# 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;
}```
```

## Merging Communities

People connect with each other in a social network. A connection between Person I and Person J is represented as . When two persons belonging to different communities connect, the net effect is the merger of both communities which I and J belongs to. At the beginning, there are N people representing N communities. Suppose person 1 and 2 connected and later 2 and 3 connected, then ,1 , 2 and 3 w

## Components in a graph

There are 2 * N nodes in an undirected graph, and a number of edges connecting some nodes. In each edge, the first value will be between 1 and N, inclusive. The second node will be between N + 1 and , 2 * N inclusive. Given a list of edges, determine the size of the smallest and largest connected components that have or more nodes. A node can have any number of connections. The highest node valu

## Kundu and Tree

Kundu is true tree lover. Tree is a connected graph having N vertices and N-1 edges. Today when he got a tree, he colored each edge with one of either red(r) or black(b) color. He is interested in knowing how many triplets(a,b,c) of vertices are there , such that, there is atleast one edge having red color on all the three paths i.e. from vertex a to b, vertex b to c and vertex c to a . Note that

## Super Maximum Cost Queries

Victoria has a tree, T , consisting of N nodes numbered from 1 to N. Each edge from node Ui to Vi in tree T has an integer weight, Wi. Let's define the cost, C, of a path from some node X to some other node Y as the maximum weight ( W ) for any edge in the unique path from node X to Y node . Victoria wants your help processing Q queries on tree T, where each query contains 2 integers, L and

## Contacts

We're going to make our own Contacts application! The application must perform two types of operations: 1 . add name, where name is a string denoting a contact name. This must store name as a new contact in the application. find partial, where partial is a string denoting a partial name to search the application for. It must count the number of contacts starting partial with and print the co

## No Prefix Set

There is a given list of strings where each string contains only lowercase letters from a - j, inclusive. The set of strings is said to be a GOOD SET if no string is a prefix of another string. In this case, print GOOD SET. Otherwise, print BAD SET on the first line followed by the string being checked. Note If two strings are identical, they are prefixes of each other. Function Descriptio