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 :



title-img


                            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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