在PHP中使用BC数学:

<?php
    $a = 12345;
    $b = bcdiv($a, 3);   
?>

MySQL(来自Oracle的采访)

> SELECT 12345 DIV 3;

帕斯卡:

a:= 12345;
b:= a div 3;

X86-64汇编语言:

mov  r8, 3
xor  rdx, rdx   
mov  rax, 12345
idiv r8

用Pascal编写程序并使用DIV操作符。

因为问题被标记为c，你可以在Pascal中编写一个函数，然后在c程序中调用它;这样做的方法是特定于系统的。

但是这里有一个在我的Ubuntu系统上运行的例子，安装了Free Pascal fp-编译器包。(我这么做完全是出于不合时宜的固执;我不敢说这是有用的。)

divide_by_3。不是:

unit Divide_By_3;
interface
    function div_by_3(n: integer): integer; cdecl; export;
implementation
    function div_by_3(n: integer): integer; cdecl;
    begin
        div_by_3 := n div 3;
    end;
end.

c:

#include <stdio.h>
#include <stdlib.h>

extern int div_by_3(int n);

int main(void) {
    int n;
    fputs("Enter a number: ", stdout);
    fflush(stdout);
    scanf("%d", &n);
    printf("%d / 3 = %d\n", n, div_by_3(n));
    return 0;
}

构建:

fpc divide_by_3.pas && gcc divide_by_3.o main.c -o main

示例执行:

$ ./main
Enter a number: 100
100 / 3 = 33

这是经典的2进制除法算法

#include <stdio.h>
#include <stdint.h>

int main()
{
  uint32_t mod3[6] = { 0,1,2,0,1,2 };
  uint32_t x = 1234567; // number to divide, and remainder at the end
  uint32_t y = 0; // result
  int bit = 31; // current bit
  printf("X=%u   X/3=%u\n",x,x/3); // the '/3' is for testing

  while (bit>0)
  {
    printf("BIT=%d  X=%u  Y=%u\n",bit,x,y);
    // decrement bit
    int h = 1; while (1) { bit ^= h; if ( bit&h ) h <<= 1; else break; }
    uint32_t r = x>>bit;  // current remainder in 0..5
    x ^= r<<bit;          // remove R bits from X
    if (r >= 3) y |= 1<<bit; // new output bit
    x |= mod3[r]<<bit;    // new remainder inserted in X
  }
  printf("Y=%u\n",y);
}

如果你提醒自己标准的学校除法方法，用二进制来做，你会发现在3的情况下，你只是在有限的一组值中除法和减法(在这种情况下，从0到5)。这些可以用switch语句处理，以摆脱算术运算符。

static unsigned lamediv3(unsigned n)
{
  unsigned result = 0, remainder = 0, mask = 0x80000000;

  // Go through all bits of n from MSB to LSB.
  for (int i = 0; i < 32; i++, mask >>= 1)
  {
    result <<= 1;
    // Shift in the next bit of n into remainder.
    remainder = remainder << 1 | !!(n & mask);

    // Divide remainder by 3, update result and remainer.
    // If remainder is less than 3, it remains intact.
    switch (remainder)
    {
    case 3:
      result |= 1;
      remainder = 0;
      break;

    case 4:
      result |= 1;
      remainder = 1;
      break;

    case 5:
      result |= 1;
      remainder = 2;
      break;
    }
  }

  return result;
}

#include <cstdio>

int main()
{
  // Verify for all possible values of a 32-bit unsigned integer.
  unsigned i = 0;

  do
  {
    unsigned d = lamediv3(i);

    if (i / 3 != d)
    {
      printf("failed for %u: %u != %u\n", i, d, i / 3);
      return 1;
    }
  }
  while (++i != 0);
}

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。

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。”