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

06 May, 2021
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
 
 
The only number that is prime, as well as even, is ‘2’.So, we need to find all the numbers within the given range that have exactly 2 divisors, i.e. prime numbers.
The only number that is prime, as well as even, is ‘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:  
Start a loop from ‘l’ to ‘r’ and check whether the number is prime(it will take more time for bigger range).If the number is prime then increment the count variable.At the end, print the value of count.
Start a loop from ‘l’ to ‘r’ and check whether the number is prime(it will take more time for bigger range).
If the number is prime then increment the count variable.
At the end, print the value of count.
An efficient approach:  
We have to count the prime numbers in range [L, R].
First, create a sieve which will help in determining whether the number is prime or not in O(1) time.
Then, create a prefix array to store the count of prime numbers where, element at index ‘i’ holds the count of the prime numbers from ‘1’ to ‘i’.
Now, if we want to find the count of prime numbers in range [L, R], the count will be (sum[R] – sum[L-1])
Finally, print the result i.e. (sum[R] – sum[L-1])
Below is the implementation of the above approach:  
C++
Java
Python 3
C#
PHP
Javascript
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 1000000 // stores whether the number is prime or notbool prime[MAX + 1]; // stores the count of prime numbers// less than or equal to the indexint sum[MAX + 1]; // create the sievevoid 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 codeint 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;}
// Java implementation of the approachclass 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);         } }
# Python 3 implementation of the approachMAX = 1000000 # stores whether the number is prime or notprime = [True] * (MAX + 1) # stores the count of prime numbers# less than or equal to the indexsum = [0] * (MAX + 1) # create the sievedef 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 codeif __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
// 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);         } }
<?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 sievefunction 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 sieveSieveOfEratosthenes(); // '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 countecho "Count: " . $c . "\n"; // This code is contributed by mits?>
<script> // Javascript implementation of the approachvar MAX = 1000000; // stores whether the number is prime or notvar prime = Array(MAX+1).fill(true); // stores the count of prime numbers// less than or equal to the indexvar sum = Array(MAX+1).fill(0); // create the sievefunction 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.     prime[1] = false;     for (var 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 (var 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 (var i = 1; i <= MAX; i++) {        if (prime[i] == true)            sum[i] = 1;         sum[i] += sum[i - 1];    }} // Driver code// create the sieveSieveOfEratosthenes();// 'l' and 'r' are the lower and upper bounds// of the rangevar l = 3, r = 9;// get the value of countvar c = (sum[r] - sum[l - 1]);// display the countdocument.write( "Count: " + c ); </script>
Count: 3
 
ihritik
Mithun Kumar
ukasp
itsok
Prime Number
sieve
Competitive Programming
Mathematical
Mathematical
Prime Number
sieve
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