title: Recursively Break Number 3 Parts Get Maximum Sum 1
signature: class Solution {
  
      public int RecursivelyBreakNumber3PartsGetMaximumSum1(int n) {
test_code: using System;
  using System.Numerics;
  using System.Diagnostics;
  using System.Collections.Generic;
  using System.Linq;
  using System.Text;
  using System.Security.Cryptography;
  using System.Web.UI;
  using System.Web.Script.Serialization;
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  namespace EvaluatePrediction {
      class TestRunner {
          public bool ValidateSolution(int actual, int expected){
              return new JavaScriptSerializer().Serialize(actual) == new JavaScriptSerializer().Serialize(expected);
          }
  
          public string driver(int n, int expected){
              try{
                PLACEHOLDER_CLS_NAME solution_cls = new PLACEHOLDER_CLS_NAME();
                int pred_result = solution_cls.PLACEHOLDER_FN_NAME(n); 
                if (ValidateSolution(
                      pred_result,
                      expected)){
                  return "PASSED";
                }
                return "FAILED";
              } 
              catch (Exception exception_obj)
              {
                return exception_obj.GetType().FullName;
              }
          }
          static void Main(string[] args) {
              TestRunner runner = new TestRunner();
              String result;
              result = runner.driver(
                  50, 
                  57
              );
              Console.WriteLine(string.Format(
                  "TEST-0...{0}",
                  result)
              );result = runner.driver(
                  83, 
                  93
              );
              Console.WriteLine(string.Format(
                  "TEST-1...{0}",
                  result)
              );result = runner.driver(
                  18, 
                  19
              );
              Console.WriteLine(string.Format(
                  "TEST-2...{0}",
                  result)
              );result = runner.driver(
                  24, 
                  27
              );
              Console.WriteLine(string.Format(
                  "TEST-3...{0}",
                  result)
              );result = runner.driver(
                  31, 
                  32
              );
              Console.WriteLine(string.Format(
                  "TEST-4...{0}",
                  result)
              );result = runner.driver(
                  38, 
                  41
              );
              Console.WriteLine(string.Format(
                  "TEST-5...{0}",
                  result)
              );result = runner.driver(
                  94, 
                  104
              );
              Console.WriteLine(string.Format(
                  "TEST-6...{0}",
                  result)
              );result = runner.driver(
                  24, 
                  27
              );
              Console.WriteLine(string.Format(
                  "TEST-7...{0}",
                  result)
              );result = runner.driver(
                  13, 
                  13
              );
              Console.WriteLine(string.Format(
                  "TEST-8...{0}",
                  result)
              );result = runner.driver(
                  53, 
                  57
              );
              Console.WriteLine(string.Format(
                  "TEST-9...{0}",
                  result)
              );
          }
      }
  }
  
source_py: def recursively_break_number_3_parts_get_maximum_sum_1(n):
      dp = ([0] * (n + 1))
      dp[0] = 0
      dp[1] = 1
      for i in range(2, (n + 1)):
          dp[i] = max(((dp[int((i / 2))] + dp[int((i / 3))]) + dp[int((i / 4))]), i)
      return dp[n]
source_cpp: int recursively_break_number_3_parts_get_maximum_sum_1(int n) {
    int dp [ n + 1 ];
    dp [ 0 ] = 0, dp [ 1 ] = 1;
    for ( int i = 2;
    i <= n;
    i ++ ) dp [ i ] = max ( dp [ i / 2 ] + dp [ i / 3 ] + dp [ i / 4 ], i );
    return dp [ n ];
  }