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  4. Recursive Digit Sum

Recursive Digit Sum

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We define super digit of an integer using the following rules:

Given an integer, we need to find the super digit of the integer.

  • If has only digit, then its super digit is .
  • Otherwise, the super digit of is equal to the super digit of the sum of the digits of .

For example, the super digit of will be calculated as:

	super_digit(9875)   	9+8+7+5 = 29 
	super_digit(29) 	2 + 9 = 11
	super_digit(11)		1 + 1 = 2
	super_digit(2)		= 2  

Example

The number is created by concatenating the string times so the initial .

    superDigit(p) = superDigit(9875987598759875)
                  9+8+7+5+9+8+7+5+9+8+7+5+9+8+7+5 = 116
    superDigit(p) = superDigit(116)
                  1+1+6 = 8
    superDigit(p) = superDigit(8)

All of the digits of sum to . The digits of sum to . is only one digit, so it is the super digit.

Function Description

Complete the function superDigit in the editor below. It must return the calculated super digit as an integer.

superDigit has the following parameter(s):

  • string n: a string representation of an integer
  • int k: the times to concatenate to make

Returns

  • int: the super digit of repeated times

Input Format

The first line contains two space separated integers, and .

Constraints

Sample Input 0

148 3

Sample Output 0

3

Explanation 0

Here and , so .

super_digit(P) = super_digit(148148148) 
               = super_digit(1+4+8+1+4+8+1+4+8)
               = super_digit(39)
               = super_digit(3+9)
               = super_digit(12)
               = super_digit(1+2)
               = super_digit(3)
               = 3

Sample Input 1

9875 4

Sample Output 1

8

Sample Input 2

123 3

Sample Output 2

9

Explanation 2

Here and , so .

super_digit(P) = super_digit(123123123) 
               = super_digit(1+2+3+1+2+3+1+2+3)
               = super_digit(18)
               = super_digit(1+8)
               = super_digit(9)
               = 9

Author

wanbo

Difficulty

Medium

Max Score

30

Submitted By

84826

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