Sum of Digits Divisibility

PROBLEM :

Check that the number can be divided by the sum of its digit.

Input:

The first line of input contains an integer T denoting the number of test cases.Then T test cases follow .Each test case consist of an integer N.

Output:

Print 1 if divisible else 0.

Constraints:

1 = T = 100
1 = N = 100000

Example:

Input
2
18
19170

Output
1
1
     
--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION :
--------------------------------------------------------------------------------

#include<iostream>
using namespace std;
int main()
 {
int t,rem,sum ;
long no,org ;
cin>>t ;
while(t--)
{
   cin>>no ;
 
   sum=0 ;
   org=no ;
 
   while(no!=0)
   {
       rem=no%10 ;
       sum=sum+rem ;
       no=no/10 ;
   }
 
   if(org%sum==0)
       cout<<1 ;
   else
       cout<<0 ;
   cout<<endl ;
}
return 0;
}

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