Project Euler > Problem 43 > Sub-string divisibility (Java Solution)

Problem:

The number, 1406357289, is a 0 to 9 pandigital number because it is made up of each of the digits 0 to 9 in some order, but it also has a rather interesting sub-string divisibility property.

Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:

* d2d3d4=406 is divisible by 2
* d3d4d5=063 is divisible by 3
* d4d5d6=635 is divisible by 5
* d5d6d7=357 is divisible by 7
* d6d7d8=572 is divisible by 11
* d7d8d9=728 is divisible by 13
* d8d9d10=289 is divisible by 17

Find the sum of all 0 to 9 pandigital numbers with this property.

Solution:

16695334890

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

public interface EulerSolution{

public String run();

}

/* 
 * Solution to Project Euler problem 43
 * By Nayuki Minase
 * 
 * http://nayuki.eigenstate.org/page/project-euler-solutions
 * https://github.com/nayuki/Project-Euler-solutions
 */


public final class p043 implements EulerSolution {
 
 public static void main(String[] args) {
  System.out.println(new p043().run());
 }
 
 
 private static int[] DIVISIBILITY_TESTS = {2, 3, 5, 7, 11, 13, 17};  // First 7 primes
 
 
 public String run() {
  long sum = 0;
  int[] digits = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
  outer:
  do {
   for (int i = 0; i < DIVISIBILITY_TESTS.length; i++) {
    if (toInteger(digits, i + 1, 3) % DIVISIBILITY_TESTS[i] != 0)
     continue outer;
   }
   sum += toInteger(digits, 0, digits.length);
  } while (Library.nextPermutation(digits));
  return Long.toString(sum);
 }
 
 
 private static long toInteger(int[] digits, int off, int len) {
  long result = 0;
  for (int i = off; i < off + len; i++)
   result = result * 10 + digits[i];
  return result;
 }
 
}