Maximum sum after repeatedly dividing N by a divisor
Last Updated :
08 Jun, 2022
Given an integer N. The task is to find the maximum possible sum of intermediate values (Including N and 1) attained after applying the below operation:
Divide N by any divisor (>1) until it becomes 1.
Examples:
Input: N = 10
Output: 16
Initially, N=10
1st Division -> N = 10/2 = 5
2nd Division -> N= 5/5 = 1
Input: N = 8
Output: 15
Initially, N=8
1st Division -> N = 8/2 = 4
2nd Division -> N= 4/2 = 2
3rd Division -> N= 2/2 = 1
Approach: Since the task is to maximize the sum of values after each step, try to maximize individual values. So, reduce the value of N by as little as possible. To achieve that, we divide N by its smallest divisor.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int smallestDivisor( int n)
{
int mx = sqrt (n);
for ( int i = 2; i <= mx; i++)
if (n % i == 0)
return i;
return n;
}
int maxSum( int n)
{
long long res = n;
while (n > 1) {
int divi = smallestDivisor(n);
n /= divi;
res += n;
}
return res;
}
int main()
{
int n = 34;
cout << maxSum(n);
return 0;
}
|
C
#include <stdio.h>
#include<math.h>
int smallestDivisor( int n)
{
int mx = sqrt (n);
for ( int i = 2; i <= mx; i++)
if (n % i == 0)
return i;
return n;
}
int maxSum( int n)
{
long long res = n;
while (n > 1) {
int divi = smallestDivisor(n);
n /= divi;
res += n;
}
return res;
}
int main()
{
int n = 34;
printf ( "%d" ,maxSum(n));
return 0;
}
|
Java
import java.io.*;
class GFG
{
static double smallestDivisor( int n)
{
double mx = Math.sqrt(n);
for ( int i = 2 ; i <= mx; i++)
if (n % i == 0 )
return i;
return n;
}
static double maxSum( int n)
{
long res = n;
while (n > 1 )
{
double divi = smallestDivisor(n);
n /= divi;
res += n;
}
return res;
}
public static void main (String[] args)
{
int n = 34 ;
System.out.println (maxSum(n));
}
}
|
Python3
from math import sqrt
def smallestDivisor(n):
mx = int (sqrt(n))
for i in range ( 2 , mx + 1 , 1 ):
if (n % i = = 0 ):
return i
return n
def maxSum(n):
res = n
while (n > 1 ):
divi = smallestDivisor(n)
n = int (n / divi)
res + = n
return res
if __name__ = = '__main__' :
n = 34
print (maxSum(n))
|
C#
using System;
class GFG
{
static double smallestDivisor( int n)
{
double mx = Math.Sqrt(n);
for ( int i = 2; i <= mx; i++)
if (n % i == 0)
return i;
return n;
}
static double maxSum( int n)
{
long res = n;
while (n > 1)
{
double divi = smallestDivisor(n);
n /= ( int )divi;
res += n;
}
return res;
}
public static void Main()
{
int n = 34;
Console.WriteLine(maxSum(n));
}
}
|
PHP
<?php
function smallestDivisor( $n )
{
$mx = sqrt( $n );
for ( $i = 2; $i <= $mx ; $i ++)
if ( $n % $i == 0)
return $i ;
return $n ;
}
function maxSum( $n )
{
$res = $n ;
while ( $n > 1)
{
$divi = smallestDivisor( $n );
$n /= $divi ;
$res += $n ;
}
return $res ;
}
$n = 34;
echo maxSum( $n );
#This code is contributed by akt_mit.
?>
|
Javascript
<script>
function smallestDivisor(n) {
var mx = Math.sqrt(n);
for (i = 2; i <= mx; i++)
if (n % i == 0)
return i;
return n;
}
function maxSum(n) {
var res = n;
while (n > 1) {
var divi = smallestDivisor(n);
n /= divi;
res += n;
}
return res;
}
var n = 34;
document.write(maxSum(n));
</script>
|
Time Complexity: O(sqrt(n)*log(n)), where n represents the given integer.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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