Project Euler Problem 44

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Problem 44 : Pentagon numbers

Pentagonal numbers are generated by the formula, P_n=n(3n−1)/2. The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, …

It can be seen that P₄ + P₇ = 22 + 70 = 92 = P₈. However, their difference, 70 − 22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, P_j and P_k, for which their sum and difference are pentagonal and D = |P_k − P_j| is minimised; what is the value of D?

#include <iostream>
#include <vector>
#include <cmath>
#include <ctime>

using namespace std;

// #define UNIT_TEST

class PE0044
{
private:
    bool checkSquare(double num);

    int  calcPentagonalNumber(int n);
    bool checkPentagonalNumber(int num, int& n);
    bool findD(int n, int max_n, int& m, int& D);

public:    
    bool getValueOfD(int& m, int& D);
};

bool PE0044::checkSquare(double num)
{
    int n = (int)sqrt(num);

    if ((double)(n*n) == num)
    {
        return true;
    } 
    return false;
}

int PE0044::calcPentagonalNumber(int n)  // calculate P(n)
{
    double Pn = (double)(n*(3*n-1) / 2.0);

    return (int)Pn;
}

bool PE0044::checkPentagonalNumber(int num, int& n)
{
    // num = n(3n-1)/2 => n = (1+sqrt(1+24*num))/6

    long double temp = (long double)(1+24*num);
    if (true == checkSquare(temp))
    {
        int n1 = (int)(1+sqrt(temp));
        if (0 == n1%6 && n1/6 > 0)
        {
            n = n1/6;
            return true;
        }
    }
    return false;
}

bool PE0044::findD(int n, int max_n, int& m, int& D)
{
    vector<int> Pn_v;
    int Pn;

    for (int i=n; i<max_n; i++)
    {
        Pn = calcPentagonalNumber(i);
        Pn_v.push_back(Pn);
    }

    int n1 = 0;   // Pn_v[k]+Pn_v[j] = P(n1)
    int m1 = 0;   // Pn_v[k]-Pn_v[j] = P(m1)
    
    D = m = 0;   // set initial value 

    for(unsigned int j=0; j<Pn_v.size()-1; j++)
    {
        for(unsigned int k=j+1;k<Pn_v.size(); k++) 
        {
            if ((true == checkPentagonalNumber(Pn_v[k]+Pn_v[j], n1)) && 
                (true == checkPentagonalNumber(Pn_v[k]-Pn_v[j], m1)))
            {
#ifdef UNIT_TEST
                cout << "P(" << n1 << ") = " << Pn_v[k]+Pn_v[j] << endl;
                cout << "P(" << m1 << ") = " << Pn_v[k]-Pn_v[j] << endl;
#endif
                D = Pn_v[k] - Pn_v[j];
                m = m1;
                return true;
            }
        }
    }

    return false;
}

bool PE0044::getValueOfD(int& m, int& D)
{
   int n=1, max_n = 3000;  
    bool found = false;

    while(false == found)
    {
        found = findD(n, max_n, m, D);
        max_n += 2000;
    }

    return found;
}

int main()
{
#ifdef UNIT_TEST
    clock_t start = clock();
#endif

    PE0044 pe0044;
    int m = 0 , D = 0;

    pe0044.getValueOfD(m, D);     // D = P(m)
    cout << "D = P(" <<  m << ") = " << D << endl;

#ifdef UNIT_TEST
    clock_t finish = clock();
    double duration = (double)(finish - start) / CLOCKS_PER_SEC;
    cout << "C/C++ application running time: " << duration << " seconds" << endl;
#endif

    return 0;
}