Sum of numbers in a range [L, R] whose count of divisors is prime

Given Q queries where each query consists of an integer range [L, R], the task is to find the sum of the integers from the given range whose count of divisors is prime.

Examples:

Input: Q[][] = {{2, 4}}
Output:
9
All the numbers in the range have only 2 divisors
which is prime.
(2 + 3 + 4) = 9.



Input: Q[][] = {{15, 17}, {2, 12}}
Output:
33
41

Approach: Find all the prime numbers and the count of divisors for each element up to a limit N using Sieve of Eratosthenes. Now create a prefix sum array sum[] where sum[i] will store the sum of elements from the range [0, i] whose count of divisors is prime using the Sieve array created earlier. Now every query can be answered in O(1) as sum[r] – sum[l – 1].

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
const int N = 100000;
  
// prime[i] stores 1 if i is prime
int prime[N];
  
// divi[i] stores the count of
// divisors of i
int divi[N];
  
// sum[i] will store the sum of all
// the integers from 0 to i whose
// count of divisors is prime
int sum[N];
  
// Function for Sieve of Eratosthenes
void SieveOfEratosthenes()
{
    // Create a boolean array "prime[0..n]" and initialize
    // all entries it as true. A value in prime[i] will
    // finally be 0 if i is Not a prime, else true.
    for (int i = 0; i < N; i++)
        prime[i] = 1;
  
    // 0 and 1 is not prime
    prime[0] = prime[1] = 0;
  
    for (int p = 2; p * p < N; p++) {
  
        // If prime[p] is not changed, then it is a prime
        if (prime[p] == 1) {
  
            // 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 (int i = p * p; i < N; i += p)
                prime[i] = 0;
        }
    }
}
  
// Function to count the divisors
void DivisorCount()
{
  
    // For each number i we will go to each of
    // the multiple of i and update the count
    // of divisor of its multiple j as i is one
    // of the factor of j
    for (int i = 1; i < N; i++) {
        for (int j = i; j < N; j += i) {
            divi[j]++;
        }
    }
}
  
// Function for pre-computation
void pre()
{
    for (int i = 1; i < N; i++) {
  
        // If count of divisors of i is prime
        if (prime[divi[i]] == 1) {
            sum[i] = i;
        }
    }
  
    // taking prefix sum
    for (int i = 1; i < N; i++)
        sum[i] += sum[i - 1];
}
  
// Driver code
int main()
{
  
    int l = 5, r = 8;
  
    // Find all the prime numbers till N
    SieveOfEratosthenes();
  
    // Update the count of divisors
    // of all the numbers till N
    DivisorCount();
  
    // Precomputation for the prefix sum array
    pre();
  
    // Perform query
    cout << sum[r] - sum[l - 1];
  
    return 0;
}

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//Java implementation of above approach 


import java.util.*; 


class GFG 


{ 


      


static int N = 100000; 


  


// prime[i] stores 1 if i is prime 


static int prime[] = new int[N]; 


  


// divi[i] stores the count of 


// divisors of i 


static int divi[] = new int[N]; 


  


// sum[i] will store the sum of all 


// the integers from 0 to i whose 


// count of divisors is prime 


static int sum[] = new int[N]; 


  


// Function for Sieve of Eratosthenes 


static void SieveOfEratosthenes() 


{ 


    // Create a boolean array "prime[0..n]" and  


    // initialize all entries it as true.  


    // A value in prime[i] will finally be 0  


    // if i is Not a prime, else true. 


    for (int i = 0; i < N; i++) 


        prime[i] = 1; 


  


    // 0 and 1 is not prime 


    prime[0] = prime[1] = 0; 


  


    for (int p = 2; p * p < N; p++)  


    { 


  


        // If prime[p] is not changed,  


        // then it is a prime 


        if (prime[p] == 1)  


        { 


  


            // 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 (int i = p * p; i < N; i += p) 


                prime[i] = 0; 


        } 


    } 


} 


  


// Function to count the divisors 


static void DivisorCount() 


{ 


  


    // For each number i we will go to each of 


    // the multiple of i and update the count 


    // of divisor of its multiple j as i is one 


    // of the factor of j 


    for (int i = 1; i < N; i++)  


    { 


        for (int j = i; j < N; j += i)  


        { 


            divi[j]++; 


        } 


    } 


} 


  


// Function for pre-computation 


static void pre() 


{ 


    for (int i = 1; i < N; i++)  


    { 


  


        // If count of divisors of i is prime 


        if (prime[divi[i]] == 1)  


        { 


            sum[i] = i; 


        } 


    } 


  


    // taking prefix sum 


    for (int i = 1; i < N; i++) 


        sum[i] += sum[i - 1]; 


} 


  


// Driver code 


public static void main(String args[]) 


{ 


    int l = 5, r = 8; 


  


    // Find all the prime numbers till N 


    SieveOfEratosthenes(); 


  


