Count of numbers upto M divisible by given Prime Numbers

Given an array arr[] of Prime Numbers and a number M, the task is to count the number of elements in the range [1, M] that are divisible by any of the given prime numbers.

Examples:

Input: arr[] = {2, 3, 5, 7} M = 100  
Output: 78  
Explanation:
In total there are 78 numbers that are divisible by either of 2 3 5 or 7.

Input: arr[] = {2, 5, 7, 11} M = 200
Output: 137
Explanation:
In total there are 137 numbers that are divisible by either of 2 5 7 or 11.

Naive Approach: The idea is iterate over the range [1, M] and check if any of the array element is divides the element in the range [1, M] then count the element else check for the next number in the range.



Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <iostream>
using namespace std;
  
// Function to count the numbers that
// are divisible by the numbers in
// the array from range 1 to M
int count(int a[], int M, int N)
{
    // Initialize the count variable
    int cnt = 0;
  
    // Iterate over [1, M]
    for (int i = 1; i <= M; i++) {
  
        // Iterate over array elements arr[]
        for (int j = 0; j < N; j++) {
  
            // Check if i is divisible by a[j]
            if (i % a[j] == 0) {
  
                // Increment the count
                cnt++;
                break;
            }
        }
    }
  
    // Return the answer
    return cnt;
}
  
// Driver code
int main()
{
    // Given array arr[]
    int arr[] = { 2, 3, 5, 7 };
  
    // Given Number M
    int m = 100;
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function Call
    cout << count(arr, m, n);
    return 0;
}

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Java

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// Java program for the above approach
class GFG{
  
// Function to count the numbers that
// are divisible by the numbers in
// the array from range 1 to M
static int count(int a[], int M, int N)
{
      
    // Initialize the count variable
    int cnt = 0;
  
    // Iterate over [1, M]
    for(int i = 1; i <= M; i++) 
    {
          
        // Iterate over array elements arr[]
        for(int j = 0; j < N; j++)
        {
              
            // Check if i is divisible by a[j]
            if (i % a[j] == 0
            {
                  
                // Increment the count
                cnt++;
                break;
            }
        }
    }
      
    // Return the answer
    return cnt;
}
  
// Driver code
public static void main(String[] args)
{
      
    // Given array arr[]
    int arr[] = { 2, 3, 5, 7 };
  
    // Given number M
    int m = 100;
    int n = arr.length;
  
    // Function call
    System.out.print(count(arr, m, n));
}
}
  
// This code is contributed by Amit Katiyar

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Python3

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# Python3 program for the above approach
  
# Function to count the numbers that 
# are divisible by the numbers in 
# the array from range 1 to M 
def count(a, M, N):
      
    # Initialize the count variable
    cnt = 0
  
    # Iterate over [1, M]
    for i in range(1, M + 1):
  
        # Iterate over array elements arr[]
        for j in range(N):
  
            # Check if i is divisible by a[j]
            if (i % a[j] == 0):
  
                # Increment the count
                cnt += 1
                break
  
    # Return the answer
    return cnt 
  
# Driver code
  
# Given list lst
lst = [ 2, 3, 5, 7 ]
  
# Given number M
m = 100
n = len(lst)
  
# Function call
print(count(lst, m, n)) 
  
# This code is contributed by vishu2908    

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Output

78

Time Complexity: O(N*M) 
Auxiliary Space: O(1) 
Another Approach: Another method to solve this problem is use Dynamic Programming and Seive. Mark all the numbers up to M that are divisible by any prime number in the array. Then count all the marked numbers and print it.

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Improved By : vishu2908