Yet another solution. This should handle all ints (including negative ints) except the min value of an int, which would need to be handled as a hard coded exception. This basically does division by subtraction but only using bit operators (shifts, xor, & and complement). For faster speed, it subtracts 3 * (decreasing powers of 2). In c#, it executes around 444 of these DivideBy3 calls per millisecond (2.2 seconds for 1,000,000 divides), so not horrendously slow, but no where near as fast as a simple x/3. By comparison, Coodey's nice solution is about 5 times faster than this one.

public static int DivideBy3(int a) {
    bool negative = a < 0;
    if (negative) a = Negate(a);
    int result;
    int sub = 3 << 29;
    int threes = 1 << 29;
    result = 0;
    while (threes > 0) {
        if (a >= sub) {
            a = Add(a, Negate(sub));
            result = Add(result, threes);
        }
        sub >>= 1;
        threes >>= 1;
    }
    if (negative) result = Negate(result);
    return result;
}
public static int Negate(int a) {
    return Add(~a, 1);
}
public static int Add(int a, int b) {
    int x = 0;
    x = a ^ b;
    while ((a & b) != 0) {
        b = (a & b) << 1;
        a = x;
        x = a ^ b;
    }
    return x;
}

这是c#，因为这是我手边的东西，但与c的区别应该很小。

使用itoa转换为以3为基数的字符串。去掉最后一个小调，转换回10进制。

// Note: itoa is non-standard but actual implementations
// don't seem to handle negative when base != 10.
int div3(int i) {
    char str[42];
    sprintf(str, "%d", INT_MIN); // Put minus sign at str[0]
    if (i>0)                     // Remove sign if positive
        str[0] = ' ';
    itoa(abs(i), &str[1], 3);    // Put ternary absolute value starting at str[1]
    str[strlen(&str[1])] = '\0'; // Drop last digit
    return strtol(str, NULL, 3); // Read back result
}

下面的脚本生成了一个C程序，可以在不使用运算符* / + - %的情况下解决这个问题:

#!/usr/bin/env python3

print('''#include <stdint.h>
#include <stdio.h>
const int32_t div_by_3(const int32_t input)
{
''')

for i in range(-2**31, 2**31):
    print('    if(input == %d) return %d;' % (i, i / 3))


print(r'''
    return 42; // impossible
}
int main()
{
    const int32_t number = 8;
    printf("%d / 3 = %d\n", number, div_by_3(number));
}
''')

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。

Yet another solution. This should handle all ints (including negative ints) except the min value of an int, which would need to be handled as a hard coded exception. This basically does division by subtraction but only using bit operators (shifts, xor, & and complement). For faster speed, it subtracts 3 * (decreasing powers of 2). In c#, it executes around 444 of these DivideBy3 calls per millisecond (2.2 seconds for 1,000,000 divides), so not horrendously slow, but no where near as fast as a simple x/3. By comparison, Coodey's nice solution is about 5 times faster than this one.

public static int DivideBy3(int a) {
    bool negative = a < 0;
    if (negative) a = Negate(a);
    int result;
    int sub = 3 << 29;
    int threes = 1 << 29;
    result = 0;
    while (threes > 0) {
        if (a >= sub) {
            a = Add(a, Negate(sub));
            result = Add(result, threes);
        }
        sub >>= 1;
        threes >>= 1;
    }
    if (negative) result = Negate(result);
    return result;
}
public static int Negate(int a) {
    return Add(~a, 1);
}
public static int Add(int a, int b) {
    int x = 0;
    x = a ^ b;
    while ((a & b) != 0) {
        b = (a & b) << 1;
        a = x;
        x = a ^ b;
    }
    return x;
}

这是c#，因为这是我手边的东西，但与c的区别应该很小。