From 2477b229d51a1fdc1e0fcd4bcbb4364dc8ff5842 Mon Sep 17 00:00:00 2001
From: Philippe Zwietering <p.zwietering@fugro.com>
Date: Tue, 9 Nov 2021 16:11:25 +0100
Subject: [PATCH] Found the solution for 066, now just have to work out the
 implementation

---
 projecteuler/066/main.cpp | 113 ++++++++++++++++++++++++++++++++++++++
 1 file changed, 113 insertions(+)
 create mode 100644 projecteuler/066/main.cpp

diff --git a/projecteuler/066/main.cpp b/projecteuler/066/main.cpp
new file mode 100644
index 0000000..fc18d82
--- /dev/null
+++ b/projecteuler/066/main.cpp
@@ -0,0 +1,113 @@
+/*
+Consider quadratic Diophantine equations of the form:
+
+x2 – Dy2 = 1
+
+For example, when D=13, the minimal solution in x is 6492 – 13×1802 = 1.
+
+It can be assumed that there are no solutions in positive integers when D is square.
+
+By finding minimal solutions in x for D = {2, 3, 5, 6, 7}, we obtain the following:
+
+32 – 2×22 = 1
+22 – 3×12 = 1
+92 – 5×42 = 1
+52 – 6×22 = 1
+82 – 7×32 = 1
+
+Hence, by considering minimal solutions in x for D ≤ 7, the largest x is obtained when D=5.
+
+Find the value of D ≤ 1000 in minimal solutions of x for which the largest value of x is obtained.
+
+-----------------------------------------------------------------------------------------------------
+
+The value of x can be found by calculating the convergent of the continued fraction of D,
+right before it repeats.
+
+So if y/x is the corresponding approximation of sqrt(D) for the value of a_n of the continued fraction right before it repeats,
+we immediately know x. So in order to solve this we just have to combine the previous two problem solution.
+
+https://mathshistory.st-andrews.ac.uk/HistTopics/Pell/
+*/
+
+#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));
+}
+
+int p(int n, int p0, int p1){
+    if(n == 0){
+        return p0;
+    } else if(n == 1){
+        return p1;
+    } else if(n > 1){
+        return 
+    }
+
+    return 0;
+}
+
+// Code yanked from problem 64
+vector<int> findCF(int n){
+    float x = sqrt((float)n);
+
+    vector<int> cf;
+
+    if(x != sqrt(n)){
+        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;
+}
+
+int main(){
+
+    cout << "Hello this is Patrick" << endl;
+    auto start = chrono::high_resolution_clock::now();
+
+    int maxX = 0;
+    for(int d = 1; d <= 1000; ++d){
+        auto cf = findCF(d);
+
+        // Insert problem 065 code once I have that fixed for big integers (cpp needs a dedicated internal library for that)
+    }
+
+    auto duration = chrono::duration_cast<chrono::milliseconds>(chrono::high_resolution_clock::now() - start);
+    cout << (float)duration.count()/1000 << endl;
+    return 0;
+}