Repetitive Addition Of Digits

PROBLEM :

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

Input:
The first line contains 'T' denoting the number of testcases. Then follows description of testcases. The next T lines contains a single integer N denoting the value of N.

Output:
Output the sum of all its digit until the result has only one digit.

Constraints:
1<=T<=30
1<=n<=10^9

Example:
Input :
2
1
98

Output :
1
8

Explanation:  For example, if we conisder 98, we get 9+8  = 17 after first addition. Then we get 1+7 = 8.  We stop at this point because we are left with one digit.

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

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

int Repetitive_Addition(long long no)
{
    long long sum=0 ;
    int rem ;
   
    while(no)
    {
        rem=no%10 ;
        sum=sum+rem ;
        no=no/10 ;
    }
   
    if(sum>=10)
        return Repetitive_Addition(sum) ;
   
    return sum ;
}

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

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

int Repetitive_Addition(long long no)
{
    if(no%9==0)
        return 9 ;
    return no%9 ;
}

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