/* Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
 *
 * 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
 *
 * By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.*/

#define _POSIX_C_SOURCE 199309L

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

#define N 4000000

int main(int argc, char **argv)
{
    int fib0 = 1, fib1 = 2, fib2, sum = 2;
    double elapsed;
    struct timespec start, end;

    clock_gettime(CLOCK_MONOTONIC, &start);

    fib2 = fib0 + fib1;

    /* Simple brute-force approach: generate every value in the Fibonacci
     * sequence smaller than 4 million and if it's even add it to the total.*/
    while(fib2 <= N)
    {
        if(fib2 % 2 == 0)
        {
            sum += fib2;
        }

        fib0 = fib1;
        fib1 = fib2;
        fib2 = fib0 + fib1;
    }

    clock_gettime(CLOCK_MONOTONIC, &end);

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

    printf("Project Euler, Problem 2\n");
    printf("Answer: %d\n", sum);

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

    return 0;
}