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);
}

首先是我想到的。

irb(main):101:0> div3 = -> n { s = '%0' + n.to_s + 's'; (s % '').gsub('   ', ' ').size }
=> #<Proc:0x0000000205ae90@(irb):101 (lambda)>
irb(main):102:0> div3[12]
=> 4
irb(main):103:0> div3[666]
=> 222

编辑:对不起，我没有注意到标签c。但是你可以使用字符串格式的想法，我猜…

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;
}

(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]);
}

#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;
}

哪里InputValue是数字除以3

SELECT AVG(NUM) 
  FROM (SELECT InputValue NUM from sys.dual
         UNION ALL SELECT 0 from sys.dual
         UNION ALL SELECT 0 from sys.dual) divby3