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);
}

很好bc:

$ num=1337; printf "scale=5;${num}\x2F3;\n" | bc
445.66666

如果我们认为__div__不是正字法上的/

def divBy3(n):
    return n.__div__(3)

print divBy3(9), 'or', 9//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。

这是Python中的，基本上，字符串比较和一个状态机。

def divide_by_3(input):
  to_do = {}
  enque_index = 0
  zero_to_9 = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
  leave_over = 0
  for left_over in (0, 1, 2):
    for digit in zero_to_9:
      # left_over, digit => enque, leave_over
      to_do[(left_over, digit)] = (zero_to_9[enque_index], leave_over)
      if leave_over == 0:
        leave_over = 1
      elif leave_over == 1:
        leave_over = 2
      elif leave_over == 2 and enque_index != 9:
        leave_over = 0
        enque_index = (1, 2, 3, 4, 5, 6, 7, 8, 9)[enque_index]
  answer_q = []
  left_over = 0
  digits = list(str(input))
  if digits[0] == "-":
    answer_q.append("-")
  digits = digits[1:]
  for digit in digits:
    enque, left_over = to_do[(left_over, int(digit))]
    if enque or len(answer_q):
      answer_q.append(enque)
  answer = 0
  if len(answer_q):
    answer = int("".join([str(a) for a in answer_q]))
  return answer

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);
}