Project Euler > Problem 115 > Counting block combinations II (Java Solution)

Problem:

NOTE: This is a more difficult version of problem 114.

A row measuring n units in length has red blocks with a minimum length of m units placed on it, such that any two red blocks (which are allowed to be different lengths) are separated by at least one black square.

Let the fill-count function, F(m, n), represent the number of ways that a row can be filled.

For example, F(3, 29) = 673135 and F(3, 30) = 1089155.

That is, for m = 3, it can be seen that n = 30 is the smallest value for which the fill-count function first exceeds one million.

In the same way, for m = 10, it can be verified that F(10, 56) = 880711 and F(10, 57) = 1148904, so n = 57 is the least value for which the fill-count function first exceeds one million.

For m = 50, find the least value of n for which the fill-count function first exceeds one million.

Solution:

137846528820

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

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 15
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p015 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p015().run());
	}
	
	
	public String run() {
		return Library.binomial(40, 20).toString();
	}
	
}