Repeated sum of digits

PROBLEM :

Given an integer N, recursively sum digits of N until we get a single digit.  The process can be described below

If N < 10  
    digSum(N) = N
Else        
    digSum(N) = Sum(digSum(N))
Example:

Input : 1234
Output : 1
Explanation : The sum of 1+2+3+4 = 10,
              digSum(x) == 10
              Hence ans will be 1+0 = 1
Input:

The first line contains an integer T, total number of test cases. Then following T lines contains an integer N.

Output:

Repeated sum of digits of N.

Constraints:

1 = T = 100

1 = N = 1000000

Example:

Input
2
123
9999

Output
6
9

--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------

#include<iostream>
using namespace std;
int Repeated_sum_of_digits(int ) ;
int sum_digits(int ) ;
int main()
 {
int t,no ;
cin>>t ;
while(t--)
{
   cin>>no ;
   cout<<Repeated_sum_of_digits(no)<<endl ;
}
return 0;
}

int Repeated_sum_of_digits(int no)
{
    if(no<10)
        return no ;
    else
        Repeated_sum_of_digits(sum_digits(no)) ;
}

int sum_digits(int no)
{
    int r,sum ;
    sum=0 ;
   
    while(no)
    {
        r=no%10 ;
        sum=sum+r ;
        no=no/10 ;
    }
    return sum ;
}

---------------------------------------------------------------------------------

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