Sum of all even factors of numbers in the range [l, r]

Given a range [l, r], the task is to find the sum of all the even factors of the numbers from the given range.

Examples:

Input: l = 6, r = 8
Output: 22
factors(6) = 1, 2, 3, 6, evenfactors(6) = 2, 6 sumEvenFactors(6) = 2 + 6 = 8
factors(7) = 1, 7, No even factors
factors(8) = 1, 2, 4, 8, evenfactors(8) = 2, 4, 8 sumEvenFactors(8) = 2 + 4 + 8 = 14
Therefore sum of all even factors = 8 + 14 = 22



Input: l = 1, r = 10
Output: 42

Approach: We can modify Sieve Of Eratosthenes to store the sum of all even factors of a number at it’s corresponding index. Then we will make a prefix array to store sum upto that index. And now each query can be answered in O(1) using prefix[r] – prefix[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;
#define ll long long int
  
const int MAX = 100000;
  
ll prefix[MAX];
  
// Function to calculate the prefix sum
// of all the even factors
void sieve_modified()
{
    for (int i = 2; i < MAX; i += 2) {
  
        // Add i to all the multiples of i
        for (int j = i; j < MAX; j += i)
            prefix[j] += i;
    }
  
    // Update the prefix sum
    for (int i = 1; i < MAX; i++)
        prefix[i] += prefix[i - 1];
}
  
// Function to return the sum of
// all the even factors of the
// numbers in the given range
ll sumEvenFactors(int L, int R)
{
    return (prefix[R] - prefix[L - 1]);
}
  
// Driver code
int main()
{
    sieve_modified();
    int l = 6, r = 10;
    cout << sumEvenFactors(l, r);
  
    return 0;
}

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Java














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// Java implementation of the approach 


class GFG { 


  


    static final int MAX = 100000; 


    static long prefix[] = new long[MAX]; 


  


    // Function to calculate the prefix sum 


    // of all the even factors 


    static void sieve_modified() 


    { 


        for (int i = 2; i < MAX; i += 2) { 


  


            // Add i to all the multiples of i 


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


                prefix[j] += i; 


        } 


  


        // Update the prefix sum 


        for (int i = 1; i < MAX; i++) 


            prefix[i] += prefix[i - 1]; 


    } 


  


    // Function to return the sum of 


    // all the even factors of the 


    // numbers in the given range 


    static long sumEvenFactors(int L, int R) 


    { 


        return (prefix[R] - prefix[L - 1]); 


    } 


  


    // Driver code 


    public static void main(String args[]) 


    { 


        sieve_modified(); 


        int l = 6, r = 10; 


        System.out.print(sumEvenFactors(l, r)); 


    } 


} 



















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Python3














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# Python3 implementation of the approach.  


  


# Function to calculate the prefix sum  


# of all the even factors  


def sieve_modified():  


  


    for i in range(2, MAX, 2):  


  


        # Add i to all the multiples of i  


        for j in range(i, MAX, i):  


            prefix[j] += i  


  


    # Update the prefix sum  


    for i in range(1, MAX):  


        prefix[i] += prefix[i - 1]  


  


# Function to return the sum of  


# all the even factors of the  


# numbers in the given range  


def sumEvenFactors(L, R):  


  


    return (prefix[R] - prefix[L - 1])  


  


# Driver code  


if __name__ == "__main__": 


      


    MAX = 100000


    prefix = [0] * MAX


    sieve_modified()  


    l, r = 6, 10


    print(sumEvenFactors(l, r))  


  


# This code is contributed by Rituraj Jain 



















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using System; 


// C# implementation of the approach 


using System; 


  


class GFG 


{ 


  


    public const int MAX = 100000; 


    public static long[] prefix = new long[MAX]; 


  


    // Function to calculate the prefix sum 


    // of all the even factors 


    public static void sieve_modified() 


    { 


        for (int i = 2; i < MAX; i += 2) 


        { 


  


            // Add i to all the multiples of i 


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


            { 


                prefix[j] += i; 


            } 


        } 


  


        // Update the prefix sum 


        for (int i = 1; i < MAX; i++) 


        { 


            prefix[i] += prefix[i - 1]; 


        } 


    } 


  


    // Function to return the sum of 


    // all the even factors of the 


    // numbers in the given range 


    public static long sumEvenFactors(int L, int R) 


    { 


        return (prefix[R] - prefix[L - 1]); 


    } 


  


    // Driver code 


    public static void Main(string[] args) 


    { 


        sieve_modified(); 


        int l = 6, r = 10; 


        Console.Write(sumEvenFactors(l, r)); 


    } 


} 


  


// This code is contributed by Shrikant13 



















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