Problem 45 : Triangular, pentagonal, and hexagonal

Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:

Triangle T_n = n(n+1)/2 1, 3, 6, 10, 15, …
Pentagonal P_n = n(3n−1)/2 1, 5, 12, 22, 35, …
Hexagonal H_n = n(2n−1) 1, 6, 15, 28, 45, …

It can be verified that T₂₈₅

Find the next triangle number that is also pentagonal and hexagonal.

#include <iostream>
#include <map>
#include <iterator>
#include <cmath>

using namespace std;

typedef unsigned long long ULL;

class PE0045
{
private:
    bool findNextTriangleNumber
 
(ULL n, ULL max_n, ULL& number);
    void printTriangleNumber(ULL number);

public:
    void getNextTriangleNumber();
};

void PE0045::printTriangleNumber(ULL number) 
{
    ULL n;

    n = (ULL)(sqrt((double long)(1 + number*8) - 1)/2.0);
    cout << "T(" << n << ") = ";

    n = (
 
ULL)(sqrt((double long)(1 + number*24) + 1)/6.0);
    cout << "P(" << n << ") = ";

    n = (ULL)(sqrt((double long)(1 + number*8) + 1)/4.0);
    cout << "H(" << n << ")  = ";

    cout << number << endl;
}

bool PE0045::findNextTriangleNumber(ULL n, ULL max_n, ULL& number)
{
    map<ULL, int> TnPnHn_mp;
    ULL Tn, Pn, Hn;

    for (ULL i = n; i<max_n; i++)
    {
        Tn = i*(i + 1) / 2;
        TnPnHn_mp[Tn]++;

        Pn = i*(3 * i - 1) / 2;
        TnPnHn_mp[Pn]++;
        
        Hn = i*(2 * i - 1);
        TnPnHn_mp[Hn]++;
    }
    
    map<ULL, int>::iterator iter;
    
    for (iter = TnPnHn_mp.begin(); iter != TnPnHn_mp.end(); iter++)
    {
        if (3 == TnPnHn_mp[iter->first] && iter->first > 40755)
        {
            number = iter->first;
            return true;
        }
    }
    return false;
}

void PE0045::getNextTriangleNumber()
{
    ULL n = 143, max_n = 50000;
    bool found = false;
    ULL  number = 0;

    while(false == found)
    {
        found = findNextTriangleNumber(n, max_n, number);
        max_n += 50000;
    }

    printTriangleNumber(number);
}

int main()
{
    PE0045 pe0045;

    pe0045.getNextTriangleNumber();

    return 0;
}