title: Recursively Break Number 3 Parts Get Maximum Sum 1
signature: pub fn recursively_break_number_3_parts_get_maximum_sum_1(n: i32) -> i32 {
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: i32, expected: i32)->bool{
  return actual==expected
  }
  pub fn driver(n: i32, expected: i32)->String{
      if validate_solution(PLACEHOLDER_FN_NAME(n), expected) {
        return "PASSED".to_string();
      };
      return "FAILED".to_string();
  
  }
  
  fn main() {
      println!("TEST-0...{}", driver(
        50, 
        57
      ));
      
      println!("TEST-1...{}", driver(
        83, 
        93
      ));
      
      println!("TEST-2...{}", driver(
        18, 
        19
      ));
      
      println!("TEST-3...{}", driver(
        24, 
        27
      ));
      
      println!("TEST-4...{}", driver(
        31, 
        32
      ));
      
      println!("TEST-5...{}", driver(
        38, 
        41
      ));
      
      println!("TEST-6...{}", driver(
        94, 
        104
      ));
      
      println!("TEST-7...{}", driver(
        24, 
        27
      ));
      
      println!("TEST-8...{}", driver(
        13, 
        13
      ));
      
      println!("TEST-9...{}", 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 ];
  }