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;

这在Setun电脑上很容易实现。

要把一个整数除3，右移1位。

不过，我不确定在这样的平台上是否有可能实现一个符合标准的C编译器。我们可能需要稍微扩展一下规则，比如将“至少8位”解释为“能够保存至少从-128到+127的整数”。

使用计数器是一个基本的解决方案:

int DivBy3(int num) {
    int result = 0;
    int counter = 0;
    while (1) {
        if (num == counter)       //Modulus 0
            return result;
        counter = abs(~counter);  //++counter

        if (num == counter)       //Modulus 1
            return result;
        counter = abs(~counter);  //++counter

        if (num == counter)       //Modulus 2
            return result;
        counter = abs(~counter);  //++counter

        result = abs(~result);    //++result
    }
}

也很容易执行一个模数函数，查看注释。

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

(注意:查看下面的编辑2以获得更好的版本!)

这并不像听起来那么棘手，因为你说“没有使用[..+[…]运营商”。如果你想禁止同时使用+字符，请参见下面。

unsigned div_by(unsigned const x, unsigned const by) {
  unsigned floor = 0;
  for (unsigned cmp = 0, r = 0; cmp <= x;) {
    for (unsigned i = 0; i < by; i++)
      cmp++; // that's not the + operator!
    floor = r;
    r++; // neither is this.
  }
  return floor;
}

然后用div_by(100,3)将100除以3。

编辑:你可以继续并替换++操作符:

unsigned inc(unsigned x) {
  for (unsigned mask = 1; mask; mask <<= 1) {
    if (mask & x)
      x &= ~mask;
    else
      return x & mask;
  }
  return 0; // overflow (note that both x and mask are 0 here)
}

编辑2:稍快的版本，不使用任何包含+、-、*、/、%字符的操作符。

unsigned add(char const zero[], unsigned const x, unsigned const y) {
  // this exploits that &foo[bar] == foo+bar if foo is of type char*
  return (int)(uintptr_t)(&((&zero[x])[y]));
}

unsigned div_by(unsigned const x, unsigned const by) {
  unsigned floor = 0;
  for (unsigned cmp = 0, r = 0; cmp <= x;) {
    cmp = add(0,cmp,by);
    floor = r;
    r = add(0,r,1);
  }
  return floor;
}

我们使用add函数的第一个参数，因为不使用*字符就不能表示指针的类型，除非在函数形参列表中，其中的语法类型[]与类型* const相同。

FWIW，你可以很容易地实现一个乘法函数使用类似的技巧使用0x55555556技巧提出的AndreyT:

int mul(int const x, int const y) {
  return sizeof(struct {
    char const ignore[y];
  }[x]);
}

你可以考虑用图或树状结构来解决这个问题。基本上生成的顶点数与要除以3的数一样多。然后继续将每个未配对的顶点与其他两个顶点配对。

粗糙的伪代码:

function divide(int num)
    while(num!=0)
        Add a new vertice to vertiexList.
        num--
    quotient = 0
    for each in vertexList(lets call this vertex A)
        if vertexList not empty
            Add an edge between A and another vertex(say B)
        else
            your Remainder is 1 and Quotient is quotient
        if vertexList not empty
            Add an edge between A and another vertex(say C)
        else
            your remainder is 2 and Quotient is quotient
        quotient++
        remove A, B, C from vertexList
    Remainder is 0 and Quotient is quotient

这显然是可以优化的，复杂度取决于你的数字有多大，但它应该工作，只要你能做++和——。 这就像数更酷的东西一样。