Numbers in range [L, R] such that the count of their divisors is both even and prime

Given a range [L, R], the task is to find the numbers from the range which have the count of their divisors as even as well as prime.
Then, print the count of the numbers found. The values of L and R are less than 10^6 and L< R.

Examples:

Input: L=3, R=9
Output: Count = 3
Explanation: The numbers are 3, 5, 7 

Input : L=3, R=17
Output : Count: 6
  1. The only number that is prime, as well as even, is ‘2’.
  2. So, we need to find all the numbers within the given range that have exactly 2 divisors,
    i.e. prime numbers.

A simple approach:



  1. Start a loop from ‘l’ to ‘r’ and check whether the number is prime(it will take more time for bigger range).
  2. If the number is prime then increment the count variable.
  3. At the end, print the value of count.

An efficient approach:

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 1000000
  
// stores whether the number is prime or not
bool prime[MAX + 1];
  
// stores the count of prime numbers
// less than or equal to the index
int sum[MAX + 1];
  
// create the sieve
void SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..n]" and initialize
    // all the entries as true. A value in prime[i] will
    // finally be false if 'i' is Not a prime, else true.
    memset(prime, true, sizeof(prime));
    memset(sum, 0, sizeof(sum));
    prime[1] = false;
  
    for (int p = 2; p * p <= MAX; p++) {
  
        // If prime[p] is not changed, then it is a prime
        if (prime[p]) {
  
            // Update all multiples of p
            for (int i = p * 2; i <= MAX; i += p)
                prime[i] = false;
        }
    }
  
    // stores the prefix sum of number
    // of primes less than or equal to 'i'
    for (int i = 1; i <= MAX; i++) {
        if (prime[i] == true)
            sum[i] = 1;
  
        sum[i] += sum[i - 1];
    }
}
  
// Driver code
int main()
{
    // create the sieve
    SieveOfEratosthenes();
  
    // 'l' and 'r' are the lower and upper bounds
    // of the range
    int l = 3, r = 9;
  
    // get the value of count
    int c = (sum[r] - sum[l - 1]);
  
    // display the count
    cout << "Count: " << c << endl;
  
    return 0;
}
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// Java implementation of the approach
class GFG
{
    static final int MAX=1000000;
      
    // stores whether the number is prime or not
    static boolean []prime=new boolean[MAX + 1];
      
    // stores the count of prime numbers
    // less than or equal to the index
    static int []sum=new int[MAX + 1];
      
    // create the sieve
    static void SieveOfEratosthenes()
    {
        // Create a boolean array "prime[0..n]" and initialize
        // all the entries as true. A value in prime[i] will
        // finally be false if 'i' is Not a prime, else true.
        for(int i=0;i<=MAX;i++)
            prime[i]=true;
              
         for(int i=0;i<=MAX;i++)
            sum[i]=0;
          
        prime[1] = false;
      
        for (int p = 2; p * p <= MAX; p++) {
      
            // If prime[p] is not changed, then it is a prime
            if (prime[p]) {
      
                // Update all multiples of p
                for (int i = p * 2; i <= MAX; i += p)
                    prime[i] = false;
            }
        }
      
        // stores the prefix sum of number
        // of primes less than or equal to 'i'
        for (int i = 1; i <= MAX; i++) {
            if (prime[i] == true)
                sum[i] = 1;
      
            sum[i] += sum[i - 1];
        }
    }
      
    // Driver code
    public static void main(String []args)
    {
        // create the sieve
        SieveOfEratosthenes();
      
        // 'l' and 'r' are the lower and upper bounds
        // of the range
        int l = 3, r = 9;
      
        // get the value of count
        int c = (sum[r] - sum[l - 1]);
      
        // display the count
        System.out.println("Count: " + c); 
      
    }
  
}
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# Python 3 implementation of the approach
MAX = 1000000
  
# stores whether the number is prime or not
prime = [True] * (MAX + 1)
  
# stores the count of prime numbers
# less than or equal to the index
sum = [0] * (MAX + 1)
  
# create the sieve
def SieveOfEratosthenes():
  
    prime[1] = False
  
    p = 2
    while p * p <= MAX:
  
