Project Euler > Problem 73 > Counting fractions in a range (Java Solution)

Problem:

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?

Solution:

4179871

Code:
The solution may include methods that will be found here: Library.java .

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 23
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p023 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p023().run());
	}
	
	
	private static final int LIMIT = 28123;
	
	private boolean[] isAbundant = new boolean[LIMIT + 1];
	
	public String run() {
		// Compute look-up table
		for (int i = 1; i < isAbundant.length; i++)
			isAbundant[i] = isAbundant(i);
		
		int sum = 0;
		for (int i = 1; i <= LIMIT; i++) {
			if (!isSumOf2Abundants(i))
				sum += i;
		}
		return Integer.toString(sum);
	}
	
	
	private boolean isSumOf2Abundants(int n) {
		for (int i = 0; i <= n; i++) {
			if (isAbundant[i] && isAbundant[n - i])
				return true;
		}
		return false;
	}
	
	
	private static boolean isAbundant(int n) {
		if (n < 1)
			throw new IllegalArgumentException();
		
		int sum = 1;  // Sum of factors less than n
		int end = Library.sqrt(n);
		for (int i = 2; i <= end; i++) {
			if (n % i == 0)
				sum += i + n / i;
		}
		if (end * end == n)
			sum -= end;
		return sum > n;
	}
	
}