/* The prime 41, can be written as the sum of six consecutive primes:
 *
 * 41 = 2 + 3 + 5 + 7 + 11 + 13
 *
 * This is the longest sum of consecutive primes that adds to a prime below one-hundred.
 *
 * The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
 *
 * Which prime, below one-million, can be written as the sum of the most consecutive primes?*/

#define _POSIX_C_SOURCE 199309L

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

#define N 1000000

int main(int argc, char **argv)
{
    int i, j, max = 0, max_p = 0, sum, count;
    int *primes;
    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 a prime i, add consecutive primes until the
     * sum exceeds the limit, every time the sum is also a prime
     * save the value and the count if the count is larger than the
     * current maximum. Repeat for all primes below N.
     * A separate loop is used for i=2, so later only odd numbers are
     * checked for primality.*/
    i = 2;
    count = 1;
    sum = i;

    for(j = i + 1; j < N && sum < N; j++)
    {
        if(primes[j])
        {
            sum += j;
            count++;

            if(sum < N && primes[sum] && count > max)
            {
                max = count;
                max_p = sum;
            }
        }
    }

    for(i = 3; i < N; i += 2)
    {
        if(primes[i])
        {
            count = 1;
            sum = i;

            for(j = i + 2; j < N && sum < N; j += 2)
            {
                if(primes[j])
                {
                    sum += j;
                    count++;

                    if(sum < N && primes[sum] && count > max)
                    {
                        max = count;
                        max_p = sum;
                    }
                }
            }
        }
    }

    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 50\n");
    printf("Answer: %d\n", max_p);

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

    return 0;
}