title: Recursively Break Number 3 Parts Get Maximum Sum 1
signature: recursivelyBreakNumber3PartsGetMaximumSum1 :: Integer -> Integer
  recursivelyBreakNumber3PartsGetMaximumSum1 n = 
test_code: import Text.Printf
  import Control.Exception (catch, SomeException)
  import Data.Typeable
  import qualified Data.List as List
  import qualified Data.Map as Map
  import qualified Data.Set as Set
  
  -- SOLUTION CODE
  -- ============================================
  
  PLACEHOLDER_CODE_BODY
  
  -- TESTING CODE 
  -- ============================================
  validateSolution :: Integer -> Integer -> Bool
  validateSolution actual expected = do
      actual == expected
  
  
  driver :: Integer -> Integer -> String
  driver n expected = do
      if (validateSolution (PLACEHOLDER_FN_NAME n) expected)
          then "PASSED"
      else "FAILED"
  
  main = do
      printf "TEST-0...%s\n" (driver (50) (57) )
  
      printf "TEST-1...%s\n" (driver (83) (93) )
  
      printf "TEST-2...%s\n" (driver (18) (19) )
  
      printf "TEST-3...%s\n" (driver (24) (27) )
  
      printf "TEST-4...%s\n" (driver (31) (32) )
  
      printf "TEST-5...%s\n" (driver (38) (41) )
  
      printf "TEST-6...%s\n" (driver (94) (104) )
  
      printf "TEST-7...%s\n" (driver (24) (27) )
  
      printf "TEST-8...%s\n" (driver (13) (13) )
  
      printf "TEST-9...%s\n" (driver (53) (57) )
  
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 ];
  }