这真的很简单。

if (number == 0) return 0;
if (number == 1) return 0;
if (number == 2) return 0;
if (number == 3) return 1;
if (number == 4) return 1;
if (number == 5) return 1;
if (number == 6) return 2;

(当然，为了简洁起见，我省略了一些程序。)如果程序员厌倦了把这些都输入，我相信他或她可以编写一个单独的程序来为他生成这些。我碰巧认识一个能大大简化他工作的人。

好吧，我想我们都同意这不是一个现实世界的问题。为了好玩，这里是如何用Ada和多线程来做这件事:

with Ada.Text_IO;

procedure Divide_By_3 is

   protected type Divisor_Type is
      entry Poke;
      entry Finish;
   private
      entry Release;
      entry Stop_Emptying;
      Emptying : Boolean := False;
   end Divisor_Type;

   protected type Collector_Type is
      entry Poke;
      entry Finish;
   private
      Emptying : Boolean := False;
   end Collector_Type;

   task type Input is
   end Input;
   task type Output is
   end Output;

   protected body Divisor_Type is
      entry Poke when not Emptying and Stop_Emptying'Count = 0 is
      begin
         requeue Release;
      end Poke;
      entry Release when Release'Count >= 3 or Emptying is
         New_Output : access Output;
      begin
         if not Emptying then
            New_Output := new Output;
            Emptying := True;
            requeue Stop_Emptying;
         end if;
      end Release;
      entry Stop_Emptying when Release'Count = 0 is
      begin
         Emptying := False;
      end Stop_Emptying;
      entry Finish when Poke'Count = 0 and Release'Count < 3 is
      begin
         Emptying := True;
         requeue Stop_Emptying;
      end Finish;
   end Divisor_Type;

   protected body Collector_Type is
      entry Poke when Emptying is
      begin
         null;
      end Poke;
      entry Finish when True is
      begin
         Ada.Text_IO.Put_Line (Poke'Count'Img);
         Emptying := True;
      end Finish;
   end Collector_Type;

   Collector : Collector_Type;
   Divisor : Divisor_Type;

   task body Input is
   begin
      Divisor.Poke;
   end Input;

   task body Output is
   begin
      Collector.Poke;
   end Output;

   Cur_Input : access Input;

   -- Input value:
   Number : Integer := 18;
begin
   for I in 1 .. Number loop
      Cur_Input := new Input;
   end loop;
   Divisor.Finish;
   Collector.Finish;
end Divide_By_3;

这是可行的:

smegma$ curl http://www.wolframalpha.com/input/?i=14+divided+by+3 2>/dev/null | gawk 'match($0, /link to /input/\?i=([0-9.+-]+)/, ary) { print substr( $0, ary[1, "start"], ary[1, "length"] )}' 4.6666666666666666666666666666666666666666666666666666

只要把你的数字换成“14”和“3”就行了。

并不是所有的答案都是面试官想听到的:

我的回答:

“我绝不会那样做，谁会为这种愚蠢的事情付出代价呢?”没有人 会有一个优势，它不是更快，它只是愚蠢。 教授设计师必须知道这一点，但这必须适用于所有数字，而不仅仅是除以3。”

使用cblas，作为OS X加速框架的一部分。

[02:31:59] [william@relativity ~]$ cat div3.c
#import <stdio.h>
#import <Accelerate/Accelerate.h>

int main() {
    float multiplicand = 123456.0;
    float multiplier = 0.333333;
    printf("%f * %f == ", multiplicand, multiplier);
    cblas_sscal(1, multiplier, &multiplicand, 1);
    printf("%f\n", multiplicand);
}

[02:32:07] [william@relativity ~]$ clang div3.c -framework Accelerate -o div3 && ./div3
123456.000000 * 0.333333 == 41151.957031

我认为正确的答案是:

为什么不用基本运算符来做基本运算呢?