Question
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.
Solution : There are some possible solutions. But here, the best solution is given. There is a pattern. If the number is divisible by 9 then the answer will be 9. If not then the answer will be n mod 9 .
Code
#include<iostream>
using namespace std;
int main()
{
//code
long long int i,j,k,l,m,n,t;
cin>>t;
for(i=1;i<=t;i++)
{
cin>>n;
if(n%9==0)
{
cout<<9<<endl;
}
else
{
cout<<n%9<<endl;
}
}
return 0;
}
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