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

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  • Difficulty Level : Medium
  • Last Updated : 13 May, 2022
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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:
 

C++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

PHP




<?php
// PHP implementation of the approach
 
$MAX = 10001;
 
$prefix = array_fill(0, $MAX, 0);
 
// Function to calculate the prefix
// sum of all the odd factors
function sieve_modified()
{
    global $prefix, $MAX;
    for ($i = 1; $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 odd factors of the
// numbers in the given range
function sumOddFactors($L, $R)
{
    global $prefix;
    return ($prefix[$R] -
            $prefix[$L - 1]);
}
 
// Driver code
sieve_modified();
$l = 6;
$r = 10;
echo sumOddFactors($l, $r);
 
// This code is contributed
// by chandan_jnu
?>

Javascript




<script>
// Javascript implementation of the approach
var MAX = 100001;
prefix = Array(MAX).fill(0)
 
// Function to calculate the prefix sum
// of all the odd factors
function sieve_modified()
{
    for (var i = 1; i < MAX; i += 2) {
 
        // Add i to all the multiples of i
        for (var j = i; j < MAX; j += i)
            prefix[j] += i;
    }
 
    // Update the prefix sum
    for (var 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
function sumOddFactors(L, R)
{
    return (prefix[R] - prefix[L - 1]);
}
 
// Driver code
sieve_modified();
var l = 6, r = 10;
document.write(sumOddFactors(l, r));
 
// This code is contributed by noob2000.
 
</script>

Output: 

32

 

Time Complexity: O(100001*100001), we are using a nested loop.

Auxiliary Space: O(100001), we are using pre[Max] extra space.


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