title: How To Print Maximum Number Of A Using Given Four Keys
signature: howToPrintMaximumNumberOfAUsingGivenFourKeys :: Integer -> Integer
  howToPrintMaximumNumberOfAUsingGivenFourKeys 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 (41) (110592) )
  
      printf "TEST-1...%s\n" (driver (94) (274877906944) )
  
      printf "TEST-2...%s\n" (driver (80) (5435817984) )
  
      printf "TEST-3...%s\n" (driver (40) (82944) )
  
      printf "TEST-4...%s\n" (driver (76) (1811939328) )
  
      printf "TEST-5...%s\n" (driver (5) (5) )
  
      printf "TEST-6...%s\n" (driver (43) (196608) )
  
      printf "TEST-7...%s\n" (driver (67) (150994944) )
  
      printf "TEST-8...%s\n" (driver (24) (1024) )
  
      printf "TEST-9...%s\n" (driver (90) (86973087744) )
  
source_py: def how_to_print_maximum_number_of_a_using_given_four_keys(N):
      if (N <= 6):
          return N
      screen = ([0] * N)
      for n in range(1, 7):
          screen[(n - 1)] = n
      for n in range(7, (N + 1)):
          screen[(n - 1)] = max((2 * screen[(n - 4)]), max((3 * screen[(n - 5)]), (4 * screen[(n - 6)])))
      return screen[(N - 1)]
source_cpp: int how_to_print_maximum_number_of_a_using_given_four_keys(int N) {
    if ( N <= 6 ) return N;
    int screen [ N ];
    int b;
    int n;
    for ( n = 1;
    n <= 6;
    n ++ ) screen [ n - 1 ] = n;
    for ( n = 7;
    n <= N;
    n ++ ) {
      screen [ n - 1 ] = max ( 2 * screen [ n - 4 ], max ( 3 * screen [ n - 5 ], 4 * screen [ n - 6 ] ) );
    }
    return screen [ N - 1 ];
  }