/* Given an integer,N, traverse its digits(d1,d2,..., dn) and determine how many digits evenly divide N(i.e.: count the number of times N divided by each digit di has a remainder of 0). Print the number of evenly divisible digits.
Note: Each digit is considered to be unique, so each occurrence of the same evenly divisible digit should be counted(i.e.: for N=111 , the answer is 3).
Input Format
The first line is an integer, T, indicating the number of test cases.
The T subsequent lines each contain an integer, N .
Constraints
1<=T<=15
0<=N<=10^9
Output Format
For every test case, count and print(on a new line) the number of digits in N that are able to evenly divide N.
Sample Input
2
12
1012
Sample Output
2
3
Explanation
The number 12 is broken into two digits,1 and 2. When 12 is divided by either of those digits, the calculation's remainder is 0 ; thus, the number of evenly-divisible digits in 12 is 2 .
The number 1012 is broken into four digits ,1 , 0, 1, and 2. 1012 is evenly divisible by its digits 1, 1, and 2, but it is not divisible by 0 as 1012 division by zero is undefined; thus, our count of evenly divisible digits 1012 is 3 .
*/
public class FindDigit
{
public void GetnoOfDigit()
{
int t = Convert.ToInt32(Console.ReadLine());
for (int a0 = 0; a0 < t; a0++)
{
int n = Convert.ToInt32(Console.ReadLine());
int temp = n, count=0;
while (temp > 0)
{
int d = temp % 10;
if (isNumberDivisiblebyDigit(n, d))
{
count++;
}
temp = temp / 10;
}
Console.WriteLine(count);
}
}
private bool isNumberDivisiblebyDigit(int n, int d)
{
return d==0 || n==d? false:n % d == 0;
}
}
}
