**Recursive Digit Sum**

### Problem Statement :

We define super digit of an integer x using the following rules: Given an integer, we need to find the super digit of the integer. If 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 sum of the digits of x. For example, the super digit of9875 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 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 n to make p Returns int: the super digit of n repeated k times Input Format The first line contains two space separated integers, n and k. Constraints 1 <= n <= 10^100000 1 <= k <= 10^5

### Solution :

` ````
Solution in C :
In C++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int k=0,flag=1;
int sum(string s){
int64_t a=0,b;
for(int i=0;i<s.size();i++){
b=s[i]-48;
a+=b;
}
return a;
}
void super(int64_t a){
if(a/10==0)
{
if(flag){
flag=0;
super(a*k);
}
else
cout<<a;
}
else{
int64_t val=0;
while(a){
val+=a%10;
a=a/10;
}
super(val);
}
}
int main() {
string s;
int64_t a;
cin>>s>>k;
a=sum(s);
super(a);
return 0;
}
In Java :
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Solution s = new Solution();
Scanner sc = new Scanner(System.in);
String str_n = sc.next();
int k = sc.nextInt();
int pSum = Integer.parseInt(s.supdig(str_n));
pSum *= k;
String sup = Integer.toString(s.supdig(pSum));
System.out.println(sup);
}
String supdig(String n) {
if(n.length() == 1) return n;
else {
int np = 0;
for(int i = 0; i < n.length(); i++) {
np += Character.getNumericValue( n.charAt(i) );
}
return supdig(Integer.toString(np));
}
}
int supdig(int n) {
if(n / 10 == 0) return n;
else {
int r = 0;
while(n > 0) {
r += n % 10;
n /= 10;
}
return supdig(r);
}
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
long long sd=0,h;
int k;
int c;
do{
c=getchar();
if(c != ' ')
sd += c -'0';
}while(c != ' ');
scanf("%d",&k);
sd *= k;
while(sd > 10){
h=0;
while(sd > 0){
h+= sd %10;
sd = sd /10;
}
sd =h;
}
printf("%lld\n",sd);
return 0;
}
In Python3 :
#!/bin/python3
def digits_sum(n):
s = 0
while n > 9:
s += n % 10
n //= 10
s += n
return s
def super_digit(n):
d = digits_sum(n)
if d < 10:
return d
return super_digit(d)
def super_digit_iter(n):
d = digits_sum(n)
while d > 9:
d = digits_sum(d)
return d
def main():
n, k = map(int, input().strip().split(' '))
print(super_digit_iter(digits_sum(n) * k))
if __name__ == "__main__":
main()
```

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