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Sum of product of x and y such that floor(n/x) = y

  • Difficulty Level : Hard
  • Last Updated : 11 May, 2021

Given a positive integer n. The task is to find the sum of product of x and y such that ⌊n/x⌋ = y (Integer Division).

Examples: 

Input : n = 5
Output : 21
Following are the possible pairs of (x, y):
(1, 5), (2, 2), (3, 1), (4, 1), (5, 1).
So, 1*5 + 2*2 + 3*1 + 4*1 + 5*1 
   = 5 + 4 + 3 + 4 + 5 
   = 21.

Input : n = 10
Output : 87

Method 1 (Brute Force): 
Iterate x from 1 to n to find y. Then add x*y to the answer over each iteration.

Below is the implementation of this approach :  

C++




// C++ program to find sum of product of x and y
// such that n/x = y (Integer Division)
#include<bits/stdc++.h>
using namespace std;
 
// Return the sum of product x*y.
int sumofproduct(int n)
{
    int ans = 0;
 
    // Iterating x from 1 to n
    for (int x = 1; x <= n; x++)
    {
        // Finding y = n/x.
        int y = n/x;
 
        // Adding product of x and y to answer.
        ans += (y * x);
    }
 
    return ans;
}
 
// Driven Program
int main()
{
    int n = 10;
    cout << sumofproduct(n) << endl;
    return 0;
}

Java




// Java program to find sum of
// product of x and y such that
// n/x = y (Integer Division)
import java.io.*;
 
class GFG {
     
// Return the sum of product x*y.
static int sumofproduct(int n)
{
    int ans = 0;
 
    // Iterating x from 1 to n
    for (int x = 1; x <= n; x++)
    {
        // Finding y = n/x.
        int y = n / x;
 
        // Adding product of x and
        // y to answer.
        ans += (y * x);
    }
 
    return ans;
}
 
    // Driver Code
    static public void main(String[] args)
    {
        int n = 10;
        System.out.println(sumofproduct(n));
    }
}
 
// This code is contributed by vt_m.

Python3




# Python3 program to find sum of
# product of x and y such that
# n/x = y (Integer Division)
 
# Return the sum of product x*y
def sumofproduct(n):
    ans = 0
 
    # Iterating x from 1 to n
    for x in range(1, n + 1):
         
        # Finding y = n/x.
        y = int(n / x)
 
        # Adding product of x and y to answer.
        ans += (y * x)
 
    return ans
 
# Driven Program
n = 10
print (sumofproduct(n))
 
#This code is Shreyanshi Arun

C#




// C# program to find sum of
// product of x and y such that
// n/x = y (Integer Division)
using System;
 
class GFG {
     
// Return the sum of product x*y.
static int sumofproduct(int n)
{
    int ans = 0;
 
    // Iterating x from 1 to n
    for (int x = 1; x <= n; x++)
    {
        // Finding y = n/x.
        int y = n / x;
 
        // Adding product of x and
        // y to answer.
        ans += (y * x);
    }
 
    return ans;
}
 
    // Driver Code
    static public void Main(String[] args)
    {
        int n = 10;
        Console.WriteLine(sumofproduct(n));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to find sum of product
// of x and y such that n/x = y
// (Integer Division)
 
// Return the sum of product x*y.
function sumofproduct($n)
{
    $ans = 0;
 
    // Iterating x from 1 to n
    for($x = 1; $x <= $n; $x++)
    {
        // Finding y = n/x.
        $y = (int)($n / $x);
 
        // Adding product of x
        // and y to answer.
        $ans += ($y * $x);
    }
 
    return $ans;
}
 
// Driver Code
$n = 10;
echo sumofproduct($n);
 
// This code is contributed by mits
?>

Javascript




<script>
 
// Javascript program to find sum of
// product of x and y such that
// n/x = y (Integer Division)
    
// Return the sum of product x*y.
function sumofproduct(n)
{
    var ans = 0;
 
    // Iterating x from 1 to n
    for(x = 1; x <= n; x++)
    {
         
        // Finding y = n/x.
        var y = parseInt(n / x);
 
        // Adding product of x and
        // y to answer.
        ans += (y * x);
    }
    return ans;
}
 
