title: How To Print Maximum Number Of A Using Given Four Keys
signature: pub fn how_to_print_maximum_number_of_a_using_given_four_keys(n: i32) -> i64 {
test_code: use std::collections::HashMap;
  use std::collections::HashSet;
  use std::cmp::*;
  use std::collections::*;
  use std::fmt::*; 
  
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  pub fn validate_solution(actual: i64, expected: i64)->bool{
  return actual==expected
  }
  pub fn driver(n: i32, expected: i64)->String{
      if validate_solution(PLACEHOLDER_FN_NAME(n), expected) {
        return "PASSED".to_string();
      };
      return "FAILED".to_string();
  
  }
  
  fn main() {
      println!("TEST-0...{}", driver(
        41, 
        110592
      ));
      
      println!("TEST-1...{}", driver(
        94, 
        274877906944
      ));
      
      println!("TEST-2...{}", driver(
        80, 
        5435817984
      ));
      
      println!("TEST-3...{}", driver(
        40, 
        82944
      ));
      
      println!("TEST-4...{}", driver(
        76, 
        1811939328
      ));
      
      println!("TEST-5...{}", driver(
        5, 
        5
      ));
      
      println!("TEST-6...{}", driver(
        43, 
        196608
      ));
      
      println!("TEST-7...{}", driver(
        67, 
        150994944
      ));
      
      println!("TEST-8...{}", driver(
        24, 
        1024
      ));
      
      println!("TEST-9...{}", 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 ];
  }