04. Write C++ programsto find GCD by using recursive and non-recursive functions (b). To find the GCD of two given integers.

((((((Don't copy and paste it. Please Practice These programs are provided to create interest in Students))))))))

#include<iostream>            //This is actual program executed in gcc compilers like Putty
#include<stdlib.h>              //similar to previous factorial program with switch case without labels
using namespace std;
int gcdr(int,int);                  //Recursive method
int gcdnr(int,int);               //Non-Recursive method
int main()
{
        int m,n,gcd;
        char ch;
        cout<<"enter 'm', 'n' value to find Greatest Common Divisor :: ";
        cin>>m>>n;
        while(n<0||m<0)                 //Validating m, n values for GCD
        {
                cout<<"Please enter valid(only positive) m, n values"<<endl;
                cin>>m>>n;
        }
        if(m==0&&n==0)
         {
               cout<<"GCD not possible. \n terminating program"<<endl;
               exit(0);
         }
        cout<<"\nenter 'r' to implement Recursive method"<<endl;
        cout<<"\nenter 'n' to implement Non-Recursive method"<<endl;
        cin>>ch    ;
        switch(ch)
        {
        case 'r':
                gcd=gcdr(m,n);
                cout<<"Executing Recursive Method"<<endl;
                cout<<"\nGCD of "<<m<<","<<n<<" is: "<<gcd;
                break;
        case 'n':
                gcd=gcdnr(m,n);
                cout<<"\nGCD of "<<m<<","<<n<<" is: "<<gcd;
                break;
        default:
                cout<<"Invalid choice"<<endl;
                cout<<"Terminating the program"<<end;
        }
        return 0;
}
int gcdnr(int m,int n)
{
       int r;
       cout<<"Executing Non-Recursive Method"<<endl;
       if(m==0||n==0)                  //GCD of Any +ve number with 0 is itself the number
       {
           return m+n;
        }
        while(n|=0)
        {
        r=m%n;                              //logic for GCD is divide greater no: by smaller and save remainder
                                                  //divide the smaller value by the above remainder value
                                                  //repeat this process until successful division (one divides another)
        m=n;
        n=r;
        }
        return m;
}
int gcdr(int m,int n)
{
        if(n==0)
                return m;
       if(m==0||n==0)
       {           return m+n;
        }
        else
                return gcdr(n,m%n);
}