/* It turns out that 12 cm is the smallest length of wire that can be bent to form an integer sided right angle triangle in exactly one way,
 * but there are many more examples.
 *
 * 12 cm: (3,4,5)
 * 24 cm: (6,8,10)
 * 30 cm: (5,12,13)
 * 36 cm: (9,12,15)
 * 40 cm: (8,15,17)
 * 48 cm: (12,16,20)
 *
 * In contrast, some lengths of wire, like 20 cm, cannot be bent to form an integer sided right angle triangle, and other lengths allow
 * more than one solution to be found; for example, using 120 cm it is possible to form exactly three different integer sided right angle triangles.
 *
 * 120 cm: (30,40,50), (20,48,52), (24,45,51)
 *
 * Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can exactly one integer sided right angle triangle be formed?*/

#define _POSIX_C_SOURCE 199309L

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

#define N 1500000

int main(int argc, char **argv)
{
    int i, a, b, c, tmpa, tmpb, tmpc, m, n, count = 0;
    int *l;
    double elapsed;
    struct timespec start, end;

    clock_gettime(CLOCK_MONOTONIC, &start);

    if((l = (int *)calloc(N + 1, sizeof(int))) == NULL)
    {
        fprintf(stderr, "Error while allocating memory\n");
        return 1;
    }

    /* Generate all Pythagorean triplets using Euclid's algorithm:
     * For m>=2 and n<m:
     * a=m*m-n*n
     * b=2*m*n
     * c=m*m+n*n
     * This gives a primitive triple if gcd(m, n)=1 and exactly one
     * of m and n is odd. To generate all the triples, generate all
     * the primitive one and multiply them by i=2,3, ..., n until the
     * perimeter is larger than the limit. The limit for m is 865, because
     * when m=866 even with the smaller n (i.e. 1) the perimeter is greater
     * than the given limit.*/
    for(m = 2; m < 866; m++)
    {
        for(n = 1; n < m; n++)
        {
            if(gcd(m, n) == 1 && ((m % 2 == 0 && n % 2 != 0) || (m % 2 != 0 && n % 2 == 0)))
            {
                a = m * m - n * n;
                b = 2 * m * n;
                c = m * m + n * n;

                if(a + b + c <= N)
                {
                    l[a+b+c]++;
                }

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

                while(tmpa + tmpb + tmpc <= N)
                {
                    l[tmpa+tmpb+tmpc]++;

                    i++;
                    tmpa = i * a;
                    tmpb = i * b;
                    tmpc = i * c;
                }
            }
        }
    }

    for(i = 0; i <= N; i++)
    {
        if(l[i] == 1)
            count++;
    }

    free(l);

    clock_gettime(CLOCK_MONOTONIC, &end);

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

    printf("Project Euler, Problem 75\n");
    printf("Answer: %d\n", count);

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

    return 0;
}