Fractions instead of decimals - c++

Question

So, I am Writing this little program in c++, it's made to compute various values with lines (sorry i am french, i don't know how to say it in English, but they are the lines with equation types Y = kx + t).
And I want my program to output fractions instead of decimals (2/3 instead of 0.666666666...).
Can anyone tell me how ?
I read online that there are some libraries for that purpose, can anyone help me on how to use them and/or how to implement them in my code ?
Thanks :)
#include "pch.h"
#include <iostream>
#include <string>
std::string mainAnswer;
bool endVar = false;
void lineEquationFromTwoPoints() {
mainAnswer.clear();
double Xa = 0;
double Ya = 0;
double Xb = 0;
double Yb = 0;
double Y = 0;
double X = 0;
double k = 0;
double t = 0;
std::cout << ("Enter the Coordinates of your first point in this format x y : ");
std::cin >> Xa >> Ya;
std::cout << ("Enter the Coordinates of your second point in this format x y : ");
std::cin >> Xb >> Yb;
if (Xb != Xa && Yb != Ya) {
k = (Yb - Ya) / (Xb - Xa);
t = -(Xa)*k + Ya;
if (k != 1 && t != 0) {
std::cout << ("Y = ") << k << ("x + ") << t << std::endl;
}
else if (k == 1) {
std::cout << ("Y = ") << ("x") << ("+") << t << std::endl;
}
else if (t == 0) {
std::cout << ("Y = ") << k << ("x") << std::endl;
}
}
else if (Xb == Xa) {
std::cout << ("Coordinates of the first point are Equal");
}
else if (Yb == Ya) {
std::cout << ("Coordinates of the second point are Equal");
}
else if (Xb == Xa && Yb == Ya) {
std::cout << ("Coordinates of both points are Equal");
}
}
void triangle() {
double Xa = 0;
double Ya = 0;
double Xb = 0;
double Yb = 0;
double Xc = 0;
double Yc = 0;
double Ym1 = 0;
double Xm1 = 0;
double km1 = 0;
double tm1 = 0;
double Ym2 = 0;
double Xm2 = 0;
double km2 = 0;
double tm2 = 0;
double Ym3 = 0;
double Xm3 = 0;
double km3 = 0;
double tm3 = 0;
std::cout << ("Work in progress. . . :-)") << std::endl;
}
void Choose() {
while (endVar != true) {
std::cout << ("Lines:") << std::endl;
std::cout << ("------") << std::endl << std::endl;
std::cout << ("Choose What Line Operations do You Want Me To Perform:") << std::endl;
std::cout << ("1.Formulas") << std::endl;
std::cout << ("2.Calculation of a Line's equation from 2 points") << std::endl;
std::cout << ("3.Calculation of all data in a triangle") << std::endl;
std::cout << ("Type Exit to Exit") << std::endl << std::endl;
std::getline(std::cin, mainAnswer);
if (mainAnswer == "exit" || mainAnswer == "Exit") {
std::exit;
endVar = true;
}
else if (mainAnswer == "1") {
std::cout << ("Formulas will be added Here once main program with main calculation functions will be finished") << std::endl;
}
else if (mainAnswer == "2") {
lineEquationFromTwoPoints();
}
else if (mainAnswer == "3") {
triangle();
}
else {
std::cout << ("Unexpected error occured. Please relaunch program.");
std::exit;
}
}
}
int main()
{
Choose();
return 0;
}

Answer 1

A nice way to approximate a float with a fraction is to used continued fractions. In the following code, epsis the desired precision. xis assumed to be strictly positive.
#include <iostream>
#include <iomanip>
#include <cmath>
#include <tuple>
#include <vector>
#include <cmath>
// Continued fraction
std::pair<int, int> fract_cont (double x, double eps = 1.0e-3) {
std::vector<int> a;
std::vector<int> b;
a.push_back(1);
b.push_back(0);
int q = int(x);
a.push_back(q);
b.push_back(1);
double err = x - q;
double e = (x != q) ? 1.0 / (x - q) : 0.0;
int i = 1;
while (std::abs(err) > eps) {
i++;
q = int (e);
e = 1.0 / (e - q);
a.push_back (q * a[i-1] + a [i-2]);
b.push_back (q * b[i - 1] + b[i-2]);
err = x - double (a[i]) / b[i];
}
return std::make_pair(a[i], b[i]);
}
int main() {
int a, b;
double x = 4 * atan(1.0);
std::tie (a,b) = fract_cont(x);
std::cout <<"Pi = " << std::setprecision(9) << x << " ~= " << a << "/" << b << "\n";
return 0;
}
Detailed information on continued fractions is available on Wikipedia for example.
If you don't need a high precision or if you assume that the denominators will be small, you can use a brute force approach instead, simply incrementing the denominator b.