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

很有趣的是，没有人回答一个泛泛的划分:

/* For the given integer find the position of MSB */
int find_msb_loc(unsigned int n)
{
    if (n == 0)
        return 0;

    int loc = sizeof(n)  * 8 - 1;
    while (!(n & (1 << loc)))
        loc--;
    return loc;
}


/* Assume both a and b to be positive, return a/b */
int divide_bitwise(const unsigned int a, const unsigned int b)
{
    int int_size = sizeof(unsigned int) * 8;
    int b_msb_loc = find_msb_loc(b);

    int d = 0; // dividend
    int r = 0; // reminder
    int t_a = a;
    int t_a_msb_loc = find_msb_loc(t_a);
    int t_b = b << (t_a_msb_loc - b_msb_loc);

    int i;
    for(i = t_a_msb_loc; i >= b_msb_loc; i--)  {
        if (t_a > t_b) {
            d = (d << 1) | 0x1;
            t_a -= t_b; // Not a bitwise operatiion
            t_b = t_b >> 1;
         }
        else if (t_a == t_b) {
            d = (d << 1) | 0x1;
            t_a = 0;
        }
        else { // t_a < t_b
            d = d << 1;
            t_b = t_b >> 1;
        }
    }

    r = t_a;
    printf("==> %d %d\n", d, r);
    return d;
}

按位加法已经在其中一个答案中给出，所以跳过它。

#!/bin/ruby

def div_by_3(i)
  i.div 3        # always return int http://www.ruby-doc.org/core-1.9.3/Numeric.html#method-i-div
end

没有反复检查这个答案是否已经发表。如果程序需要扩展到浮点数，可以将这些数字乘以所需精度的10*数，然后可以再次应用下面的代码。

#include <stdio.h>

int main()
{
    int aNumber = 500;
    int gResult = 0;

    int aLoop = 0;

    int i = 0;
    for(i = 0; i < aNumber; i++)
    {
        if(aLoop == 3)
        {
           gResult++;
           aLoop = 0;
        }  
        aLoop++;
    }

    printf("Reulst of %d / 3 = %d", aNumber, gResult);

    return 0;
}

#include <stdio.h>

typedef struct { char a,b,c; } Triple;

unsigned long div3(Triple *v, char *r) {
  if ((long)v <= 2)  
    return (unsigned long)r;
  return div3(&v[-1], &r[1]);
}

int main() {
  unsigned long v = 21; 
  int r = div3((Triple*)v, 0); 
  printf("%ld / 3 = %d\n", v, r); 
  return 0;
}

这应该适用于任何除数，而不仅仅是3。目前仅适用于unsigned，但将其扩展到signed应该没有那么困难。

#include <stdio.h>

unsigned sub(unsigned two, unsigned one);
unsigned bitdiv(unsigned top, unsigned bot);
unsigned sub(unsigned two, unsigned one)
{
unsigned bor;
bor = one;
do      {
        one = ~two & bor;
        two ^= bor;
        bor = one<<1;
        } while (one);
return two;
}

unsigned bitdiv(unsigned top, unsigned bot)
{
unsigned result, shift;

if (!bot || top < bot) return 0;

for(shift=1;top >= (bot<<=1); shift++) {;}
bot >>= 1;

for (result=0; shift--; bot >>= 1 ) {
        result <<=1;
        if (top >= bot) {
                top = sub(top,bot);
                result |= 1;
                }
        }
return result;
}

int main(void)
{
unsigned arg,val;

for (arg=2; arg < 40; arg++) {
        val = bitdiv(arg,3);
        printf("Arg=%u Val=%u\n", arg, val);
        }
return 0;
}