/*
Find the number of continued fractions for the square roots of n <= 10000 which have an odd period
*/

#include <bits/stdc++.h>

using namespace std;

int gcd(int a, int b){
    if(b == 0){
        return a;
    }
    return gcd(b, a % b);
}

int gcd(int a, int b, int c){
    return gcd(a, gcd(b, c));
}

// Pseudocode found in the video on continued fraction of square roots of integers https://www.youtube.com/watch?v=GFJsU9QsytM

bool check(int n){
    float x = sqrt((float)n);

    if(x != sqrt(n)){
        vector<int> cf;
        int a, b = 1, c = 1, d = 0;
        int bn, cn, dn, g;
        int b1, c1, d1;

        for(int i = 0; ; ++i){
            a = floor((floor(b * x) + d) / c);
            cf.push_back(a);

            bn = b*c;
            cn = b*b*n - d*d - a*a*c*c + 2*a*c*d;
            dn = a*c*c - c*d;

            g = gcd(bn, cn, dn);

            b = bn / g;
            c = cn / g;
            d = dn / g;

            if(i == 0){
                b1 = b;
                c1 = c;
                d1 = d;
            } else if(b1 == b && c1 == c && d1 == d){
                break;
            }
        }

        return cf.size() % 2 == 0;
    }

    return false;
}

int main(){

    cout << "Hello this is Patrick" << endl;
    auto start = chrono::high_resolution_clock::now();

    int result = 0;

    for(int n = 1; n <= 10000; ++n){
        result += check(n);
    }

    cout << result << endl;

    auto duration = chrono::duration_cast<chrono::milliseconds>(chrono::high_resolution_clock::now() - start);
    cout << (float)duration.count()/1000 << endl;
    return 0;
}