Sum of numbers in a range [ L , R ] whose count of divisors is prime | C ++ implementation of the approach ; prime [ i ] stores 1 if i is prime ; divi [ i ] stores the count of divisors of i ; sum [ i ] will store the sum of all the integers from 0 to i whose count of divisors is prime ; Function for Sieve of Eratosthenes ; 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 . ; 0 and 1 is not prime ; If prime [ p ] is not changed , then it is a prime ; 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 . ; Function to count the divisors ; 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 ; Function for pre - computation ; If count of divisors of i is prime ; taking prefix sum ; Driver code ; Find all the prime numbers till N ; Update the count of divisors of all the numbers till N ; Precomputation for the prefix sum array ; Perform query
#include <bits/stdc++.h> NEW_LINE using namespace std ; const int N = 100000 ; int prime [ N ] ; int divi [ N ] ; int sum [ N ] ; void SieveOfEratosthenes ( ) { for ( int i = 0 ; i < N ; i ++ ) prime [ i ] = 1 ; prime [ 0 ] = prime [ 1 ] = 0 ; for ( int p = 2 ; p * p < N ; p ++ ) { if ( prime [ p ] == 1 ) { for ( int i = p * p ; i < N ; i += p ) prime [ i ] = 0 ; } } } void DivisorCount ( ) { for ( int i = 1 ; i < N ; i ++ ) { for ( int j = i ; j < N ; j += i ) { divi [ j ] ++ ; } } } void pre ( ) { for ( int i = 1 ; i < N ; i ++ ) { if ( prime [ divi [ i ] ] == 1 ) { sum [ i ] = i ; } } for ( int i = 1 ; i < N ; i ++ ) sum [ i ] += sum [ i - 1 ] ; } int main ( ) { int l = 5 , r = 8 ; SieveOfEratosthenes ( ) ; DivisorCount ( ) ; pre ( ) ; cout << sum [ r ] - sum [ l - 1 ] ; return 0 ; }