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

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

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++

// 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; 
} 

Java

// 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)); 
    } 
} 

Python3

# 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 

C#

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 

PHP

<?php 
// PHP implementation of the approach 
$MAX = 10000; 
  
$prefix = array_fill(0, $MAX, 0); 
  
// Function to calculate the prefix sum 
// of all the even factors 
function sieve_modified() 
{ 
    global $MAX, $prefix; 
    for ($i = 2; $i < $MAX; $i += 2)  
    { 
  
        // Add i to all the multiples of i 
        for ($j = $i; $j < $MAX; $j += $i) 
            $prefix[$j] += $i; 
    } 
  
    // Update the prefix sum 
    for ($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 
function sumEvenFactors($L, $R) 
{ 
    global $MAX, $prefix; 
    return ($prefix[$R] - $prefix[$L - 1]); 
} 
  
// Driver code 
sieve_modified(); 
$l = 6; 
$r = 10; 
echo sumEvenFactors($l, $r); 
  
// This code is contributed by mits 
?> 

Output:

My Personal Notes

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

C++

Java

Python3

C#

PHP

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