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# Find elements of an array which are divisible by N using STL in C++

Given an array and an integer N, find elements which are divisible by N, using STL in C++

Examples:

```Input: a[] = {1, 2, 3, 4, 5, 10}, N = 2
Output: 3
Explanation:
As 2, 4, and 10 are divisible by 2
Therefore the output is 3

Input:a[] = {4, 3, 5, 9, 11}, N = 5
Output: 1
Explanation:
As only 5 is divisible by 5
Therefore the output is 3
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: This can be achieved using count_if() method in C++

Syntax:

```count_if(lower_bound, upper_bound, function)
```

where function takes the element of given sequence one by one as a parameter and returns a boolean value on the basis of condition specified in that function.

In this case, the function will be:

```bool isDiv(int i)
{
if (i % N == 0)
return true;
else
return false;
}
```

Below is the implementation of the above approach:

 `// C++ simple program to ` `// find elements which are ` `// divisible by N ` ` `  `#include ` `using` `namespace` `std; ` ` `  `int` `N; ` ` `  `// Function to check ` `// if the element is divisible by N ` `bool` `isDiv(``int` `i) ` `{ ` `    ``if` `(i % N == 0) ` `        ``return` `true``; ` `    ``else` `        ``return` `false``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `a[] = { 1, 2, 6, 3, 4, 5 }; ` `    ``N = 2; ` ` `  `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]); ` ` `  `    ``int` `count = count_if(a, a + n, isDiv); ` ` `  `    ``cout << ``"Elements divisible by "` `         ``<< N << ``": "` `<< count; ` ` `  `    ``return` `0; ` `} `

Output:

```Elements divisible by 2: 3
```
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