GCD function for C

Question

Q 1. Problem 5 (evenly divisible) I tried the brute force method but it took time, so I referred few sites and found this code:

#include
int gcd(i         


        

Answer 1

This problem can also be solved in a very clean way with recursion:

int gcd(int a, int b) {
    int remainder = a % b;

    if (remainder == 0) {
        return b;
    }

    return gcd(b, remainder);
}

Answer 2

Using a bit of recursion and Objective-C

-(int)euclid:(int)numA numB:(int)numB
{
    if (numB == 0)
        return numA;
    else
        return ([self euclid:numB numB:numA % numB]);
}

Answer 3

I executed this statements for GCD :

#include<stdio.h>
#include<conio.h>
int main(){
   int l, s,r;

   printf("\n\tEnter value : ");
   scanf("%d %d",&l,&s);

  while(l%s!=0){
    r=l%s;
    l=s;
    s=r;
  }
  printf("\n\tGCD = %d",s);
  getch();
}

Answer 4

To your second question: The GCD function uses Euclid's Algorithm. It computes A mod B, then swaps A and B with an XOR swap. A more readable version might look like this:

int gcd(int a, int b)
{
    int temp;
    while (b != 0)
    {
        temp = a % b;

        a = b;
        b = temp;
    }
    return a;
}