title: How To Print Maximum Number Of A Using Given Four Keys
signature: function howToPrintMaximumNumberOfAUsingGivenFourKeys($n) {
test_code: <?php
  
  
  // SOLUTION CODE 
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  function validateSolution($actual, $expected){
      return serialize($actual) == serialize($expected);
  }
  
  function driver($n, $expected){
      try{
        if (validateSolution(PLACEHOLDER_FN_NAME($n), $expected)){
          return "PASSED";
        }
        return "FAILED";
      }
      catch (Exception $exception_obj) {
        return get_class($exception_obj);
      }
  }
  
  function main(){
      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));
      
  }
  
  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 ];
  }