Question

Assignment: Euclid’s algorithm for finding the greatest common divisor (gdc) of two numbers is fairly simple, but there are some possible problems that we will guard against.

The algorithm: given two numbers, n1 and n2:

1. Divide n1 by n2 and let r be the remainder.
2. If the remainder r is 0, the algorithm is finished and the answer is n2. (If the remainder is 1, the numbers are mutually prime-see below.)
3. Set n1 to the value of n2, set n2 to the value of remainder r and go back to step 1.

There are some things to watch out for:

• For the purposes of this exercise use integer values for n1 and n2.
• Entering 0 for one of the values is bad. It should work for the other value, but you have to figure out which is OK and which is bad.
  • Catch this problem as it happens and make the user enter another value until they enter an acceptable one.
  • Give an appropriate error message if this happens.
• The user should be prompted to enter values for n1 and n2.
  • For our purposes entering either n1 or n2 less than zero is an error:
    • Catch these errors and make the user enter another value until they enter an acceptable one for the value in error.
    • Give an appropriate error message if this happens.
    • Hint: the program code for catching errors in the entry of n1 and n2 is different.
• If the remainder is 1, the two numbers are mutually prime:
  • This means that they have no common divisors except 1.
  • In this case print a message informing the user of that circumstance
• If the two numbers have a greatest common divisor other than 1, print the two numbers and their gcd in an informative message.

Answer 1

************************************CODE*********************************

#include <bits/stdc++.h>
using namespace std;
int gcd_n1_n2(int a, int b)
{
if (b == 0)
return a;
return gcd_n1_n2(b, a % b);
  
}
int main()
{
int n1,n2;
cout<<"Enter the first number\n";
cin>>n1;
while(n1<=0)
{
cout<<"Enter the first number again as n1 <= 0.\n";
cin>>n1;
}
  
cout<<"Enter the second number\n";
cin>>n2;
while(n2 <= 0)
{
cout<<"Enter the second number again as n2 <= 0.\n";
cin>>n2;
}
  
int gcd = gcd_n1_n2(n1,n2);
if(gcd == 1)
cout<<"Two numbers "<<n1 <<" and "<<n2<<" are mutually prime.";
else
cout<<"GCD of two numbers "<<n1 <<" and "<<n2<<" is: "<<gcd;
return 0;
}

CODE SCREENSHOT:

OUTPUT SCREENSHOT:

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