title: Maximum Value Choice Either Dividing Considering
signature: function maximumValueChoiceEitherDividingConsidering($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(
        3, 
        3));
      
      printf("TEST-1...%s\n",driver(
        19, 
        24));
      
      printf("TEST-2...%s\n",driver(
        39, 
        57));
      
      printf("TEST-3...%s\n",driver(
        89, 
        152));
      
      printf("TEST-4...%s\n",driver(
        96, 
        192));
      
      printf("TEST-5...%s\n",driver(
        68, 
        114));
      
      printf("TEST-6...%s\n",driver(
        48, 
        83));
      
      printf("TEST-7...%s\n",driver(
        5, 
        5));
      
      printf("TEST-8...%s\n",driver(
        3, 
        3));
      
      printf("TEST-9...%s\n",driver(
        4, 
        4));
      
  }
  
  main();
  ?>
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 ];
  }