project euler -- problem 72

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 21 elements in this set.

How many elements would be contained in the set of reduced proper fractions for d ≤ 1,000,000?

This problem is relatively easy if problem 70 is solved. The number of reduced proper fraction of d is equal to number less than d and is relatively prime to d. This can be calculated by the euler's totient function.

The *totient* function defined in problem 70, was used here to calculate φ(d).

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#include<iostream>
#include<cmath>
 
using namespace std;
 
const int N = 1e6+1;
 
int totient(int n);
 
int main() {
  long int cnt=0;
  for (int d = 2; d < N; d++) {
    cnt += totient(d);
  }
  cout << "Answers to PE 72: " << cnt << endl;
}
 
int totient(int n) {
  static int phis[N] = {0, 1, 1};
  if (phis[n] != 0) {
    return(phis[n]);
  }
  for (int p=2; p <= sqrt(n); p++) {
    if ( n % p == 0 ) {
      int pk = p;
      while ( (n % (pk * p) == 0) ) {
	pk *= p;
      }
      if ( pk == n ) {
	// phis[n] = p^k
	phis[n] = n - n/p;
	return phis[n];
      }
      if ( phis[pk] == 0 ) {
	phis[pk] = pk - pk/p;
      }
      phis[n] = totient(pk) * totient(n/pk);
      return phis[n];
    }
  }
  phis[n] = n-1;
  return phis[n];
}

Runs < 2 mins.

Answers to PE 72: 303963552391
   user  system elapsed 
  1.188   0.010   1.201 

Related Posts

• January 8, 2013 -- project euler — problem 71 (0)
• January 10, 2013 -- project euler — problem 73 (0)
• December 31, 2012 -- project euler — problem 70 (0)
• December 14, 2012 -- project euler — problem 69 (0)
• July 31, 2012 -- project euler — problem 51 (0)