/* Pentagonal numbers are generated by the formula, Pn=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, Pj and Pk, for which their sum and difference are pentagonal and D = |Pk − Pj| is minimised;
 * what is the value of D?*/

#define _POSIX_C_SOURCE 199309L

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>
#include "projecteuler.h"

int main(int argc, char **argv)
{
    int n, m, pn, pm, found = 0;
    double elapsed;
    struct timespec start, end;

    clock_gettime(CLOCK_MONOTONIC, &start);

    n = 2;

    /* Check all couples of pentagonal numbers until the right one
     * is found. Use the function implemented in projecteuler.c to
     * check if the sum and difference ot the two numbers is pentagonal.*/
    while(!found)
    {
        pn = n * (3 * n - 1) / 2;

        for(m = 1; m < n; m++)
        {
            pm = m * (3 * m - 1) / 2;

            if(is_pentagonal(pn+pm) && is_pentagonal(pn-pm))
            {
                found = 1;
                break;
            }
        }
        n++;
    }

    clock_gettime(CLOCK_MONOTONIC, &end);

    elapsed = (end.tv_sec - start.tv_sec) + (double)(end.tv_nsec - start.tv_nsec) / 1000000000;

    printf("Project Euler, Problem 44\n");
    printf("Answer: %d\n", pn-pm);

    printf("Elapsed time: %.9lf seconds\n", elapsed);

    return 0;
}