title: How To Print Maximum Number Of A Using Given Four Keys
signature: function how_to_print_maximum_number_of_a_using_given_four_keys(n)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return actual == expected
  end
  
  function driver(n, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(n), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver(41, 110592);
      @printf("TEST-0...%s\n",result);
      
      result = driver(94, 274877906944);
      @printf("TEST-1...%s\n",result);
      
      result = driver(80, 5435817984);
      @printf("TEST-2...%s\n",result);
      
      result = driver(40, 82944);
      @printf("TEST-3...%s\n",result);
      
      result = driver(76, 1811939328);
      @printf("TEST-4...%s\n",result);
      
      result = driver(5, 5);
      @printf("TEST-5...%s\n",result);
      
      result = driver(43, 196608);
      @printf("TEST-6...%s\n",result);
      
      result = driver(67, 150994944);
      @printf("TEST-7...%s\n",result);
      
      result = driver(24, 1024);
      @printf("TEST-8...%s\n",result);
      
      result = driver(90, 86973087744);
      @printf("TEST-9...%s\n",result);
      
  end
  
  main()
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 ];
  }