/* The decimal number, 585 = 1001001001_2 (binary), is palindromic in both bases.
 *
 * Find the sum of all numbers, less than one million, which are palindromic in base 10 and base 2.
 *
 * (Please note that the palindromic number, in either base, may not include leading zeros.)*/

#define _POSIX_C_SOURCE 199309L

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

#define N 1000000

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

    clock_gettime(CLOCK_MONOTONIC, &start);

    /* Brute force approach. For every number below 1 million,
     * check if they're palindrome in base 2 and 10 using the
     * function implemented in projecteuler.c.*/
    for(i = 1; i < N; i += 2)
    {
        if(is_palindrome(i, 10) && is_palindrome(i, 2))
        {
            sum += 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 36\n");
    printf("Answer: %d\n", sum);

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

    return 0;
}