89 --> 8¹ + 9² = 89 * 1 得出原本的數的多少倍數等於 分拆次方數
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
Solution:
using System;
using System.Linq;
public class DigPow {
public static long digPow(int n, int p) {
Array squares = Enumerable.Range(p, n.ToString().Length).ToArray();
long s =
long.Parse(string.Concat(
n.ToString().Select(
(x, i) => Math.Pow(double.Parse(x.ToString()), double.Parse(squares.GetValue(i).ToString()))
).Sum()
));
if (s % n == 0)
return s / n;
return -1;
}
}
先宣告一個Array來存放我需要的次方升冪range
之後就是把數字拆開來分別去做次方運算 然後加總 之後再去做除法得出答案