Related Articles
Numbers in range [L, R] such that the count of their divisors is both even and prime
• Last Updated : 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```

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:

• 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++

 `// C++ implementation of the approach``#include ``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;``}`

## Java

 `// 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);``    ` `    ``}` `}`

## Python 3

 `# 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`

## 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

 ``

## Javascript

 ``
Output:
`Count: 3`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price.

In case you wish to attend live classes with industry experts, please refer Geeks Classes Live and Geeks Classes Live USA

My Personal Notes arrow_drop_up