/* The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual in two ways: (i) each of the three terms are prime,
 * and, (ii) each of the 4-digit numbers are permutations of one another.
 *
 * There are no arithmetic sequences made up of three 1-, 2-, or 3-digit primes, exhibiting this property, but there is one other 4-digit
 * increasing sequence.
 *
 * What 12-digit number do you form by concatenating the three terms in this sequence?*/

#define _POSIX_C_SOURCE 199309L

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

#define N 10000

int is_permutation(int a, int b);

int *primes;

int main(int argc, char **argv)
{
    int i = 1489, j, found = 0;
    double elapsed;
    struct timespec start, end;

    clock_gettime(CLOCK_MONOTONIC, &start);

    if((primes = sieve(N)) == NULL)
    {
        fprintf(stderr, "Error! Sieve function returned NULL\n");
        return 1;
    }

    /* Starting from i=1489 (bigger than the first number in the sequence given in the problem),
     * check odd numbers. If they're prime, loop on even numbers j (odd+even=odd, odd+odd=even and
     * we need odd numbers because we're looking for primes) up to 4254 (1489+2*4256=10001 which has
     * 5 digits.*/
    while(i < N)
    {
        if(primes[i])
        {
            for(j = 2; j < 4255; j += 2)
            {
                /* If i, i+j and i+2*j are all primes and they have
                 * all the same digits, the result has been found.*/
                if(i + 2 * j < N && primes[i+j] && primes[i+2*j] &&
                        is_permutation(i, i+j) && is_permutation(i, i+2*j))
                {
                    found = 1;
                    break;
                }
            }
        }
        if(found)
        {
            break;
        }
        i += 2;
    }

    free(primes);

    clock_gettime(CLOCK_MONOTONIC, &end);

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

    printf("Project Euler, Problem 49\n");
    printf("Answer: %d%d%d\n", i, i+j, i+2*j);

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

    return 0;
}

int is_permutation(int a, int b)
{
    int i;
    int digits1[10] = {0}, digits2[10] = {0};

    /* Get digits of a.*/
    while(a > 0)
    {
        digits1[a%10]++;
        a /= 10;
    }

    /* Get digits of b.*/
    while(b > 0)
    {
        digits2[b%10]++;
        b /= 10;
    }

    /* If they're not the same, return 0.*/
    for(i = 0; i < 10; i++)
    {
        if(digits1[i] != digits2[i])
        {
            return 0;
        }
    }

    return 1;
}