/* Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.
 *
 * If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:
 *
 * 1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
 *
 * It can be seen that there are 3 fractions between 1/3 and 1/2.
 *
 * How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for d ≤ 12,000?*/

#define _POSIX_C_SOURCE 199309L

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

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

    clock_gettime(CLOCK_MONOTONIC, &start);

    /* For each denominator q, we need to find the fractions p/q for which
     * 1/3<p/q<1/2. For the lower limit, if p=q/3, then p/q=1/3, so we take
     * p=q/3+1. For the upper limit, if p=q/2 p/q=1/2, so we take p=(q-1)/2.*/
    for(i = 2; i <= 12000; i++)
    {
        limit = (i - 1) / 2;

        for(j = i / 3 + 1; j <= limit; j++)
        {
            /* Increment the counter if the current fraction is reduced.*/
            if(gcd(j, i) == 1)
            {
                count++;
            }
        }
    }

    clock_gettime(CLOCK_MONOTONIC, &end);

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

    printf("Project Euler, Problem 73\n");
    printf("Answer: %d\n", count);

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

    return 0;
}