Find elements of an array which are divisible by N using STL in C++
Last Updated :
18 Dec, 2019
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
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:
#include <bits/stdc++.h>
using namespace std;
int N;
bool isDiv( int i)
{
if (i % N == 0)
return true ;
else
return false ;
}
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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