# 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; ` `} ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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". ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` `?> ` |

*chevron_right*

*filter_none*

**Output :**

MaxSum = 106

## Recommended Posts:

- Maximum sum after repeatedly dividing N by a divisor
- Choice of Area
- Longest dividing subsequence
- Count of divisors having more set bits than quotient on dividing N
- Ways of dividing a group into two halves such that two elements are in different groups
- Minimum integer such that it leaves a remainder 1 on dividing with any element from the range [2, N]
- Count number of digits after decimal on dividing a number
- Maximum value of |arr[i] - arr[j]| + |i - j|
- Find maximum among x^(y^2) or y^(x^2) where x and y are given
- Find maximum value of x such that n! % (k^x) = 0
- Maximum value of |arr[0] - arr[1]| + |arr[1] - arr[2]| + ... +|arr[n - 2] - arr[n - 1]| when elements are from 1 to n
- Pair with maximum GCD from two arrays
- Maximum GCD of N integers with given product
- Maximum Primes whose sum is equal to given N
- Maximum array sum that can be obtained after exactly k changes

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.