// Driver Code
var n = 10;
document.write(sumofproduct(n));
 
// This code is contributed by Princi Singh
 
</script>

Output :



87

Time Complexity : O(n)
Method 2 (Efficient Approach): 
Let’s solve for n = 10, so 
x = 1, y = 10 
x = 2, y = 5 
x = 3, y = 3 
x = 4, y = 2 
x = 5, y = 2 
x = 6, y = 1 
x = 7, y = 1 
x = 8, y = 1 
x = 9, y = 1 
x = 10, y = 1
So, our answer would be 1*10 + 2*5 + 3*3 + 4*2 + 5*2 + 6*1 + 7*1 + 8*1 + 9*1 + 10*1.
Now, observe some value of y is repeating. Also, observe that they are repeating for some range of consecutive value of x like y = 1 is repeating for x = 6 to 10.

So, instead of finding the value of y for all the value of x (1 to n) as done in method 1, try to find the lower and higher value of x for which the value of possible value of y like for y = 1 try to find lower value of x = 6 and higher value of x = 10. Now, observe lower value will be (n/(y+1)) + 1 and higher value will be (n/y). Find the sum of range of x and multiply with y and add to the answer.
How to find the possible value of y? 

Observe, y has all values from 1 to sqrt(n) when y is smaller than or equal to x. So for y = 1 to sqrt(n), find the lower and higher limits of x for each y. For n = 10, 
y = 1, lo = 6 and hi = 10, ans += (6 + 7 + 8 + 9 + 10)*1 
y = 2, lo = 4 and hi = 5, ans += (4 + 5)*2 
y = 3, lo = 3 and hi = 3, ans += (3)*3

For other values to be added (for y = 10 and 5 in n = 10), observe they can be found in above steps, for each y, add y * (n/y) in the answer. 
For n = 10, 
y = 1, ans += 1 * (10/1) 
y = 2, ans += 2 * (10/2).

Below is the implementation of this approach:  

C++




// C++ program to find sum of product of x and y
// such that n/x = y (Integer Division)
#include<bits/stdc++.h>
using namespace std;
 
// Return the sum of natural number in a range.
int sumOfRange(int a, int b)
{
    // n*(n+1)/2.
    int i = (a * (a+1)) >> 1;
    int j = (b * (b+1)) >> 1;
    return (i - j);
}
 
// Return the sum of product x*y.
int sumofproduct(int n)
{
    int sum = 0;
 
    // Iterating i from 1 to sqrt(n)
    int root = sqrt(n);
    for (int i=1; i<=root; i++)
    {
        // Finding the upper limit.
        int up = n/i;
 
        // Finding the lower limit.
        int low = max(n/(i+1), root);
 
        sum += (i * sumOfRange(up, low));
        sum += (i * (n/i));
    }
 
    return sum;
}
 
// Driven Program
int main()
{
    int n = 10;
    cout << sumofproduct(n) << endl;
    return 0;
}

Java




// Java program to find sum of
// product of x and y such that
// n / x = y (Integer Division)
import java.io.*;
 
class GFG {
     
// Return the sum of natural number in a range.
static int sumOfRange(int a, int b)
{
    // n * (n + 1) / 2.
    int i = (a * (a + 1)) >> 1;
    int j = (b * (b + 1)) >> 1;
    return (i - j);
}
 
// Return the sum of product x*y.
static int sumofproduct(int n)
{
    int sum = 0;
 
    // Iterating i from 1 to sqrt(n)
    int root = (int)Math.sqrt(n);
    for (int i = 1; i <= root; i++)
    {
        // Finding the upper limit.
        int up = n / i;
 
        // Finding the lower limit.
        int low = Math.max(n / (i + 1), root);
 
        sum += (i * sumOfRange(up, low));
        sum += (i * (n / i));
    }
 
    return sum;
}
 
    // Driver Code
    static public void main(String[] args)
    {
        int n = 10;
        System.out.println(sumofproduct(n));
    }
}
 
// This code is contributed by vt_m.

