Hello!
I have been working on a program that can add two fractions together. But I have some trouble with my code and how to make a code that can shorten a fraction. I have tried to search on the internet for solutions for the fraction shortening code, but non of them really worked.
I have tried the best I could to explain my code, but I'm still learning to improve it and I would be more then happy for feedback on my code :)
#include <iostream>
int main()
{
/*"menu" for the code*/
int fraction_choice;
std::cout << "1 - add two fractions\n" << std::endl;
std::cout << "2 - takeaway two fractions\n" << std::endl;
std::cout << "Please enter your choise\n" << std::endl;
std::cin >> fraction_choice; //input from user
switch (fraction_choice)
{
case 1://adds tow fractions together
int a1, b1, a2, b2; //decalring variables,
// A stands for the numerator and B stands for the denominator. The number is there to show what fraction it is. (vv guide under vv)
/*
first fraction secound fraction
a1 = x a2 = x
--- + ---
b1 = y b2 = y
*/
/*input from user*/
std::cout << "please enter the numerator and the denominator of the first fraction\n";
std::cin >> a1;
std::cout << "/\n";
std::cin >> b1;
std::cout << "plase enter the numerator and the denominator of the secound fraction\n";
std::cin >> a2;
std::cout << "/\n";
std::cin >> b2;
int result_1, result_2; //decalring variables for result of the fraction
/*
result 1 is the numerator
result 2 is the denominator
*/
if (b1 == b2) // is the denominator the same? if yes then just taking the sum of the two numerator
{
result_1 = a1 + a2; // sum of the two numerator
result_2 = b1; //asagin result2 to value of denominator
}
else // is the denominators not the same?
{
int F1, F2; //variables for "intercompany"
F1 = a1 * b2; // F1 stores the result of the numerator for the first fraction times the denominator for the secound fraction.
F2 = b1 * a2; // F2 stores the result of the numerator for the secound fraction times the denominator for the first fraction.
result_1 = F1 + F2; //takes the sum of F1 and F2.
result_2 = b1 * b2; //takes the product of both denominator in each fraction.
}
std::cout << "The result is: " << result_1 << "/" << result_2 << std::endl; //output for result
int Shorten; //secound menu for shorening the found fraction
std::cout << "do you wish to shorten the fraction?\n";
std::cout << "1 - Yes\n";
std::cout << "2 - No\n";
std::cout << "Please enter your choise\n" << std::endl;
std::cin >> Shorten;
// Dont want to run the code from now on??
switch (Shorten)
{
case 1:
{
// no idea how to make a code for shortening a fraction
}
break;
case 2: // stops the program
{
break;
}
default:
std::cout << "!!!!!!!!!!!!!!!!!!!!!!!!!!!\n";
std::cout << "Not a Valid Choice. \n";
std::cout << "Choose again.\n";
std::cout << "!!!!!!!!!!!!!!!!!!!!!!!!!!!\n";
std::cin >> Shorten;
break;
}
//case 2: *not sartet*
}
return 0;
}
Geez... It's "numerator" and "denominator", FFS. You spelled each word differently each time!
A fraction of the form n/m can be reduced by dividing n and m by their greatest common divisor. The GCD of two numbers is found by by Euclidean algorithm.
(Obviously the fraction is irreducible if the GCD is 1.)
