Comment simplifier une fraction

• Calculer le plus grand diviseur commun de x et y
• Diviser à la fois par le PGCD

D'euclide l'algorithme est un moyen facile de calculer le PGCD.

     #include<iostream>
using namespace std;
struct fraction
{
int n1, d1, n2, d2, s1, s2;
};
void simplification(int a,int b)
{
bool e = true;
int t; int z;
for (int i = (a*b); i > 1;i--)
{ if ((a%i==0)&&(b%i==0))
{
t = a / i;
z = b / i;
}
else
{
e = false;
}
}
cout << "simplest form=" << t << "/" << z << endl;
}
void sum(int num1, int deno1, int num2, int deno2)
{
int k,y;
k = num1* deno2 + num2*deno1;
y = deno2*deno1;
cout << "addition of given fraction = " << k << "/" << y << endl;
simplification(k, y);
}
void sub(int num1, int deno1, int num2, int deno2)
{              
int k, y;
k = num1*deno2 - num2*deno1;
y = deno1*deno2;
cout << "Substraction of given fraction = " << k << "/" << y << endl;
}
void mul(int num1, int deno1, int num2, int deno2)
{
int k, y;
k = num1*num2;
y = deno1*deno2;
cout << "multiplication of given fration= " << k<< "/" <<y;                                        cout<< endl;
simplification(k, y);
}
void div(int num1, int deno1, int num2, int deno2)
{
int k, y;
;
k = num1*deno1;
y = deno1*num2;
cout << "division of given fraction" << k << "/" << y << endl;
simplification(k, y);
}
int main()
{    fraction a;
cout << "enter numirator of f1=";cin >> a.n1;
cout << "enter denominator of f1=";cin >> a.d1;
cout << "enter numirator of f2=";cin >> a.n2;
cout << "enter denominator of f2=";cin >> a.d2;
cout << "f1= " << a.n1 << "/" << a.d1 << endl;
cout << "f2= " << a.n2 << "/" << a.d2 << endl;
mul(a.n1, a.d1, a.n2, a.d2);
div(a.n1, a.d1, a.n2, a.d2); 
sub(a.n1, a.d1, a.n2, a.d2);
sum(a.n1, a.d1, a.n2, a.d2);
system("pause");
}