Scala: the best solution
QUESTION:
We define the supervalue of an integer x using the following rules:
Iff x has only 1 digit, then its super digit is x.
Otherwise, the super digit of x is equal to the super digit of the digit-sum of x. Here, digit-sum of a number is defined as the sum of its digits.
For example, super digit of 9875 will be calculated as:
super-digit(9875) = super-digit(9+8+7+5)
= super-digit(29)
= super-digit(2+9)
= super-digit(11)
= super-digit(1+1)
= super-digit(2)
= 2.
You are given two numbers - n k. You have to calculate the super digit of P.
P is created when number n is concatenated k times. That is, if n = 123 and k = 3, then P = 123123123.
Input Format
Input will contain two space separated integers, n and k.
Output Format
Output the super digit of P, where P is created as described above.
Constraint
1≤n<10100000
1≤k≤105
Sample Input
148 3
Sample Output
3
Explanation
Here n = 148 and k = 3, so P = 148148148.
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.
I wrote the following program to solve the above problem, but how to solve it even efficiently and works with a string than a math operation? and for multiple inputs it takes a long time like
861568688536788 100000
object SuperDigit {
def main(args: Array[String]) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution
*/
def generateString (no:String,re:BigInt , tot:BigInt , temp:String):String = {
if(tot-1>re) generateString(no+temp,re+1,tot,temp)
else no
}
def totalSum(no:List[Char]):BigInt = no match {
case x::xs => x.asDigit+totalSum(xs)
case Nil => '0'.asDigit
}
def tot(no:List[Char]):BigInt = no match {
case _ if no.length == 1=> no.head.asDigit
case no => tot(totalSum(no).toString.toList)
}
var list = readLine.split(" ");
var one = list.head.toString();
var two = BigInt(list(1));
//println(generateString("148",0,3,"148"))
println(tot(generateString(one,BigInt(0),two,one).toList))
}
}
+3
source to share
1 answer
One solution is to understand that you do not need to concatenate a number considered as a string k times, but can start with k * qs (n) (where qs is a function that maps a number to its sum of digits, i.e. i.e. qs (123) = 1 + 2 + 3). Here's a more functional programming style. I don't know if it can be done faster than this.
object Solution {
def qs(n: BigInt): BigInt = n.toString.foldLeft(BigInt(0))((n, ch) => n + (ch - '0').toInt)
def main(args: Array[String]) {
val input = scala.io.Source.stdin.getLines
val Array(n, k) = input.next.split(" ").map(BigInt(_))
println(Stream.iterate(k * qs(n))(qs(_)).find(_ < 10).get)
}
}
+4
source to share