Question

C++

Problem 1

Write a function to calculate the greatest common divisor (GCD) of two integers using Euclid’s algorithm (also known as the Euclidean algorithm). Write a main () function that requests two integers from the user, calls your function to compute the GCD, and outputs the return value of the function (all user input and output should be done in main ()). In particular, you will find this pseudocode for calculating the GCD, which should be useful to you:

function gcd(a, b)

                while b ≠ 0

                t := b

         b := a mod b

         a := t

    return a

Problem 2

Write a program that works with fractions. Your program should be able to add, subtract, multiply, and divide two fractions. Specifically, your program must request two fractions from the user by getting the numerator and denominator separately for each fraction. It must also get the operation to perform (add, subtract, multiply, or divide) as a single character. That is, the user should enter '+', '-', '*', or '/' for the operation. Be sure to validate the input for the operation. Your program will then compute the resulting fraction, keeping the numerator and denominator separate, and output the result. The resulting fraction must be simplified. For example, (4 / 6) should be displayed as (2 / 3), and (8 / 8) should be displayed as (1 / 1).

You will need to use the GCD function from Problem 1 to simplify the answer, so be sure to test it thoroughly before using it here. You must also write a separate function for each fraction operation. That is, you must write a function that adds two fractions, another that subtracts two fractions, a third that multiplies two fractions, and a fourth that divides two fractions. All input and output should happen in main ( ). These functions are only for handling fractional math. Note that each of these functions should have six arguments: two for the first fraction, two for the second fraction, and two for the answer. The resulting fraction must therefore be passed into and out of the function using call by reference arguments.

You must compute the resulting fraction using fraction-based math (working with numerators and denominators). Do not convert the fractions to double values (like 1.5), do the math, and convert back to a fraction. There are numerous resources on the Internet if you need a refresher on how to add, subtract, multiply, and divide fractions.

Answer 1

Program:

#include <iostream>

using namespace std;

int gcd(int a, int b)
{
while(b!=0)
{
int t = b;
b = a % b;
a = t;
}

return a;
}

//simplify
void simplify(int &n, int &d)
{
   int i, j, k;

   j = abs(n);
   k = abs(d);

   i = gcd(j, k);

   n = n/i;
   d = d/i;

   if((n<0 && d<0) || d<0)
   {
       n = -n;
       d = -d;
   }
}

//addition
void add(int n1, int d1, int n2, int d2, int& n3, int& d3)
{
d3 = d2 * d1;
n3 = n2 * d1 + d2 * n1;
simplify(n3, d3);
}

//addition
void subtract(int n1, int d1, int n2, int d2, int& n3, int& d3)
{
d3 = d2 * d1;
n3 = n2 * d1 - d2 * n1;
simplify(n3, d3);
}

//multiplication
void multiply(int n1, int d1, int n2, int d2, int& n3, int& d3)
{
d3 = d2 * d1;
n3 = n2 * n1;
simplify(n3, d3);
}

//division
void divide(int n1, int d1, int n2, int d2, int& n3, int& d3)
{
d3 = n2 * d1;
n3 = d2 * n1;
simplify(n3, d3);
}

//main function
int main()
{
int n1, d1, n2, d2, n3, d3;
char c;

cout<<"Enter the numerator and denominator: ";
cin>>n1>>d1;

cout<<"Enter the numerator and denominator: ";
cin>>n2>>d2;

cout<<"Enter the operation (+ , -, * or /): ";
cin>>c;

switch(c)
{
case '+': add(n1, d1, n2, d2, n3, d3);
cout<<n1<<"/"<<d1<<" + "<<n2<<"/"<<d2<<" = ";
cout<<n3<<"/"<<d3<<endl;
break;
case '-': subtract(n1, d1, n2, d2, n3, d3);
cout<<n1<<"/"<<d1<<" - "<<n2<<"/"<<d2<<" = ";
cout<<n3<<"/"<<d3<<endl;
break;
case '*': multiply(n1, d1, n2, d2, n3, d3);
cout<<n1<<"/"<<d1<<" * "<<n2<<"/"<<d2<<" = ";
cout<<n3<<"/"<<d3<<endl;
break;
case '/': divide(n1, d1, n2, d2, n3, d3);
cout<<n1<<"/"<<d1<<" / "<<n2<<"/"<<d2<<" = ";
cout<<n3<<"/"<<d3<<endl;
break;
default:
cout<<"Wrong input!!"<<endl;
}

return 0;
}

Output: