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# Count subsets that satisfy the given condition in C++

Given an array of numbers and an integer x as input. The goal is to find all the subsets of arr[] such that individual elements of that set as well as the sum of them fully divisible by x.

**For Example**

## Input

arr[] = {1,2,3,4,5,6} x=3

## Output

Count of subsets that satisfy the given condition :3

## Explanation

The subsets will be: [3], [6], [3,6]

## Input

arr[] = {1,2,3,4,5,6} x=4

## Output

Count of subsets that satisfy the given condition :1

## Explanation

The subsets will be: [4]

**Approach used in the below program is as follows** −

In this approach we will count the elements of arr[] that are fully divisible by x and then return 2^{count}−1 as required number of subsets.

Take an integer array arr[].

Take x as input.

Function count(int arr[], int n, int x) takes an array and x and returns count of subsets that satisfy the given condition.

If x is 1 then it divides all elements, so return

## Example

#include <bits/stdc++.h> #define ll long long int using namespace std; int sub_sets(int arr[], int size, int val){ int count = 0; if (val == 1){ count = pow(2, size) − 1; return count; } for (int i = 0; i < size; i++){ if (arr[i] % val == 0){ count++; } } count = pow(2, count) − 1; return count; } int main(){ int arr[] = { 4, 6, 1, 3, 8, 10, 12 }, val = 4; int size = sizeof(arr) / sizeof(arr[0]); cout<<"Count of sub−sets that satisfy the given condition are: "<<sub_sets(arr, size, val); return 0; }

## Output

If we run the above code it will generate the following output −

Count of sub−sets that satisfy the given condition are: 7

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