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:
// C++ simple program to // find elements which are // divisible by N #include <bits/stdc++.h> 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