Python3




# Python3 program to find sum
# of product of x and y such
# that n/x = y (Integer Division)
import math
 
# Return the sum of natural
# number in a range.
def sumOfRange(a, b):
    # n*(n+1)/2.
    i = (a * (a + 1)) >> 1;
    j = (b * (b + 1)) >> 1;
    return (i - j);
 
# Return the sum of product x*y.
def sumofproduct(n):
    sum = 0;
 
    # Iterating i from 1 to sqrt(n)
    root = int(math.sqrt(n));
    for i in range(1, root + 1):
        # Finding the upper limit.
        up = int(n / i);
 
        # Finding the lower limit.
        low = max(int(n / (i + 1)), root);
 
        sum += (i * sumOfRange(up, low));
        sum += (i * int(n / i));
 
    return sum;
 
# Driven Code
n = 10;
print(sumofproduct(n));
     
# This code is contributed by mits

C#




// C# program to find sum of
// product of x and y such that
// n / x = y (Integer Division)
using System;
 
class GFG {
     
// Return the sum of natural number in a range.
static int sumOfRange(int a, int b)
{
    // n * (n + 1) / 2.
    int i = (a * (a + 1)) >> 1;
    int j = (b * (b + 1)) >> 1;
    return (i - j);
}
 
// Return the sum of product x*y.
static int sumofproduct(int n)
{
    int sum = 0;
 
    // Iterating i from 1 to sqrt(n)
    int root = (int)Math.Sqrt(n);
    for (int i = 1; i <= root; i++)
    {
        // Finding the upper limit.
        int up = n / i;
 
        // Finding the lower limit.
        int low = Math.Max(n / (i + 1), root);
 
        sum += (i * sumOfRange(up, low));
        sum += (i * (n / i));
    }
 
    return sum;
}
 
    // Driver Code
    static public void Main(String[] args)
    {
        int n = 10;
        Console.WriteLine(sumofproduct(n));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// PHP program to find sum of
// product of x and y such
// that n/x = y (Integer Division)
 
// Return the sum of natural
// number in a range.
function sumOfRange($a, $b)
{
    // n*(n+1)/2.
    $i = ($a * ($a + 1)) >> 1;
    $j = ($b * ($b + 1)) >> 1;
    return ($i - $j);
}
 
// Return the sum of product x*y.
function sumofproduct($n)
{
    $sum = 0;
 
    // Iterating i from 1 to sqrt(n)
    $root = sqrt($n);
    for ($i = 1; $i <= $root; $i++)
    {
        // Finding the upper limit.
        $up = (int)($n / $i);
 
        // Finding the lower limit.
        $low = max((int)($n / ($i + 1)), $root);
 
        $sum += ($i * sumOfRange($up, $low));
        $sum += ($i * (int)($n / $i));
    }
 
    return $sum;
}
 
// Driven Code
$n = 10;
echo sumofproduct($n) . "\n";
 
// This code is contributed
// by Akanksha Rai(Abby_akku)
?>

Javascript




<script>
 
// Javascript program to implement
// the above approach
 
// Return the sum of natural number in a range.
function sumOfRange(a, b)
{
    // n * (n + 1) / 2.
    let i = (a * (a + 1)) >> 1;
    let j = (b * (b + 1)) >> 1;
    return (i - j);
}
  
// Return the sum of product x*y.
function sumofproduct(n)
{
    let sum = 0;
  
    // Iterating i from 1 to sqrt(n)
    let root = Math.floor(Math.sqrt(n));
    for (let i = 1; i <= root; i++)
    {
        // Finding the upper limit.
        let up = Math.floor(n / i);
  
        // Finding the lower limit.
        let low = Math.max(Math.floor(n / (i + 1)), root);
  
        sum += (i * sumOfRange(up, low));
        sum += (i * Math.floor(n / i));
    }
  
    return sum;
}
 
    // Driver Code
         
    let n = 10;
    document.write(sumofproduct(n));
         
</script>

Output: 

87

Time Complexity: O((√n)
Source: 
https://www.quora.com/What-is-the-fastest-way-to-solve-the-problem-that-states-given-a-number-N-find-the-sum-of-all-products-x*y-such-that-N-x-y-integer-division-Here-N-would-be-N-10
This article is contributed by Anuj Chauhan(APC). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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