project euler -- problem 73

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?

As I have solved problem 71, this problem is very simple. Just narrow down the n to satisfy n/d in (1/3, 1/2), and count it if hcf(n,d) = 1.

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#include<iostream>
#include<cmath>
 
using namespace std;
 
const int N = 12000;
 
int hcf(int n, int d);
 
int main() {
  double min = (double) 1/3;
  double max = (double) 1/2;
  int cnt = 0;
 
  for (int d=3; d <= N; d++ ) {
    int left = d * min + 1;
    int right = d * max;
    if ( (double) right/d == max)
      right -= 1;
    if ( right < left)
      continue;
    for (int n=left; n <= right; n++) {
      if (hcf(n, d) == 1)
	cnt++;
    }
  }
 
  cout << "Answers to PE 73: " << cnt << endl;
}
 
int hcf(int n, int d) {
  int m = d % n;
  if (m == 0)
    return n;
  if (m == 1)
    return 1;
  while(m != 0) {
    d = n;
    n = m;
    m = d % n;
  }
  return n;
}

Answers to PE 73: 7295372
   user  system elapsed 
  1.197   0.002   1.198 

Related Posts

• January 8, 2013 -- project euler — problem 71 (0)
• January 9, 2013 -- project euler — problem 72 (0)
• July 30, 2012 -- project euler — problem 52 (0)
• November 14, 2012 -- project euler-problem 63 (0)
• July 31, 2012 -- project euler — problem 51 (0)