title: Recursively Break Number 3 Parts Get Maximum Sum 1
signature: function recursivelyBreakNumber3PartsGetMaximumSum1($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(
        50, 
        57));
      
      printf("TEST-1...%s\n",driver(
        83, 
        93));
      
      printf("TEST-2...%s\n",driver(
        18, 
        19));
      
      printf("TEST-3...%s\n",driver(
        24, 
        27));
      
      printf("TEST-4...%s\n",driver(
        31, 
        32));
      
      printf("TEST-5...%s\n",driver(
        38, 
        41));
      
      printf("TEST-6...%s\n",driver(
        94, 
        104));
      
      printf("TEST-7...%s\n",driver(
        24, 
        27));
      
      printf("TEST-8...%s\n",driver(
        13, 
        13));
      
      printf("TEST-9...%s\n",driver(
        53, 
        57));
      
  }
  
  main();
  ?>
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 ];
  }