/*
	Matthew Giampiccolo
			&
	Jason M. Snouffer

	Computer Science & Programming 
	Wednesday, October 30, 2002

	Lab 9 Problem 1

	This program takes a numerator and denominator from the user, ensures that the denominator is nonzero;
	then displays the fraction and its simplest form.


	Test Suite:

	Input: 2/4
	Output: The fraction is 2/4
			The fraction is 1/2
	
	Input: 3/9
	Output: The fraction is 3/9
			The fraction is 1/3

	Input: 5/6
	Output: The fraction is 5/6
			The fraction is 5/6

	Input: 14/16
	Output: The fraction is 14/16
			The fraction is 7/8

	Input: 90/99
	Output: The fraction is 90/99
			The fraction is 10/11

	Input: 10/10
	Output: The fraction is 10/10
			The fraction is 1/1

	Input: 88/100
	Output: The fraction is 88/100
			The fraction is 22/25

	Input: 4/10
	Output: The fraction is 4/10
			The fraction is 2/5

  	Input: 16/30
	Output: The fraction is 16/30
			The fraction is 8/15

	All of the sample inputs within this test suite were tested and performed as expected.
*/



#include<iostream>
#include<cstdlib>

using namespace std;


/*
The function reads from the user the numerator and denominator of a fraction, insisting that the 
denominator not be 0.

preconditions: none

postconditions: recieves two integers from user, ensures that the second int is nonzero, and 
modifies arguments that are referenced by the function's 2 pass-by-reference varaiables.
*/
void get_fraction (int & numerator, int & denominator);


/*
The function displays the numerator and denominator of a fraction, with a '/' in between them.

preconditions: denominator must be nonzero

postconditions: prints value of first argument, a slash, and the value of the second argument.
*/
void display_fraction (int numerator, int denominator); 


/*
The function simplifies the input fraction and returns to its caller with the corresponding values of 
numerator and denominator.

preconditions: denominator must be nonzero

postconditions: modifies pass-by-reference variables
*/
void simplify_fraction (int & numerator, int & denominator);



int main ()
{
	int numerator, denominator;

	get_fraction(numerator, denominator);

	display_fraction (numerator, denominator);

	simplify_fraction (numerator, denominator);
	
	display_fraction (numerator, denominator);

	return EXIT_SUCCESS;
}


/*
The function reads from the user the numerator and denominator of a fraction, insisting that the 
denominator not be 0.

preconditions: none

postconditions: recieves two integers from user, ensures that the second int is nonzero, and modifies arguments that
are referenced by the function's 2 pass-by-reference varaiables.
*/
void get_fraction (int & numerator, int & denominator)
{
	cout<<"Please enter a numerator: ";
	cin>>numerator;

	do
	{
		cout<<"Please enter a nonzero denominator: ";
		cin>>denominator;
	} while(denominator == 0);
}


/*
The function displays the numerator and denominator of a fraction, with a '/' in between them.

preconditions: denominator must be nonzero

postconditions: prints value of first argument, a slash, and the value of the second argument.
*/
void display_fraction (int numerator, int denominator)
{
	cout<<"The fraction is: "
		<<numerator
		<<'/'
		<<denominator
		<<endl;
}


/*
The function simplifies the input fraction and returns to its caller with the corresponding values of 
numerator and denominator.

preconditions: denominator must be nonzero

postconditions: modifies pass-by-reference variables
*/
void simplify_fraction (int & numerator, int & denominator)
{
	int new_numerator, new_denominator;

	if ( (numerator < 0) && (denominator < 0 ) )
	{
		numerator = abs(numerator);
		denominator = abs(denominator);
	}

	int i;
	for (i = 1; i <= denominator; i++)
	{
		if( (numerator%i == 0) && (denominator%i == 0) )
		{
			new_numerator = numerator/i;
			new_denominator = denominator/i;
		}
	}

	numerator = new_numerator;
	denominator = new_denominator;

}