Count of numbers below N whose sum of prime divisors is K

Last Updated : 18 Mar, 2022

Given two integers K and N, the task is to find the count of integers from the range [2, N - 1] whose sum of prime divisors is K
Example:

Input: N = 20, K = 7
Output: 2
7 and 10 are the only valid numbers.
sumPFactors(7) = 7
sumPFactors(10) = 2 + 5 = 7
Input: N = 25, K = 5
Output: 5

Approach: Create an array sumPF[] where sumPF[i] stores the sum of prime divisors of i which can be easily calculated using the approach used in this article. Now, initialise a variable count = 0 and run a loop from 2 to N - 1 and for every element i if sumPF[i] = K then increment the count.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

#define MAX 1000001

// Function to return the count of numbers
// below N whose sum of prime factors is K
int countNum(int N, int K)
{
    // To store the sum of prime factors
    // for all the numbers
    int sumPF[MAX] = { 0 };

    for (int i = 2; i < N; i++) {

        // If i is prime
        if (sumPF[i] == 0) {

            // Add i to all the numbers
            // which are divisible by i
            for (int j = i; j < N; j += i) {
                sumPF[j] += i;
            }
        }
    }

    // To store the count of required numbers
    int count = 0;
    for (int i = 2; i < N; i++) {
        if (sumPF[i] == K)
            count++;
    }

    // Return the required count
    return count;
}

// Driver code
int main()
{
    int N = 20, K = 7;

    cout << countNum(N, K);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG 
{
static int MAX = 1000001;

// Function to return the count of numbers
// below N whose sum of prime factors is K
static int countNum(int N, int K)
{
    // To store the sum of prime factors
    // for all the numbers
    int []sumPF = new int[MAX];

    for (int i = 2; i < N; i++) 
    {

        // If i is prime
        if (sumPF[i] == 0) 
        {

            // Add i to all the numbers
            // which are divisible by i
            for (int j = i; j < N; j += i) 
            {
                sumPF[j] += i;
            }
        }
    }

    // To store the count of required numbers
    int count = 0;
    for (int i = 2; i < N; i++)
    {
        if (sumPF[i] == K)
            count++;
    }

    // Return the required count
    return count;
}

// Driver code
public static void main(String[] args) 
{
    int N = 20, K = 7;

    System.out.println(countNum(N, K));
}
} 

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach 
MAX = 1000001

# Function to return the count of numbers 
# below N whose sum of prime factors is K 
def countNum(N, K) : 

    # To store the sum of prime factors 
    # for all the numbers 
    sumPF = [0] * MAX; 

    for i in range(2, N) :
        
        # If i is prime 
        if (sumPF[i] == 0) : 

            # Add i to all the numbers 
            # which are divisible by i 
            for j in range(i, N, i) :
                sumPF[j] += i; 

    # To store the count of required numbers 
    count = 0; 
    for i in range(2, N) :
        if (sumPF[i] == K) :
            count += 1; 

    # Return the required count 
    return count; 

# Driver code 
if __name__ == "__main__" : 

    N = 20; K = 7;
    
    print(countNum(N, K)); 

# This code is contributed by AnkitRai01

// C# implementation of the approach
using System;
    
class GFG 
{
static int MAX = 1000001;

// Function to return the count of numbers
// below N whose sum of prime factors is K
static int countNum(int N, int K)
{
    // To store the sum of prime factors
    // for all the numbers
    int []sumPF = new int[MAX];

    for (int i = 2; i < N; i++) 
    {

        // If i is prime
        if (sumPF[i] == 0) 
        {

            // Add i to all the numbers
            // which are divisible by i
            for (int j = i; j < N; j += i) 
            {
                sumPF[j] += i;
            }
        }
    }

    // To store the count of required numbers
    int count = 0;
    for (int i = 2; i < N; i++)
    {
        if (sumPF[i] == K)
            count++;
    }

    // Return the required count
    return count;
}

// Driver code
public static void Main(String[] args) 
{
    int N = 20, K = 7;

    Console.WriteLine(countNum(N, K));
}
}

// This code is contributed by 29AjayKumar

JavaScript

<script>

// Javascript implementation of the approach

const MAX = 1000001;

// Function to return the count of numbers
// below N whose sum of prime factors is K
function countNum(N, K)
{
    // To store the sum of prime factors
    // for all the numbers
    let sumPF = new Array(MAX).fill(0);

    for (let i = 2; i < N; i++) {

        // If i is prime
        if (sumPF[i] == 0) {

            // Add i to all the numbers
            // which are divisible by i
            for (let j = i; j < N; j += i) {
                sumPF[j] += i;
            }
        }
    }

    // To store the count of required numbers
    let count = 0;
    for (let i = 2; i < N; i++) {
        if (sumPF[i] == K)
            count++;
    }

    // Return the required count
    return count;
}

// Driver code
    let N = 20, K = 7;

    document.write(countNum(N, K));

</script>

Output:

Time Complexity: O(N²)

Auxiliary Space: O(MAX)

Sum of all the prime divisors of a number

Samdare B

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