Maximum value with the choice of either dividing or considering as it is

We are given a number n, we need to find the maximum sum possible with the help of following function:
F(n) = max( (F(n/2) + F(n/3) + F(n/4) + F(n/5)), n). To calculate F(n, ) we may either have n as our result or we can further break n into four part as in given function definition. This can be done as much time as we can. Find the maximum possible sum you can get from a given N. Note : 1 can not be break further so F(1) = 1 as a base case.

Examples :

Input : n = 10
Output : MaxSum = 12
Explanation: 
f(10) = f(10/2) + f(10/3) + f(10/4) + f(10/5)
      = f(5)  +   f(3)  +  f(2)   +  f(2)
      = 12
5, 3 and 2 cannot be further divided.

Input : n = 2
Output : MaxSum = 2


Approach : This problem can be solve with recursive approach but that will cost us a high complexity because of its overlapping sub problems. So we apply dynamic programming concept to solve this question in bottom up manner as:

C++

// CPP program for maximize result when
// we have choice to divide or consider
// as it is.
#include <bits/stdc++.h>
using namespace std;
  
// function for calculating max possible result
int maxDP(int n)
{
    int res[n + 1];
    res[0] = 0;
    res[1] = 1;
  
    // Compute remaining values in bottom
    // up manner.
    for (int i = 2; i <= n; i++)
        res[i] = max(i, (res[i / 2] + res[i / 3] + res[i / 4] + res[i / 5]));
  
    return res[n];
}
  
// driver program
int main()
{
    int n = 60;
    cout << "MaxSum =" << maxDP(n);
    return 0;
}

Java

// Java program for maximize result when
// we have choice to divide or consider
// as it is.
import java.io.*;
  
class GFG {
  
    // function for calculating max
    // possible result
    static int maxDP(int n)
    {
        int res[] = new int[n + 1];
        res[0] = 0;
        res[1] = 1;
  
        // Compute remaining values in
        // bottom up manner.
        for (int i = 2; i <= n; i++)
            res[i] = Math.max(i, (res[i / 2] + res[i / 3] + res[i / 4] + res[i / 5]));
  
        return res[n];
    }
  
    // driver program
    public static void main(String[] args)
    {
        int n = 60;
        System.out.println("MaxSum = " + maxDP(n));
    }
}
  
// This code is contributed by vt_m

Python3

# Python3 code to maximize result when
# we have choice to divide or consider
# as it is.
  
# function for calculating max 
# possible result
def maxDP (n):
    res = list()
    res.append(0)
    res.append(1)
      
    # Compute remaining values in 
    # bottom up manner.
    i = 2
    while i<n + 1:
        res.append(max(i, (res[int(i / 2)] + res[int(i / 3)] +
                          res[int(i / 4)]+ res[int(i / 5)])))
        i = i + 1
      
    return res[n]
  
# driver code
n = 60
print("MaxSum =", maxDP(n))
  
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program for maximize result when
// we have choice to divide or consider
// as it is.
using System;
  
class GFG {
  
    // function for calculating max
    // possible result
    static int maxDP(int n)
    {
        int[] res = new int[n + 1];
        res[0] = 0;
        res[1] = 1;
  
        // Compute remaining values in
        // bottom up manner.
        for (int i = 2; i <= n; i++)
            res[i] = Math.Max(i, (res[i / 2] + res[i / 3] + res[i / 4] + res[i / 5]));
  
        return res[n];
    }
  
    // Driver program
    public static void Main()
    {
        int n = 60;
        Console.WriteLine("MaxSum = " + maxDP(n));
    }
}
  
// This code is contributed by vt_m

PHP

<?php
  
// PHP program to maximize 
// result when we have choice
// to divide or consider as it is.
  
// function for calculating
// max possible result
  
function maxDP ($n)
{
  
    $res[0] = 0;
    $res[1] = 1;
  
    // Compute remaining values  
    // in bottom up manner.
    for ($i = 2; $i <= $n; $i++)
    $res[$i] = max($i, ($res[$i / 2] + 
                        $res[$i / 3] + 
                        $res[$i / 4] + 
                        $res[$i / 5]));
  
    return $res[$n];
}
  
// Driver Code
$n = 60;
echo "MaxSum =", maxDP ($n);
  
// This code is contributed by aj_36
?>


Output :

MaxSum = 106


My Personal Notes arrow_drop_up


Discovering ways to develop a plane for soaring career goals

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : jit_t