/* The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, are themselves prime.
 *
 * There are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97.
 *
 * How many circular primes are there below one million?*/

#define _POSIX_C_SOURCE 199309L

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

#define N 1000000

int is_circular_prime(int n);

int *primes;

int main(int argc, char **argv)
{
    int i, count = 13;
    double elapsed;
    struct timespec start, end;

    clock_gettime(CLOCK_MONOTONIC, &start);

    /* Calculate all primes below one million, then check if they're circular.*/
    if((primes = sieve(N)) == NULL)
    {
        fprintf(stderr, "Error! Sieve function returned NULL\n");
        return 1;
    }

    /* Starting from 101 because we already know that there are 13 circular primes below 100.*/
    for(i = 101; i < 1000000; i += 2)
    {
        if(is_circular_prime(i))
        {
            count++;
        }
    }

    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 35\n");
    printf("Answer: %d\n", count);

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

    return 0;
}

int is_circular_prime(int n)
{
    int i, tmp, count;

    /* If n is not prime, it's obviously not a circular prime.*/
    if(primes[n] == 0)
    {
        return 0;
    }

    /* The primes below 10 are circular primes.*/
    if(primes[n] == 1 && n < 10)
    {
        return 1;
    }

    tmp = n;
    count = 0;

    while(tmp > 0)
    {
        /* If the number has one or more even digits, it can't be a circular prime.
         * because at least one of the rotations will be even.*/
        if(tmp % 2 == 0)
        {
            return 0;
        }
        /* Count the number of digits.*/
        count++;
        tmp /= 10;
    }

    for(i = 1; i < count; i++)
    {
        /* Generate rotations and check if they're prime.*/
        n = n % (int)pow(10, count-1) * 10 + n / (int)pow(10, count-1);
        if(primes[n] == 0)
        {
            return 0;
        }
    }

    return 1;
}