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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++ 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 = 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 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 = 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 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 =  * (MAX + 1)    # create the sieve def SieveOfEratosthenes():        prime = 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# 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 = 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);             }    }



Output:
Count: 3

Second year Department of Information Technology Jadavpur University

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