不使用加减乘除求余操作符求一个数的1/3

内容主要来源：http://stackoverflow.com/questions/11694546/divide-a-number-by-3-without-using-operators 。根据此论坛进行整理。

问题：

在不使用*、/、+、-、%操作符的情况下，如何求一个数的1/3？

解决方法：

第1种方法：使用位操作符并实现“+”操作

There is a simple function I found here. But it's using the+ operator, so you have to add the values with the bit-operators:

// replaces the + operator
int add(int x, int y) {
    int a, b;
    do {
        a = x & y;
        b = x ^ y;
        x = a << 1;
        y = b;
    } while (a);
    return b;
}

int divideby3 (int num) {
    int sum = 0;
    while (num > 3) {
        sum = add(num >> 2, sum);
        num = add(num >> 2, num & 3);
    }
    if (num == 3)
        sum = add(sum, 1);
    return sum; 
}

As Jim commented this works because:

• n = 4 * a + b
• n / 3 = a + (a + b) / 3
• So sum += a, n = a + b, and iterate
• When a == 0 (n < 4), sum += floor(n / 3); i.e. 1, if n == 3, else 0
  

原理：n = 4 * a + b; n / 3 = a + (a + b) / 3; 然后 sum += a, n = a + b 并迭代; 当 a == 0 (n < 4)时，sum += floor(n / 3); i.e. 1, if n == 3, else 0

第2种方法

Idiotic conditions call for an idiotic solution:

#include <stdio.h>
#include <stdlib.h>

int main()
{
    FILE * fp=fopen("temp.dat","w+b");
    int number=12346;
    int divisor=3;
    char * buf = calloc(number,1);
    fwrite(buf,number,1,fp);
    rewind(fp);
    int result=fread(buf,divisor,number,fp);
    printf("%d / %d = %d", number, divisor, result);
    free(buf);
    fclose(fp);
    return 0;
}

If also the decimal part is needed, just declareresult asdouble and add to it the result offmod(number,divisor).

Explanation of how it works

1. The fwrite writes number bytes (number being 123456 in the example above).
2. rewind resets the file pointer to the front of the file.
3. fread reads a maximum ofnumber "records" that aredivisor in length from the file, and returns the number of elements it read.

If you write 30 bytes then read back the file in units of 3, you get 10 "units". 30 / 3 = 10

第3种方法

log(pow(exp(number),0.33333333333333333333)) /* :-) */

Improved version:

log(pow(exp(number),sin(atan2(1,sqrt(8)))))

第4种方法

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
    int num = 1234567;
    int den = 3;
    div_t r = div(num,den); // div() is a standard C function.
    printf("%d\n", r.quot);
    return 0;
}

第5种方法:使用内联汇编

Use inline assembler: 

#include <stdio.h>

int main() {
  int dividend = -42, divisor = 3, quotient, remainder;

  __asm__ ( "movl   %2, %%edx;"
            "sarl  $31, %%edx;"
            "movl   %2, %%eax;"
            "movl   %3, %%ebx;"
            "idivl      %%ebx;"
          : "=a" (quotient), "=d" (remainder)
          : "g"  (dividend), "g"  (divisor)
          : "ebx" );

  printf("%i / %i = %i, remainder: %i\n", dividend, divisor, quotient, remainder);
}

第6种方法

Use itoa to convert to base 3 string.Drop last trit and convert back to base 10.

// 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) str[0] = ' ';       // remove sign if positive
  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
}

第7种方法

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;
}

Edit: go on an replace the ++ operator as well(替换掉上面算法的++运算符):

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)
}

Edit2:Slightly faster Version(这个版本更快一些：):

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;
}

第8种方法 实现乘法

(implement a multiplication function using a similar trick to use the0x55555556 trick proposed)

int mul(int const x, int const y) {
  return sizeof(struct {
    char const ignore[y];
  }[x]);
}

第9种方法:极限

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);
}

First, note that

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

Now, the rest is simple!

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 + ...

Now all we have to do is add together these bit shifted values of a! Oops! We can't add though, so instead, we'll have to write an add function using bit-wise operators! If you're familiar with bit-wise operators, my solution should look fairly simple... but just in-case you aren't, I'll walk through an example at the end.

Another thing to note is that first I shift left by 30! This is to make sure that the fractions don't get rounded off.

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!

It's simply carry addition that you learned as a child!

111
 1011
+0110
-----
10001

This implementation failed because we can not add all terms of the equation:

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

Suppose the reslut of div_by_3(a) = x, thenx <= floor(f(a, i)) < a / 3. Whena = 3k, we get wrong answer.

第10种方法：

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;
}

This is c# because that's what I had handy, but differences from c should be minor.

第11种方法：

Using Ada and multithreading:

with Ada.Text_IO;

procedure Divide_By_3 is

   protected type Divisor_Type is
      entry Poke;
      entry Finish;
   private
      entry Release;
      entry Stop_Emptying;
      Emptying : Boolean := False;
   end Divisor_Type;

   protected type Collector_Type is
      entry Poke;
      entry Finish;
   private
      Emptying : Boolean := False;
   end Collector_Type;

   task type Input is
   end Input;
   task type Output is
   end Output;

   protected body Divisor_Type is
      entry Poke when not Emptying and Stop_Emptying'Count = 0 is
      begin
         requeue Release;
      end Poke;
      entry Release when Release'Count >= 3 or Emptying is
         New_Output : access Output;
      begin
         if not Emptying then
            New_Output := new Output;
            Emptying := True;
            requeue Stop_Emptying;
         end if;
      end Release;
      entry Stop_Emptying when Release'Count = 0 is
      begin
         Emptying := False;
      end Stop_Emptying;
      entry Finish when Poke'Count = 0 and Release'Count < 3 is
      begin
         Emptying := True;
         requeue Stop_Emptying;
      end Finish;
   end Divisor_Type;

   protected body Collector_Type is
      entry Poke when Emptying is
      begin
         null;
      end Poke;
      entry Finish when True is
      begin
         Ada.Text_IO.Put_Line (Poke'Count'Img);
         Emptying := True;
      end Finish;
   end Collector_Type;

   Collector : Collector_Type;
   Divisor : Divisor_Type;

   task body Input is
   begin
      Divisor.Poke;
   end Input;

   task body Output is
   begin
      Collector.Poke;
   end Output;

   Cur_Input : access Input;

   -- Input value:
   Number : Integer := 18;
begin
   for I in 1 .. Number loop
      Cur_Input := new Input;
   end loop;
   Divisor.Finish;
   Collector.Finish;
end Divide_By_3;

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