package kata_011;

/**
 * Some numbers have funny properties. For example:
 * 
 * 89 --> 8¹ + 9² = 89 * 1
 * 
 * 695 --> 6² + 9³ + 5⁴= 1390 = 695 * 2
 * 
 * 46288 --> 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51
 * 
 * Given a positive integer n written as abcd... (a, b, c, d... being digits)
 * and a positive integer p we want to find a positive integer k, if it exists,
 * such as the sum of the digits of n taken to the successive powers of p is
 * equal to k * n. In other words:
 * 
 * Is there an integer k such as : (a ^ p + b ^ (p+1) + c ^(p+2) + d ^ (p+3) + ...) = n * k
 * 
 * If it is the case we will return k, if not return -1.
 * 
 * Note: n, p will always be given as strictly positive integers.
 * 
 * digPow(89, 1) should return 1 since 8¹ + 9² = 89 = 89 * 1 digPow(92, 1)
 * should return -1 since there is no k such as 9¹ + 2² equals 92 * k
 * digPow(695, 2) should return 2 since 6² + 9³ + 5⁴= 1390 = 695 * 2
 * digPow(46288, 3) should return 51 since 4³ + 6⁴+ 2⁵ + 8⁶ + 8⁷ = 2360688 = 46288 * 51
 * 
 * @author SeeClanUkyo
 *
 */
public class DigPow {
    public static void main(String[] args) {

        System.out.println(digPow(46288, 3));
    }

    public static long digPow(int n, int p) {
        // your code
        if (n > 0) {
            String nstr = n + "";
            int nlen = nstr.length();

            long sum = 0;
            for (int i = 0; i < nlen; i++) {
                sum += Math.pow(Integer.parseInt(nstr.substring(i, i + 1)), (p + i));
                if (sum % n == 0) {
                    return sum / n;
                }
            }

        }
        return -1;
    }
}