        # If prime[p] is not changed, 
        # then it is a prime
        if (prime[p]):
  
            # Update all multiples of p
            i = p * 2
            while i <= MAX:
                prime[i] = False
                i += p
                  
        p += 1
  
    # stores the prefix sum of number
    # of primes less than or equal to 'i'
    for i in range(1, MAX + 1):
        if (prime[i] == True):
            sum[i] = 1
  
        sum[i] += sum[i - 1]
  
# Driver code
if __name__ == "__main__":
      
    # create the sieve
    SieveOfEratosthenes()
  
    # 'l' and 'r' are the lower and 
    # upper bounds of the range
    l = 3
    r = 9
  
    # get the value of count
    c = (sum[r] - sum[l - 1])
  
    # display the count
    print("Count:", c)
  
# This code is contributed by ita_c
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// C# implementation of the approach
  
  
using System;
class GFG
{
    static int MAX=1000000;
      
    // stores whether the number is prime or not
    static bool []prime=new bool[MAX + 1];
      
    // stores the count of prime numbers
    // less than or equal to the index
    static int []sum=new int[MAX + 1];
      
    // create the sieve
    static void SieveOfEratosthenes()
    {
        // Create a boolean array "prime[0..n]" and initialize
        // all the entries as true. A value in prime[i] will
        // finally be false if 'i' is Not a prime, else true.
        for(int i=0;i<=MAX;i++)
            prime[i]=true;
              
         for(int i=0;i<=MAX;i++)
            sum[i]=0;
          
        prime[1] = false;
      
        for (int p = 2; p * p <= MAX; p++) {
      
            // If prime[p] is not changed, then it is a prime
            if (prime[p]) {
      
                // Update all multiples of p
                for (int i = p * 2; i <= MAX; i += p)
                    prime[i] = false;
            }
        }
      
        // stores the prefix sum of number
        // of primes less than or equal to 'i'
        for (int i = 1; i <= MAX; i++) {
            if (prime[i] == true)
                sum[i] = 1;
      
            sum[i] += sum[i - 1];
        }
    }
      
    // Driver code
    public static void Main()
    {
        // create the sieve
        SieveOfEratosthenes();
      
        // 'l' and 'r' are the lower and upper bounds
        // of the range
        int l = 3, r = 9;
      
        // get the value of count
        int c = (sum[r] - sum[l - 1]);
      
        // display the count
        Console.WriteLine("Count: " + c); 
      
    }
  
}
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<?php
// PHP implementation of the approach
$MAX = 100000;
  
// Create a boolean array "prime[0..n]"
// and initialize all the entries as 
// true. A value in prime[i] will finally
// be false if 'i' is Not a prime, else true.
  
// stores whether the number 
// is prime or not
$prime = array_fill(0, $MAX + 1, true);
  
// stores the count of prime numbers
// less than or equal to the index
$sum = array_fill(0, $MAX + 1, 0);
  
// create the sieve
function SieveOfEratosthenes()
{
    global $MAX, $sum, $prime;
    $prime[1] = false;
  
    for ($p = 2; $p * $p <= $MAX; $p++)
    {
  
        // If prime[p] is not changed, 
        // then it is a prime
        if ($prime[$p]) 
        {
  
            // Update all multiples of p
            for ($i = $p * 2; $i <= $MAX; $i += $p)
                $prime[$i] = false;
        }
    }
  
    // stores the prefix sum of number
    // of primes less than or equal to 'i'
    for ($i = 1; $i <= $MAX; $i++)
    {
        if ($prime[$i] == true)
            $sum[$i] = 1;
  
        $sum[$i] += $sum[$i - 1];
    }
}
  
// Driver code
  
// create the sieve
SieveOfEratosthenes();
  
// 'l' and 'r' are the lower 
// and upper bounds of the range
$l = 3;
$r = 9;
  
// get the value of count
$c = ($sum[$r] - $sum[$l - 1]);
  
// display the count
echo "Count: " . $c . "\n";
  
// This code is contributed by mits
?>
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Output:
Count: 3



Second year Department of Information Technology Jadavpur University

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