/* 145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145.
 *
 * Find the sum of all numbers which are equal to the sum of the factorial of their digits.
 *
 * Note: as 1! = 1 and 2! = 2 are not sums they are not included.*/

#define _POSIX_C_SOURCE 199309L

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

int main(int argc, char **argv)
{
    int i;
    unsigned long int digit;
    double elapsed;
    struct timespec start, end;
    mpz_t a, b, q, sum_f, sum, factorials[10];

    clock_gettime(CLOCK_MONOTONIC, &start);

    mpz_init_set_ui(a, 10);
    mpz_init_set_ui(sum, 0);
    mpz_inits(b, q, sum_f, sum, NULL);

    for(i = 0; i < 10; i++)
    {
        mpz_init_set_ui(factorials[i], 1);
    }

    /* Pre-calculate factorials of each digit from 0 to 9.*/
    for(i = 2; i < 10; i++)
    {
        mpz_fac_ui(factorials[i], i);
    }

    /* 9!*7<9999999, so 9999999 is certainly un upper bound.*/
    while(mpz_cmp_ui(a, 9999999) < 0)
    {
        mpz_set(b, a);
        mpz_set_ui(sum_f, 0);

        while(mpz_cmp_ui(b, 0))
        {
            digit = mpz_fdiv_qr_ui(b, q, b, 10);
            mpz_add(sum_f, sum_f, factorials[digit]);
        }

        if(!mpz_cmp(a, sum_f))
        {
            mpz_add(sum, sum, a);
        }

        mpz_add_ui(a, a, 1);
    }

    mpz_clears(a, b, q, sum_f, NULL);

    for(i = 0; i < 10; i++)
    {
        mpz_clear(factorials[i]);
    }

    clock_gettime(CLOCK_MONOTONIC, &end);

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

    printf("Project Euler, Problem 34\n");
    gmp_printf("Answer: %Zd\n", sum);

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

    mpz_clear(sum);

    return 0;
}