Project Euler > Problem 126 > Cuboid layers (Java Solution)

Problem:

The minimum number of cubes to cover every visible face on a cuboid measuring 3 x 2 x 1 is twenty-two.


If we then add a second layer to this solid it would require forty-six cubes to cover every visible face, the third layer would require seventy-eight cubes, and the fourth layer would require one-hundred and eighteen cubes to cover every visible face.

However, the first layer on a cuboid measuring 5 x 1 x 1 also requires twenty-two cubes; similarly the first layer on cuboids measuring 5 x 3 x 1, 7 x 2 x 1, and 11 x 1 x 1 all contain forty-six cubes.

We shall define C(n) to represent the number of cuboids that contain n cubes in one of its layers. So C(22) = 2, C(46) = 4, C(78) = 5, and C(118) = 8.

It turns out that 154 is the least value of n for which C(n) = 10.

Find the least value of n for which C(n) = 1000.

Solution:

983

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

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 26
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */

import java.util.HashMap;
import java.util.Map;


public final class p026 implements EulerSolution {
	
	public static void main(String[] args) {
		System.out.println(new p026().run());
	}
	
	
	public String run() {
		int bestNumber = 0;
		int bestLength = 0;
		for (int i = 1; i <= 1000; i++) {
			int len = getCycleLength(i);
			if (len > bestLength) {
				bestNumber = i;
				bestLength = len;
			}
		}
		return Integer.toString(bestNumber);
	}
	
	
	private static int getCycleLength(int n) {
		Map<Integer,Integer> stateToIter = new HashMap<Integer,Integer>();
		int state = 1;
		int iter = 0;
		while (!stateToIter.containsKey(state)) {
			stateToIter.put(state, iter);
			state = state * 10 % n;
			iter++;
		}
		return iter - stateToIter.get(state);
	}
	
}