/* The cube, 41063625 (345^3), can be permuted to produce two other cubes: 56623104 (384^3) and 66430125 (405^3).
 * In fact, 41063625 is the smallest cube which has exactly three permutations of its digits which are also cube.
 *
 * Find the smallest cube for which exactly five permutations of its digits are cube.*/

#define _POSIX_C_SOURCE 199309L

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

#define N 10000

int is_permutation(long int a, long int b);
int count_digits(long int a);

int main(int argc, char **argv)
{
    int count;
    long int i, j, cubes[N];
    double elapsed;
    struct timespec start, end;

    clock_gettime(CLOCK_MONOTONIC, &start);

    /* Calculate i^3 for all i smaller than 10000.*/
    for(i = 0; i < N; i++)
    {
        cubes[i] = i * i * i;
    }

    /* For each cube, check if there are four other cubes which are also
     * a permutation of the first cube.*/
    for(i = 0; i < N - 5; i++)
    {
        count = 1;

        /* Stop when the limit has been reached, when 5 values have been found or
         * when j^3 has more digits than i^3 (if they don't have the same digits,
         * they can't be permutations).*/
        for(j = i + 1; j < N && count_digits(cubes[j]) == count_digits(cubes[i]); j++)
        {
            if(is_permutation(cubes[i], cubes[j]))
            {
                count++;
            }

            if(count == 5)
            {
                break;
            }
        }

        if(count == 5)
        {
            break;
        }
    }

    clock_gettime(CLOCK_MONOTONIC, &end);

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

    printf("Project Euler, Problem 62\n");
    printf("Answer: %ld\n", cubes[i]);

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

    return 0;
}

int count_digits(long int a)
{
    int count = 0;

    if(a == 0)
    {
        return 1;
    }

    while(a > 0)
    {
        count++;
        a /= 10;
    }

    return count;
}

int is_permutation(long int a, long 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;
}