    // Update the count of divisors 


    // of all the numbers till N 


    DivisorCount(); 


  


    // Precomputation for the prefix sum array 


    pre(); 


  


    // Perform query 


    System.out.println( sum[r] - sum[l - 1]); 


} 


} 


  


// This code is contributed by Arnab Kundu 



















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Python3














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# Python3 implementation of the approach  


from math import sqrt 


  


N = 100000;  


  


# Create a boolean array "prime[0..n]" and  


# initialize all entries it as true.  


# A value in prime[i] will finally be 0 if 


# i is Not a prime, else true.  


# prime[i] stores 1 if i is prime  


prime = [1] * N;  


  


# divi[i] stores the count of  


# divisors of i  


divi = [0] * N;  


  


# sum[i] will store the sum of all  


# the integers from 0 to i whose  


# count of divisors is prime  


sum = [0] * N;  


  


# Function for Sieve of Eratosthenes  


def SieveOfEratosthenes() : 


  


    for i in range(N) : 


        prime[i] = 1;  


  


    # 0 and 1 is not prime  


    prime[0] = prime[1] = 0;  


  


    for p in range(2, int(sqrt(N)) + 1) :  


  


        # If prime[p] is not changed,  


        # then it is a prime  


        if (prime[p] == 1) : 


  


            # 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 i in range(p * p, N, p) : 


                prime[i] = 0;  


  


# Function to count the divisors  


def DivisorCount() : 


  


    # For each number i we will go to each of  


    # the multiple of i and update the count  


    # of divisor of its multiple j as i is one  


    # of the factor of j  


    for i in range(1, N) : 


        for j in range(i, N , i) : 


            divi[j] += 1;  


  


# Function for pre-computation  


def pre() : 


  


    for i in range(1, N) : 


  


        # If count of divisors of i is prime  


        if (prime[divi[i]] == 1) : 


            sum[i] = i;  


  


    # taking prefix sum  


    for i in range(1, N) :  


        sum[i] += sum[i - 1];  


  


# Driver code  


if __name__ == "__main__" :  


  


    l = 5; r = 8;  


  


    # Find all the prime numbers till N  


    SieveOfEratosthenes();  


  


    # Update the count of divisors  


    # of all the numbers till N  


    DivisorCount();  


  


    # Precomputation for the prefix sum array  


    pre();  


  


    # Perform query  


    print(sum[r] - sum[l - 1]);  


  


# This code is contributed by AnkitRai01 



















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C#














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// C# implementation of the approach 


using System; 


using System.Collections.Generic; 


      


class GFG 


{ 


static int N = 100000; 


  


// prime[i] stores 1 if i is prime 


static int []prime = new int[N]; 


  


// divi[i] stores the count of 


// divisors of i 


static int []divi = new int[N]; 


  


// sum[i] will store the sum of all 


// the integers from 0 to i whose 


// count of divisors is prime 


static int []sum = new int[N]; 


  


// Function for Sieve of Eratosthenes 


static void SieveOfEratosthenes() 


{ 


    // Create a boolean array "prime[0..n]"  


    // and initialize all entries it as true.  


    // A value in prime[i] will finally be 0  


    // if i is Not a prime, else true. 


    for (int i = 0; i < N; i++) 


        prime[i] = 1; 


  


    // 0 and 1 is not prime 


    prime[0] = prime[1] = 0; 


  


    for (int p = 2; p * p < N; p++)  


    { 


  


        // If prime[p] is not changed,  


        // then it is a prime 


        if (prime[p] == 1)  


        { 


  


            // 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 (int i = p * p; i < N; i += p) 


                prime[i] = 0; 


        } 


    } 


} 


  


// Function to count the divisors 


static void DivisorCount() 


{ 


  


    // For each number i we will go to each of 


    // the multiple of i and update the count 


    // of divisor of its multiple j as i is one 


    // of the factor of j 


    for (int i = 1; i < N; i++)  


    { 


        for (int j = i; j < N; j += i)  


        { 


            divi[j]++; 


        } 


    } 


} 


  


// Function for pre-computation 


static void pre() 


{ 


    for (int i = 1; i < N; i++)  


    { 


  


        // If count of divisors of i is prime 


        if (prime[divi[i]] == 1)  


        { 


            sum[i] = i; 


        } 


    } 


  


    // taking prefix sum 


    for (int i = 1; i < N; i++) 


        sum[i] += sum[i - 1]; 


} 


  


// Driver code 


public static void Main(String []args) 


{ 


    int l = 5, r = 8; 


  


    // Find all the prime numbers till N 


    SieveOfEratosthenes(); 


  


    // Update the count of divisors 


    // of all the numbers till N 


    DivisorCount(); 


  


    // Precomputation for the prefix sum array 


    pre(); 


  


    // Perform query 


    Console.WriteLine( sum[r] - sum[l - 1]); 


} 


} 


  


// This code is contributed by 29AjayKumar 


  



















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