使用fma()库函数的解决方案，适用于任何正数:

#include <stdio.h>
#include <math.h>

int main()
{
    int number = 8;//Any +ve no.
    int temp = 3, result = 0;
    while(temp <= number){
        temp = fma(temp, 1, 3); //fma(a, b, c) is a library function and returns (a*b) + c.
        result = fma(result, 1, 1);
    } 
    printf("\n\n%d divided by 3 = %d\n", number, result);
}

请看我的另一个答案。

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

(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]);
}

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。

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。

以下是我的解决方案:

public static int div_by_3(long a) {
    a <<= 30;
    for(int i = 2; i <= 32 ; i <<= 1) {
        a = add(a, a >> i);
    }
    return (int) (a >> 32);
}

public static long add(long a, long b) {
    long carry = (a & b) << 1;
    long sum = (a ^ b);
    return carry == 0 ? sum : add(carry, sum);
}

首先，请注意

1/3 = 1/4 + 1/16 + 1/64 + ...

现在，剩下的很简单!

a/3 = a * 1/3  
a/3 = a * (1/4 + 1/16 + 1/64 + ...)
a/3 = a/4 + a/16 + 1/64 + ...
a/3 = a >> 2 + a >> 4 + a >> 6 + ...

现在我们要做的就是把a的这些位移位值加在一起!哦!但是我们不能做加法，所以我们必须使用位操作符来编写一个加法函数!如果您熟悉逐位操作符，那么我的解决方案应该看起来相当简单……但以防你不懂，我会在最后讲一个例子。

另一件需要注意的事情是，首先我左移30!这是为了确保分数不会四舍五入。

11 + 6

1011 + 0110  
sum = 1011 ^ 0110 = 1101  
carry = (1011 & 0110) << 1 = 0010 << 1 = 0100  
Now you recurse!

1101 + 0100  
sum = 1101 ^ 0100 = 1001  
carry = (1101 & 0100) << 1 = 0100 << 1 = 1000  
Again!

1001 + 1000  
sum = 1001 ^ 1000 = 0001  
carry = (1001 & 1000) << 1 = 1000 << 1 = 10000  
One last time!

0001 + 10000
sum = 0001 ^ 10000 = 10001 = 17  
carry = (0001 & 10000) << 1 = 0

Done!

这就是你小时候学过的简单加法!

111
 1011
+0110
-----
10001

这个实现失败了，因为我们不能把方程的所有项相加:

a / 3 = a/4 + a/4^2 + a/4^3 + ... + a/4^i + ... = f(a, i) + a * 1/3 * 1/4^i
f(a, i) = a/4 + a/4^2 + ... + a/4^i

假设div_by_3(a) = x的结果，则x <= floor(f(a, i)) < a / 3。当a = 3k时，我们得到错误的答案。