title: Maximum Value Choice Either Dividing Considering
signature: function maximum_value_choice_either_dividing_considering(n)
test_code: using Printf
  
  # SOLUTION CODE
  # ============================================
  PLACEHOLDER_CODE_BODY
  
  # TESTING CODE 
  # ============================================
  function validate_solution(actual, expected)
      return actual == expected
  end
  
  function driver(n, expected)
      try
  
          if validate_solution(PLACEHOLDER_FN_NAME(n), expected)
              return "PASSED";
          end
          return "FAILED";
      catch exception_obj
          return string(Base.typename(typeof(exception_obj)).wrapper);
      end
  end
  
  function main()
      
      result = driver(3, 3);
      @printf("TEST-0...%s\n",result);
      
      result = driver(19, 24);
      @printf("TEST-1...%s\n",result);
      
      result = driver(39, 57);
      @printf("TEST-2...%s\n",result);
      
      result = driver(89, 152);
      @printf("TEST-3...%s\n",result);
      
      result = driver(96, 192);
      @printf("TEST-4...%s\n",result);
      
      result = driver(68, 114);
      @printf("TEST-5...%s\n",result);
      
      result = driver(48, 83);
      @printf("TEST-6...%s\n",result);
      
      result = driver(5, 5);
      @printf("TEST-7...%s\n",result);
      
      result = driver(3, 3);
      @printf("TEST-8...%s\n",result);
      
      result = driver(4, 4);
      @printf("TEST-9...%s\n",result);
      
  end
  
  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 ];
  }