Find sum of even factors of a number
Given a number n, the task is to find the even factor sum of a number.
Examples:
Input : 30
Output : 48
Even dividers sum 2 + 6 + 10 + 30 = 48
Input : 18
Output : 26
Even dividers sum 2 + 6 + 18 = 26
Prerequisite : Sum of factors
As discussed in above mentioned previous post, sum of factors of a number is
Let p1, p2, … pk be prime factors of n. Let a1, a2, .. ak be highest powers of p1, p2, .. pk respectively that divide n, i.e., we can write n as n = (p1a1)*(p2a2)* … (pkak).
Sum of divisors = (1 + p1 + p12 ... p1a1) *
(1 + p2 + p22 ... p2a2) *
...........................
(1 + pk + pk2 ... pkak)
If number is odd, then there are no even factors, so we simply return 0.
If number is even, we use above formula. We only need to ignore 20. All other terms multiply to produce even factor sum. For example, consider n = 18. It can be written as 2132 and sum of all factors is (20 + 21)*(30 + 31 + 32). if we remove 20 then we get the
Sum of even factors (2)*(1+3+32) = 26.
To remove odd number in even factor, we ignore then 20 which is 1. After this step, we only get even factors. Note that 2 is the only even prime.
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
int sumofFactors( int n)
{
if (n % 2 != 0)
return 0;
int res = 1;
for ( int i = 2; i <= sqrt (n); i++) {
int count = 0, curr_sum = 1, curr_term = 1;
while (n % i == 0) {
count++;
n = n / i;
if (i == 2 && count == 1)
curr_sum = 0;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
if (n >= 2)
res *= (1 + n);
return res;
}
int main()
{
int n = 18;
cout << sumofFactors(n);
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
public class GfG{
public static int sumofFactors( int n)
{
if (n % 2 != 0 )
return 0 ;
int res = 1 ;
for ( int i = 2 ; i <= Math.sqrt(n); i++)
{
int count = 0 , curr_sum = 1 ;
int curr_term = 1 ;
while (n % i == 0 )
{
count++;
n = n / i;
if (i == 2 && count == 1 )
curr_sum = 0 ;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
if (n >= 2 )
res *= ( 1 + n);
return res;
}
public static void main(String argc[]){
int n = 18 ;
System.out.println(sumofFactors(n));
}
}
|
Python3
import math
def sumofFactors(n) :
if (n % 2 ! = 0 ) :
return 0
res = 1
for i in range ( 2 , ( int )(math.sqrt(n)) + 1 ) :
count = 0
curr_sum = 1
curr_term = 1
while (n % i = = 0 ) :
count = count + 1
n = n / / i
if (i = = 2 and count = = 1 ) :
curr_sum = 0
curr_term = curr_term * i
curr_sum = curr_sum + curr_term
res = res * curr_sum
if (n > = 2 ) :
res = res * ( 1 + n)
return res
n = 18
print (sumofFactors(n))
|
C#
using System;
public class GfG {
public static int sumofFactors( int n)
{
if (n % 2 != 0)
return 0;
int res = 1;
for ( int i = 2; i <= Math.Sqrt(n); i++)
{
int count = 0, curr_sum = 1;
int curr_term = 1;
while (n % i == 0)
{
count++;
n = n / i;
if (i == 2 && count == 1)
curr_sum = 0;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
if (n >= 2)
res *= (1 + n);
return res;
}
public static void Main() {
int n = 18;
Console.WriteLine(sumofFactors(n));
}
}
|
PHP
<?php
function sumofFactors( $n )
{
if ( $n % 2 != 0)
return 0;
$res = 1;
for ( $i = 2; $i <= sqrt( $n ); $i ++) {
$count = 0;
$curr_sum = 1;
$curr_term = 1;
while ( $n % $i == 0) {
$count ++;
$n = floor ( $n / $i );
if ( $i == 2 && $count == 1)
$curr_sum = 0;
$curr_term *= $i ;
$curr_sum += $curr_term ;
}
$res *= $curr_sum ;
}
if ( $n >= 2)
$res *= (1 + $n );
return $res ;
}
$n = 18;
echo sumofFactors( $n );
?>
|
Javascript
<script>
function sumofFactors(n)
{
if (n % 2 != 0)
return 0;
let res = 1;
for (let i = 2; i <= Math.sqrt(n); i++)
{
let count = 0, curr_sum = 1;
let curr_term = 1;
while (n % i == 0)
{
count++;
n = n / i;
if (i == 2 && count == 1)
curr_sum = 0;
curr_term *= i;
curr_sum += curr_term;
}
res *= curr_sum;
}
if (n >= 2)
res *= (1 + n);
return res;
}
let n = 18;
document.write(sumofFactors(n));
</script>
|
Output:
26
Time Complexity: O(?n log n)
Auxiliary Space: O(1)
Please suggest if someone has a better solution which is more efficient in terms of space and time.
Last Updated :
25 Oct, 2022
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...