/* 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 2/5 is the fraction immediately to the left of 3/7.
 *
 * By listing the set of reduced proper fractions for d ≤ 1,000,000 in ascending order of size, find the numerator
 * of the fraction immediately to the left of 3/7.*/

#define _POSIX_C_SOURCE 199309L

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

#define N 1000000

int main(int argc, char **argv)
{
    int i, j, n, d, max_n;
    double elapsed, max = 0.0;
    struct timespec start, end;

    clock_gettime(CLOCK_MONOTONIC, &start);

    /* For each denominator q, we need to find the biggest numerator p for which
     * p/q<a/b, where a/b is 3/7 for this problem. So:
     * pb<aq
     * pb<=aq-1
     * p<=(aq-1)/b
     * So we can set p=(3*q-1)/7 (using integer division).*/
    for(i = 2; i <= N; i++)
    {
        j = (3 * i - 1) / 7;

        if((double)j / i > max)
        {
            n = j;
            d = i;
            max = (double)n / d;

            /* Reduce the fraction if it's not already reduced.*/
            if(gcd(i, j) > 1)
            {
                n /= gcd(i, j);
                d /= gcd(i, j);
            }

            max_n = n;
        }
    }

    clock_gettime(CLOCK_MONOTONIC, &end);

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

    printf("Project Euler, Problem 71\n");
    printf("Answer: %d\n", max_n);

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

    return 0;
}