**Find Digits**

### Problem Statement :

An integer d is a divisor of an integer n if the remainder of n/d = 0. Given an integer, for each digit that makes up the integer determine whether it is a divisor. Count the number of divisors occurring within the integer. Example n = 124 Check whether 1, 2 and 4 are divisors of 124. All 3 numbers divide evenly into 124 so return 3. n = 111 Check whether 1, 1, 1 and are divisors of 111. All 3 numbers divide evenly into 111 so return 3. n = 10 Check whether 1 and 0 are divisors of 10. 1 is, but 0 is not. Return 1. Function Description Complete the findDigits function in the editor below. findDigits has the following parameter(s): int n: the value to analyze Returns int: the number of digits in n that are divisors of n Input Format The first line is an integer, t, the number of test cases. The t subsequent lines each contain an integer, n. Constraints 1 <= t <= 15 0 < n < 10^9

### Solution :

` ````
Solution in C :
python 3 :
def func(A):
return len([1 for i in str(A) if i !='0' and A%int(i)==0])
for t in range(int(input())):
print(func(int(input())))
java :
import java.util.*;
class Solution
{
public static void main(String args[])
{
Scanner in=new Scanner(System.in);
int t,ans,d;
long n,m;
t=in.nextInt();
while(t-->0)
{
ans=0;
n=in.nextLong();
m=n;
while(m!=0)
{
d=(int)m%10;
m=m/10;
if(d==0)
continue;
if(n%d==0)
ans++;
}
System.out.println(ans);
}
}
}
C ++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int getR(string n, int q) {
if(!q) return 0;
int r = 0;
for(int i = 0;i < n.length();++i) {
r *= 10;
r += (n[i] - '0');
r %= q;
}
if(!r) return 1;
return 0;
}
int main() {
int T;
string n;
int res = 0;
cin >> T;
while(T--) {
cin >> n;
res = 0;
for(int i = 0;i < n.length();++i)
res += getR(n, n[i] - '0');
cout << res << endl;
}
return 0;
}
C :
#include <stdio.h>
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int N,M,d,c=0;
scanf("%d",&N); M=N;
while(N)
{
d=N%10;
N=N/10;
if(d && M%d==0) c++;
}
printf("%d\n",c);
}
return 0;
}
```

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