/* Euler discovered the remarkable quadratic formula:
 *
 * n^2+n+41
 *
 * It turns out that the formula will produce 40 primes for the consecutive integer values 0≤n≤39. However, when n=40,402+40+41=40(40+1)+41 is
 * divisible by 41, and certainly when n=41,412+41+41 is clearly divisible by 41.
 *
 * The incredible formula n^2−79n+1601 was discovered, which produces 80 primes for the consecutive values 0≤n≤79.
 * The product of the coefficients, −79 and 1601, is −126479.
 *
 * Considering quadratics of the form:
 *
 * n^2+an+b, where |a|<1000 and |b|≤1000
 *
 * where |n| is the modulus/absolute value of n
 * e.g. |11|=11 and |−4|=4
 *
 * Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n, starting with n=0.*/

#define _POSIX_C_SOURCE 199309L

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

int main(int argc, char **argv)
{
    int a, b, n, p, count, max = 0, save_a, save_b;
    double elapsed;
    struct timespec start, end;

    clock_gettime(CLOCK_MONOTONIC, &start);

    /* Brute force approach, optimized by checking only values of b where b is prime.*/
    for(a = -999; a < 1000; a++)
    {
        for(b = 2; b <= 1000; b++)
        {
            /* For n=0, n^2+an+b=b, so b must be prime.*/
            if(is_prime(b))
            {
                n = 0;
                count = 0;

                while(1)
                {
                    p = n * n + a * n + b;

                    if(p > 1 && is_prime(p))
                    {
                        count++;
                        n++;
                    }
                    else
                    {
                        break;
                    }
                }

                if(count > max)
                {
                    max = count;
                    save_a = a;
                    save_b = b;
                }
            }
        }
    }

    clock_gettime(CLOCK_MONOTONIC, &end);

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

    printf("Project Euler, Problem 27\n");
    printf("Answer: %d\n", save_a * save_b);

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

    return 0;
}