Question

dev c+

Write the function called simplicity, which makes the denominator and denominator values of the simple fraction sent to it as simple as possible, together with a main function where the denominator and denominator information are entered from the keyboard. Print the values of variables that hold the numerator and denominator values entered from the keyboard before sending them to the function and their changed states after they are sent to the screen. (The global variable should not be used and should also return 1 if the simplify function was done, and 0 if it could not be done. Write this program using a reference (pseudonym) without using a pointer. SAMPLE STUDY Enter numerator and denominator: 48 Previous value Simplified: 1

Answer 1

Note: Could you plz go through this code and let me know if u need any changes in this.Thank You
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#include <iostream>
#include <iomanip>
#include <string>

using namespace std;

int simplicity(int &numerator,int &denominator);
int main()
{
int numerator,denominator;
  
   cout<<"Enter numerator and denominator :";
   cin>>numerator>>denominator;
  
   cout<<"---------------------------------"<<endl;
   cout<<"Prevoius value :"<<numerator<<"/"<<denominator<<endl;
   int res=simplicity(numerator,denominator);
if(res==1)
{
   cout<<"Simplified :"<<numerator<<"/"<<denominator<<endl;
   }
   else
   {
       cout<<"Fraction :"<<numerator<<"/"<<denominator<<endl;
   }

   return 0;
}
int simplicity(int &numerator,int &denominator)
{
   int a,b,gcd;
   a=numerator;
   b=denominator;
  
   int divi = (a > b ? a : b);
int div = (a < b ? a : b);
int rem = divi % div;
while (rem != 0)
{
divi = div;
div = rem;
rem = divi % div;
}
  
gcd=div;
//Dividing the Fraction class Numerator and Denominator with GCD to reduce to lowest terms
if (gcd != 0) {
numerator /= gcd;
denominator /= gcd;
return 1;
}
return 0;

  

}

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Output:

============================Thank You