/*
Triangle, square, pentagonal, hexagonal, heptagonal, and octagonal numbers are all figurate (polygonal) numbers and are generated by the following formulae:

Triangle	 	P3,n=n(n+1)/2	 	1, 3, 6, 10, 15, ...
Square	 	P4,n=n2	 	1, 4, 9, 16, 25, ...
Pentagonal	 	P5,n=n(3n−1)/2	 	1, 5, 12, 22, 35, ...
Hexagonal	 	P6,n=n(2n−1)	 	1, 6, 15, 28, 45, ...
Heptagonal	 	P7,n=n(5n−3)/2	 	1, 7, 18, 34, 55, ...
Octagonal	 	P8,n=n(3n−2)	 	1, 8, 21, 40, 65, ...
The ordered set of three 4-digit numbers: 8128, 2882, 8281, has three interesting properties.

The set is cyclic, in that the last two digits of each number is the first two digits of the next number (including the last number with the first).
Each polygonal type: triangle (P3,127=8128), square (P4,91=8281), and pentagonal (P5,44=2882), is represented by a different number in the set.
This is the only set of 4-digit numbers with this property.
Find the sum of the only ordered set of six cyclic 4-digit numbers for which each polygonal type: triangle, square, pentagonal, hexagonal, heptagonal, and octagonal, is represented by a different number in the set.
*/

#include <iostream>
#include <chrono>
#include <vector>
#include <algorithm>
#include <map>

using namespace std;

uint triangle(uint n){
    return n * (n + 1) / 2;
}

uint square(uint n){
    return n * n;
}

uint pentagonal(uint n){
    return n * (3 * n - 1) / 2;
}

uint hexagonal(uint n){
    return n * (2 * n - 1);
}

uint heptagonal(uint n){
    return n * (5 * n - 3) / 2;
}

uint octogonal(uint n){
    return n * (3 * n - 2);
}

bool validEnd(uint n){
    return n % 100 >= 10;
}

void fillVector(std::vector<uint> & toFill, const uint bitmask, uint (*calc)(uint n)){
    uint value = calc(0);

    for(uint i = 0; value < 10000; ++i){
        if(value >= 1000 && validEnd(value)){
            toFill[value] |= bitmask;
        }
        value = calc(i + 1);
    }
}

// Non-recursive version of Heap's algorithm for permutations
// Source: https://en.wikipedia.org/wiki/Heap%27s_algorithm
vector<vector<uint>> permutations(uint n, vector<uint> & A){
    vector<vector<uint>> result;
    vector<uint> c(n, 0);

    result.push_back(A);

    uint i = 1;
    while(i < n){
        if(c[i] < i){
            if(i % 2 == 0){
                swap(A[0], A[i]);
            } else{
                swap(A[c[i]], A[i]);
            }
            result.push_back(A);
            c[i] += 1;
            i = 1;
        } else{
            c[i] = 0;
            i++;
        }
    }
    return result;
}

bool checkCyclicness(const vector<uint> & v){
    for(int i = 0; i < 5; ++i){
        uint end = v[i] % 100;
        uint start = v[i + 1] / 100;

        if(end != start){
            return false;
        }
    }
    if(v[5] % 100 != v[0] / 100){
        return false;
    }

    return true;
}

void search(vector<uint> & sequence, uint mask = 0){
    
}

int main(){

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

    // creating big vector with the numeric values and their corresponding mask
    uint finalMask = 0b111111000;

    // Finally learned what exactly it means when 1 << n, which is nice
    vector<uint> all(10000, 0);
    fillVector(all, 1 << 3, triangle);
    fillVector(all, 1 << 4, square);
    fillVector(all, 1 << 5, pentagonal);
    fillVector(all, 1 << 6, hexagonal);
    fillVector(all, 1 << 7, heptagonal);
    fillVector(all, 1 << 8, octogonal);
    


    cout << "Cycle not found" << endl;
    auto duration = chrono::duration_cast<chrono::milliseconds>(chrono::high_resolution_clock::now() - start);
    cout << (float)duration.count()/1000 << endl;
    return 0;
}