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#49 Playing with digits

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

import java.util.stream.IntStream;

public class DigPow {
	
	public static long digPow(int n, int p) {
		// your code
		int input = n;
	    int result = 0;
	    p += String.valueOf(n).length()-1;

	    while (n != 0) {
	        result += Math.pow((n % 10), p--);
	        n /= 10;
	    }
	    if(result%input != 0)
	    	return -1;
	    return result/input;
	}
	
	public static long digPow2(int n, int p) {
		// your code
	    int[] digits = String.valueOf(n).chars().map(Character::getNumericValue).toArray();
	    int sum = IntStream.range(0, digits.length).map(i -> (int) Math.pow(digits[i], i + p)).sum();
	    return sum%n != 0 ? -1 : sum/n;
	}
	
}