title: Recursively Break Number 3 Parts Get Maximum Sum
signature: function recursivelyBreakNumber3PartsGetMaximumSum($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(
        39, 
        41));
      
      printf("TEST-1...%s\n",driver(
        79, 
        87));
      
      printf("TEST-2...%s\n",driver(
        7, 
        7));
      
      printf("TEST-3...%s\n",driver(
        76, 
        87));
      
      printf("TEST-4...%s\n",driver(
        48, 
        57));
      
      printf("TEST-5...%s\n",driver(
        18, 
        19));
      
      printf("TEST-6...%s\n",driver(
        58, 
        63));
      
      printf("TEST-7...%s\n",driver(
        17, 
        17));
      
      printf("TEST-8...%s\n",driver(
        36, 
        41));
      
      printf("TEST-9...%s\n",driver(
        5, 
        5));
      
  }
  
  main();
  ?>
source_py: def recursively_break_number_3_parts_get_maximum_sum(n):
      if ((n == 0) or (n == 1)):
          return n
      return max(((recursively_break_number_3_parts_get_maximum_sum((n // 2)) + recursively_break_number_3_parts_get_maximum_sum((n // 3))) + recursively_break_number_3_parts_get_maximum_sum((n // 4))), n)
source_cpp: int recursively_break_number_3_parts_get_maximum_sum(int n) {
    if ( n == 0 || n == 1 ) return n;
    return max ( ( recursively_break_number_3_parts_get_maximum_sum ( n / 2 ) + recursively_break_number_3_parts_get_maximum_sum ( n / 3 ) + recursively_break_number_3_parts_get_maximum_sum ( n / 4 ) ), n );
  }