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.

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

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.

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())):

java  :

import java.util.*;
class Solution
	public static void main(String args[])
		Scanner in=new Scanner(;
		int t,ans,d;
		long n,m;

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;
        int N,M,d,c=0;
        scanf("%d",&N); M=N;
            if(d && M%d==0) c++;
    return 0;

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