n2 n1 percentage n2)
else
return n1
}
}

Answer 2

Answer:

#include <stdio.h>

int hcf(int n1, int n2);

int main() {

   int n1, n2;

   printf("Enter two positive integers: ");

   scanf("%d %d", &n1, &n2);

   printf("G.C.D of %d and %d is %d.", n1, n2, hcf(n1, n2));

   return 0;

}

int hcf(int n1, int n2) {

   if (n2 != 0)

       return hcf(n2, n1 % n2);

   else

       return n1;

}

//c++ programming

#include<iostream>

using namespace std;

int hcf(int n1, int n2);

int main() {

   int n1, n2;

   cin>>n1>>n2;

   cout<<"G.C.D of "<<n1<<" and "<<n2<<" = "<< hcf(n1, n2);

   return 0;

}

int hcf(int n1, int n2) {

   if (n2 != 0)

       return hcf(n2, n1 % n2);

   else

       return n1;

}

Explanation: