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

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

Examples:

Input: l = 6, r = 8
Output: 32
factors(6) = 1, 2, 3, 6, oddfactors(6) = 1, 3 sum_Odd_Factors(6) = 1 + 3 = 4
factors(7) = 1, 7, oddfactors(6) = 1 7, sum_Odd_Factors(7) = 1 + 7 = 8
factors(8) = 1, 2, 4, 8, oddfactors(6) = 1, sum_Odd_Factors(8) = 1 = 1
Therefore sum of all odd factors = 4 + 8 + 1 = 13

Input: l = 1, r = 10
Output: 45

Approach: We can modify Sieve Of Eratosthenes to store sum of all odd 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:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int
  
const int MAX = 100001;
  
ll prefix[MAX];
  
// Function to calculate the prefix sum
// of all the odd factors
void sieve_modified()
{
    for (int i = 1; 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 odd factors of the
// numbers in the given range
ll sumOddFactors(int L, int R)
{
    return (prefix[R] - prefix[L - 1]);
}
  
// Driver code
int main()
{
    sieve_modified();
    int l = 6, r = 10;
    cout << sumOddFactors(l, r);
    return 0;
}
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// Java implementation of the approach
import java.io.*;
  
class GFG 
{
      
static int MAX = 100001;
static int prefix[] = new int[MAX];
  
// Function to calculate the prefix sum
// of all the odd factors
static void sieve_modified()
{
    for (int i = 1; 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 odd factors of the
// numbers in the given range
static int sumOddFactors(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.println (sumOddFactors(l, r));
    }
}
  
// This code is contributed by jit_t
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# Python3 implementation of the approach
MAX = 100001;
  
prefix = [0] * MAX;
  
# Function to calculate the prefix sum
# of all the odd factors
def sieve_modified():
  
    for i in range(1, 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 odd factors of the
# numbers in the given range
def sumOddFactors(L, R):
  
    return (prefix[R] - prefix[L - 1]);
  
# Driver code
sieve_modified();
l = 6;
r = 10;
print(sumOddFactors(l, r));
  
# this code is contributed by chandan_jnu
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// C# implementation of the approach 
using System;
  
class GFG
{
  
public static int MAX = 100001;
public static int[] prefix = new int[MAX];
  
// Function to calculate the prefix sum 
// of all the odd factors 
public static void sieve_modified()
{
    for (int i = 1; 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 odd factors of the 
// numbers in the given range 
public static int sumOddFactors(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.WriteLine(sumOddFactors(l, r));
}
}
  
// This code is contributed by Shrikant13
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Output:
32



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Improved By : jit_t, shrikanth13, Chandan_Kumar



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