题目

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 and 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

代码

public class DigPow {
    public static long digPow(int n, int p) {
    string s=n.ToString();
    double sum=0;
    for(int i=0;i<s.Length;i++){
      int num=s[i]-'0';
      double d=System.Math.Pow(num,p+i);
      sum+=d;
    }// your code
    if(sum%n==0){return (long)sum/n;}
        else{return -1;}
    }
}

解题思路

本题需要考虑以下操作：
1. 将整数每一位拆分为一个个的数字
2. 将字符char转换为int整数（直接-‘0’）
3. 注意数据类型（Math.Pow()是double）