/* If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120.
 *
 * {20,48,52}, {24,45,51}, {30,40,50}
 *
 * For which value of p ≤ 1000, is the number of solutions maximised?*/

#define _POSIX_C_SOURCE 199309L

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main(int argc, char **argv)
{
    int p, m, n, a, b, c, count, max = 0, res = 0, tmpa, tmpb, tmpc, i;
    int savedc[1000] = {0};
    double elapsed;
    struct timespec start, end;

    clock_gettime(CLOCK_MONOTONIC, &start);

    /* Start with p=12 (the smallest pythagorean triplet is (3,4,5) and 3+4+5=12.*/
    for(p = 12; p <= 1000; p++)
    {
        count = 0;
        a = 0;
        b = 0;
        c = 0;

        /* Generate pythagorean triplets.*/
        for(m = 2; m * m < p; m++)
        {
            for(n = 1; n < m; n++)
            {
                a = m * m - n * n;
                b = 2 * m * n;
                c = m * m + n * n;

                /* Increase counter if a+b+c=p and the triplet is new,
                 * then save the value of c to avoid counting the same
                 * triplet more than once.*/
                if(a + b + c == p && !savedc[c])
                {
                    savedc[c] = 1;
                    count++;
                }

                i = 2;
                tmpa = a;
                tmpb = b;
                tmpc = c;

                /* Check all the triplets obtained multiplying a, b and c
                 * for integer numbers, until the perimeters exceeds p.*/
                while(tmpa + tmpb + tmpc < p)
                {
                    tmpa = a * i;
                    tmpb = b * i;
                    tmpc = c * i;

                    /* Increase counter if the new a, b and c give a perimeter=p.*/
                    if(tmpa + tmpb + tmpc == p && !savedc[tmpc])
                    {
                        savedc[tmpc] = 1;
                        count++;
                    }

                    i++;
                }
            }
        }

        /* If the current value is greater than the maximum,
         * save the new maximum and the value of p.*/
        if(count > max)
        {
            max = count;
            res = p;
        }
    }

    clock_gettime(CLOCK_MONOTONIC, &end);

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

    printf("Project Euler, Problem 39\n");
    printf("Answer: %d\n", res);

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

    return 0;
}