Find Digits

Problem Statement

You are given an integer N. Find the digits in this number that exactly divide N (division that leaves 0 as remainder) and display their count. For N=24, there are 2 digits (2 & 4). Both of these digits exactly divide 24. So our answer is 2.

Note

  • If the same number is repeated twice at different positions, it should be counted twice, e.g., For N=122, 2 divides 122 exactly and occurs at ones’ and tens’ position. So for this case, our answer is 3.
  • Division by 0 is undefined.

Input Format

The first line contains T (the number of test cases), followed by T lines (each containing an integer N).

Constraints
1≤T≤15
0<N<1010

Output Format

For each test case, display the count of digits in N that exactly divide N in a separate line.

Sample Input

2
12
1012

Sample Output

2
3

Explanation

2 digits in the number 12 divide it exactly. The first digit, 1, divides it exactly in twelve parts, and the second digit, 2 divides 12 equally in six parts.

1 divides 1012 exactly (and it occurs twice), and 2 also divides it exactly. Division by 0 is undefined, therefore it will not be counted.

This challenge was part of Pragyan 12.

Solution

  
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */ 
    vector vec;
    int t,n,a,count;
    cin>>t;
    while(t--)
        {
        count = 0;
        cin>>n;
        a=n;
        while(n!=0)
            {
            int temp = n%10;
            vec.push_back(temp);
            n/=10;
            }
        reverse(vec.begin(),vec.end());
        for(std::vector::iterator it = vec.begin(); it != vec.end(); ++it)
            {
                if(*it==0);
                else if((a%*it)==0)
                   count++;
                   
            }
        cout<<count<<endl;
        vec.clear();
    }
    return 0;
}

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