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Sum of Semi-Prime Numbers less than or equal to N
  • Last Updated : 19 Jul, 2019

Given an integer N, the task is to find the sum of semi-prime numbers which are less than or equal to N. A Semi prime number is a number which is a multiple of two prime numbers.

Examples:

Input: N = 6
Output: 10
4 and 6 are the semi primes ≤ 6
4 + 6 = 10

Input: N = 10000000
Output: 9322298311255

Approach:



  1. First Calculate the primes less than or equal to N using Sieve and store them in a vector in sorted order.
  2. Iterate over the vector of primes. Fix one of the primes and starting checking the value of the product of all primes with this fixed prime.
  3. As the primes are arranged in sorted order, once we find a prime for which the product exceeds N, then it would exceed for all remaining primes. Hence, break the nested loop here.
  4. Add the product value to the answer variable for all valid pairs.

Below is the implementation of the above approach:

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
  
// Vector to store the primes
vector<ll> pr;
  
// Create a boolean array "prime[0..n]"
bool prime[10000000 + 1];
void sieve(ll n)
{
  
    // Initialize all prime values to be true
    for (int i = 2; i <= n; i += 1) {
        prime[i] = 1;
    }
  
    for (ll p = 2; (ll)p * (ll)p <= n; p++) {
  
        // If prime[p] is not changed then it is a prime
        if (prime[p] == true) {
  
            // Update all multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked
            for (ll i = (ll)p * (ll)p; i <= n; i += p)
                prime[i] = false;
        }
    }
  
    // Print all prime numbers
    for (ll p = 2; p <= n; p++)
        if (prime[p])
            pr.push_back(p);
}
  
// Function to return the semi-prime sum
ll SemiPrimeSum(ll N)
{
  
    // Variable to store the sum of semi-primes
    ll ans = 0;
  
    // Iterate over the prime values
    for (int i = 0; i < pr.size(); i += 1) {
  
        for (int j = i; j < pr.size(); j += 1) {
  
            // Break the loop once the product exceeds N
            if ((ll)pr[i] * (ll)pr[j] > N)
                break;
  
            // Add valid products which are less than
            // or equal to N
            // each product is a semi-prime number
            ans += (ll)pr[i] * (ll)pr[j];
        }
    }
    return ans;
}
  
// Driver code
int main()
{
  
    ll N = 6;
  
    sieve(N);
  
    cout << SemiPrimeSum(N);
  
    return 0;
}

Java




// Java implementation of the above approach
import java.util.*;
  
class GFG
{
  
// Vector to store the primes
static Vector<Long> pr = new Vector<>();
  
// Create a boolean array "prime[0..n]"
static boolean prime[] = new boolean[10000000 + 1];
static void sieve(long n)
{
  
    // Initialize along prime values to be true
    for (int i = 2; i <= n; i += 1
    {
        prime[i] = true;
    }
  
    for (int p = 2; (int)p * (int)p <= n; p++)
    {
  
        // If prime[p] is not changed then it is a prime
        if (prime[p] == true)
        {
  
            // Update along multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked
            for (int i = (int)p * (int)p; i <= n; i += p)
                prime[i] = false;
        }
    }
  
    // Print all prime numbers
    for (int p = 2; p <= n; p++)
        if (prime[p])
            pr.add((long)p);
}
  
// Function to return the semi-prime sum
static long SemiPrimeSum(long N)
{
  
    // Variable to store the sum of semi-primes
    long ans = 0;
  
    // Iterate over the prime values
    for (int i = 0; i < pr.size(); i += 1
    {
  
        for (int j = i; j < pr.size(); j += 1
        {
  
            // Break the loop once the product exceeds N
            if ((long)pr.get(i) * (long)pr.get(j) > N)
                break;
  
            // Add valid products which are less than
            // or equal to N
            // each product is a semi-prime number
            ans += (long)pr.get(i) * (long)pr.get(j);
        }
    }
    return ans;
}
  
// Driver code
public static void main(String[] args)
{
    long N = 6;
  
    sieve(N);
  
    System.out.println(SemiPrimeSum(N));
}
}
  
// This code is contributed by Princi Singh

Python3




# Python3 implementation of the approach
  
# Vector to store the primes
pr=[]
  
# Create a boolean array "prime[0..n]"
prime = [1 for i in range(10000000 + 1)]
def sieve(n):
  
  
    for p in range(2, n):
  
        if p * p > n:
            break
  
        # If prime[p] is not changed then it is a prime
        if (prime[p] == True):
  
            # Update amultiples of p greater than or
            # equal to the square of it
            # numbers which are multiple of p and are
            # less than p^2 are already been marked
            for i in range(2 * p, n + 1, p):
                prime[i] = False
      
  
    # Praprime numbers
    for p in range(2, n + 1):
        if (prime[p]):
            pr.append(p)
  
  
# Function to return the semi-prime sum
def SemiPrimeSum(N):
  
    # Variable to store the sum of semi-primes
    ans = 0
  
    # Iterate over the prime values
    for i in range(len(pr)):
  
        for j in range(i,len(pr)):
  
            # Break the loop once the product exceeds N
            if (pr[i] * pr[j] > N):
                break
  
            # Add valid products which are less than
            # or equal to N
            # each product is a semi-prime number
            ans += pr[i] * pr[j]
      
    return ans
  
# Driver code
N = 6
  
sieve(N)
  
print(SemiPrimeSum(N))
  
# This code is contributed by mohit kumar 29

C#




// C# implementation of the above approach
using System;
using System.Collections.Generic;
  
class GFG
{
  
// Vector to store the primes
static List<long> pr = new List<long>();
  
// Create a boolean array "prime[0..n]"
static bool []prime = new bool[10000000 + 1];
static void sieve(long n)
{
  
    // Initialize along prime values to be true
    for (int i = 2; i <= n; i += 1) 
    {
        prime[i] = true;
    }
  
    for (int p = 2; (int)p * (int)p <= n; p++)
    {
  
        // If prime[p] is not changed then it is a prime
        if (prime[p] == true)
        {
  
            // Update along multiples of p greater than or
            // equal to the square of it
            // numbers which are multiple of p and are
            // less than p^2 are already been marked
            for (int i = (int)p * (int)p; i <= n; i += p)
                prime[i] = false;
        }
    }
  
    // Print all prime numbers
    for (int p = 2; p <= n; p++)
        if (prime[p])
            pr.Add((long)p);
}
  
// Function to return the semi-prime sum
static long SemiPrimeSum(long N)
{
  
    // Variable to store the sum of semi-primes
    long ans = 0;
  
    // Iterate over the prime values
    for (int i = 0; i < pr.Count; i += 1) 
    {
  
        for (int j = i; j < pr.Count; j += 1) 
        {
  
            // Break the loop once the product exceeds N
            if ((long)pr[i] * (long)pr[j] > N)
                break;
  
            // Add valid products which are less than
            // or equal to N
            // each product is a semi-prime number
            ans += (long)pr[i] * (long)pr[j];
        }
    }
    return ans;
}
  
// Driver code
public static void Main(String[] args)
{
    long N = 6;
  
    sieve(N);
  
    Console.WriteLine(SemiPrimeSum(N));
}
}
  
// This code is contributed by Rajput-Ji
Output:
10

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