Recursive Digit Sum - c++

Question

I was trying to solve this problem on hackerrank. But I got some problem. Specific problem is:
For example:
The sum of digits 9875 will be calculate as: sum(9875) = 9+8+7+5 = 29. sum(29) = 11. sum(11) = 2. (Using recursive function).
In my test case, (n ='9875', k=4) the number p is created by concatenating the string n k times so the initial p = 9875987598759875 ( the string '9875' repeat 4 times ).
But when i code this test case, it doesn't work. Here is my source code:
int SuperDigit(long n){
long sum =0;
if(n==0) return 0;
else{
return sum= sum +(n%10 + SuperDigit(n/10));
}
if(sum>10){
return (sum%10 + SuperDigit(sum/10));
}
}
int main(){
string n;cin>>n;
int T;cin>>T;
string repeat;
for(int i=0; i <T;i++){
repeat += n;
}
cout<<repeat;
long x=0;
stringstream geek(repeat);
geek>>x;
long sum = SuperDigit(x);
printf("\n%ld ",sum);
for(int i=0;i<10;i++){
if(sum>=10){
sum = SuperDigit(sum);
}
else{
break;
}
}
printf("\n%ld ",sum);
}
If i try: n = '123' and k =3 (Expected output: 9)
My output will be correct, here is my output for this test case:
123 3
123123123
18
9
But when i try n = '9875' and k = 4 (Expected output: 8)
My output will be wrong:
9875 4
9875987598759875
46
1
As you can see in this test case, the first sum of all digits must be 116. But mine only show 46. Can anyone explain for me? Thanks a lot!

Answer 1

In your current code you return prematurely in
if(n==0) return 0;
else{
return sum= sum +(n%10 + SuperDigit(n/10));
}
Imagine that n == 89 so n%10 returns 9 and SuperDigit(n/10) returns 8 and you have 17 as an answer (when 8 is expected).
You can put it as
int SuperDigit(long n) {
int result = 0;
/* We compute digital root (sum of digits) */
for (long number = n; number != 0; number /= 10)
result += (int) (number % 10);
/* if result is out of range [-9..9]
we compute digital root again from the answer */
if (result < -9 || result > 9)
result = SuperDigit(result);
return result;
}

Answer 2

You can simplify your program as shown below. Since you want to find the sum recursively, the below program shows one possible way of doing it.
Version 1: Using recursive function
#include <iostream>
int findDigit(int passed_num, int currentSum)
{
int lastDigit;
if (passed_num == 0) {
return currentSum;
}
// find the last didit
lastDigit = passed_num % 10;
currentSum+= lastDigit;
//call findDigit() repeatedly
currentSum = findDigit(passed_num / 10, currentSum);
std::cout<<lastDigit<<" ";
return currentSum;
}
int main()
{
std::cout << "Enter a number: ";
int input_num, sum;
std::cin>>input_num;
sum = findDigit(input_num, 0);
std::cout<<"sum is: "<<sum<<std::endl;
std::cout << "Enter another number: ";
std::cin>>input_num;
sum = findDigit(input_num, 0);
std::cout<<"sum is: "<<sum<<std::endl;
return 0;
}
Note there are simpler(other) ways of finding the sum without recursively. One such way is shown below:
Version 2: Using loop
#include <string>
#include <iostream>
int main()
{
std::cout << "Enter a number: ";
int individual_number = 0, sum = 0;//these are local built in types so initialize them
std::string input_num;
std::cin >> input_num;
for(char c : input_num)
{
individual_number = c -'0';
std::cout<<individual_number<<" ";
sum+= individual_number;
}
std::cout<<"total amount: "<<sum<<std::endl;
// std::cout<<"The sum comes out to be: "<<sum<<std::endl;
return 0;
}