54 lines
1.6 KiB
C
54 lines
1.6 KiB
C
/* 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;
|
|
}
|