Trailing zeroes in factorial

PROBLEM :

For an integer n find number of trailing zeroes in n! .

Input:
The first line contains an integer 'T' denoting the total number of test cases. In each test cases, it contains an integer 'N'.


Output:
In each seperate line output the answer to the problem.


Constraints:
1<=T<=100
1<=N<=1000


Example:
Input:
1
9

Output:
1

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

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

int Trailing_zeroes(int no)
{
   int i , count ;
   count=0 ;
 
   for(i=5;no/i>=1;i*=5)
   {
       count+=no/i ;
   }
   return count ;
}

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

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