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

Difficulty Level : Medium
Last Updated : 21 Apr, 2021

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

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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++

// 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;
}

Java

//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

Python3

# 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

C#

// 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

Javascript

<script>
 
// Javascript implementation of above approach  
var N = 100000;
 
// prime[i] stores 1 if i is prime
var prime = Array.from({length: N}, (_, i) => 0);
 
// divi[i] stores the count of
// divisors of i
var divi = Array.from({length: N}, (_, i) => 0);
 
// sum[i] will store the sum of all
// the integers from 0 to i whose
// count of divisors is prime
var sum = Array.from({length: N}, (_, i) => 0);
 
// Function for Sieve of Eratosthenes
function 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 (i = 0; i < N; i++)
        prime[i] = 1;
 
    // 0 and 1 is not prime
    prime[0] = prime[1] = 0;
 
    for (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 (i = p * p; i < N; i += p)
                prime[i] = 0;
        }
    }
}
 
// Function to count the divisors
function 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 = 1; i < N; i++)
    {
        for (j = i; j < N; j += i)
        {
            divi[j]++;
        }
    }
}
 
// Function for pre-computation
function pre()
{
    for (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 (i = 1; i < N; i++)
        sum[i] += sum[i - 1];
}
 
// Driver code
 
var 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
document.write( sum[r] - sum[l - 1]);
 
 
// This code is contributed by Amit Katiyar
 
</script>

Output:

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Table of Contents

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

C++

Java

Python3

C#

Javascript

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