Вычисление цепных дробей

int N = 0;
cin >> N;
double result = 1;
for( int n = 2; n <= N; ++n )
    result += 1./(double)n;
 
cout << result;

#include <iostream>
#include <sstream>
#include <string>
 
double solve(int n)
{
    if (n <= 0)
        return 0.0;
    
    double result = n;
    
    for (int i = n - 1; i >= 1; --i)
        result = i + 1.0 / result;
    
    return result;
}
 
std::string int_to_str(int value)
{
    std::ostringstream ostr;
    
    ostr << value;
    
    return ostr.str();
}
 
std::string build_formula(int n)
{
    if (n == 1)
        return "1";
    
    std::string formula = int_to_str(n);
    
    for (int i = n - 1; i > 1; --i)
        formula = "(" + int_to_str(i) + " + 1 / " + formula + ")";
    
    return "1 + 1 / " + formula;
}
 
int main()
{
    std::cout << build_formula(1) << " = " << solve(1) << std::endl;
    std::cout << build_formula(2) << " = " << solve(2) << std::endl;
    std::cout << build_formula(3) << " = " << solve(3) << std::endl;
    std::cout << build_formula(4) << " = " << solve(4) << std::endl;
    
    return 0;
}

#include<iostream>
 
void F(int *a, int n, int &p, int &q)
{
   int i, p0, p1, q0, q1;
   p0 = 1; p = p1 = a[0];
   q0 = 0; q = q1 = 1;
   for(i = 1; i < n; i++)
   {
       p = p0 + p1*a[i];
       p0 = p1; p1 = p;
       q = q0 + q1*a[i];
       q0 = q1; q1 = q;
   }
}
 
int main()
{
   int p, q, n, a[] = {1, 2, 3};
   n = sizeof(a) / sizeof(*a);
   F(a, n, p, q);
   std::cout << p << "/" << q << "\n";
   return 0;
}