小于N且主除数之和为K的数字的计数

给定两个整数K和N ，任务是从素数和为K的[2，N – 1]范围内找到整数的数目

例子：

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

的方法：创建一个数组sumPF []其中sumPF [I]存储我的素因数，可以使用本文中所使用的方法来容易地计算出的总和。现在，初始化一个变量count = 0并运行一个从2到N – 1的循环，如果sumPF [i] = K，则对每个元素i递增计数。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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#

// 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

输出：

2