title: Maximum Value Choice Either Dividing Considering
signature: maximumValueChoiceEitherDividingConsidering(n) {
test_code: import 'dart:convert';
  
  // SOLUTION CODE
  // ============================================
  PLACEHOLDER_CODE_BODY
  
  // TESTING CODE 
  // ============================================
  bool validateSolution(int actual, int expected){
      return actual.toString() == expected.toString();
  }
  
  String driver(n, int expected){
      try
      {
          if (validateSolution(PLACEHOLDER_FN_NAME(n), expected)){
              return "PASSED";
          }
          return "FAILED";
      }catch(exception_object){
          return exception_object.runtimeType.toString();
      }
  }
  
  int main() {
  
      print("TEST-0...${driver(
          3, 
          3)
      }");
  
      print("TEST-1...${driver(
          19, 
          24)
      }");
  
      print("TEST-2...${driver(
          39, 
          57)
      }");
  
      print("TEST-3...${driver(
          89, 
          152)
      }");
  
      print("TEST-4...${driver(
          96, 
          192)
      }");
  
      print("TEST-5...${driver(
          68, 
          114)
      }");
  
      print("TEST-6...${driver(
          48, 
          83)
      }");
  
      print("TEST-7...${driver(
          5, 
          5)
      }");
  
      print("TEST-8...${driver(
          3, 
          3)
      }");
  
      print("TEST-9...${driver(
          4, 
          4)
      }");
  
      return 0;
  }
source_py: def maximum_value_choice_either_dividing_considering(n):
      res = list()
      res.append(0)
      res.append(1)
      i = 2
      while (i < (n + 1)):
          res.append(max(i, (((res[int((i / 2))] + res[int((i / 3))]) + res[int((i / 4))]) + res[int((i / 5))])))
          i = (i + 1)
      return res[n]
source_cpp: int maximum_value_choice_either_dividing_considering(int n) {
    int res [ n + 1 ];
    res [ 0 ] = 0;
    res [ 1 ] = 1;
    for ( int i = 2;
    i <= n;
    i ++ ) res [ i ] = max ( i, ( res [ i / 2 ] + res [ i / 3 ] + res [ i / 4 ] + res [ i / 5 ] ) );
    return res [ n ];
  }