Divide a number by 3 without using *, /, +, -, % o

2019-01-03 03:56发布

How would you divide a number by 3 without using *, /, +, -, %, operators?

The number may be signed or unsigned.

30条回答
不美不萌又怎样
2楼-- · 2019-01-03 04:09

I think the right answer is:

Why would I not use a basic operator to do a basic operation?

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来,给爷笑一个
3楼-- · 2019-01-03 04:10

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 declare result as double and add to it the result of fmod(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 of number "records" that are divisor 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

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萌系小妹纸
4楼-- · 2019-01-03 04:12

This should work for any divisor, not only three. Currently only for unsigned, but extending it to signed should not be that difficult.

#include <stdio.h>

unsigned sub(unsigned two, unsigned one);
unsigned bitdiv(unsigned top, unsigned bot);
unsigned sub(unsigned two, unsigned one)
{
unsigned bor;
bor = one;
do      {
        one = ~two & bor;
        two ^= bor;
        bor = one<<1;
        } while (one);
return two;
}

unsigned bitdiv(unsigned top, unsigned bot)
{
unsigned result, shift;

if (!bot || top < bot) return 0;

for(shift=1;top >= (bot<<=1); shift++) {;}
bot >>= 1;

for (result=0; shift--; bot >>= 1 ) {
        result <<=1;
        if (top >= bot) {
                top = sub(top,bot);
                result |= 1;
                }
        }
return result;
}

int main(void)
{
unsigned arg,val;

for (arg=2; arg < 40; arg++) {
        val = bitdiv(arg,3);
        printf("Arg=%u Val=%u\n", arg, val);
        }
return 0;
}
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ら.Afraid
5楼-- · 2019-01-03 04:12

first:

x/3 = (x/4) / (1-1/4)

then figure out how to solve x/(1 - y):

x/(1-1/y)
  = x * (1+y) / (1-y^2)
  = x * (1+y) * (1+y^2) / (1-y^4)
  = ...
  = x * (1+y) * (1+y^2) * (1+y^4) * ... * (1+y^(2^i)) / (1-y^(2^(i+i))
  = x * (1+y) * (1+y^2) * (1+y^4) * ... * (1+y^(2^i))

with y = 1/4:

int div3(int x) {
    x <<= 6;    // need more precise
    x += x>>2;  // x = x * (1+(1/2)^2)
    x += x>>4;  // x = x * (1+(1/2)^4)
    x += x>>8;  // x = x * (1+(1/2)^8)
    x += x>>16; // x = x * (1+(1/2)^16)
    return (x+1)>>8; // as (1-(1/2)^32) very near 1,
                     // we plus 1 instead of div (1-(1/2)^32)
}

although it uses +, but somebody already implements add by bitwise op

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干净又极端
6楼-- · 2019-01-03 04:13

This one is the classical division algorithm in base 2:

#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);
}
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forever°为你锁心
7楼-- · 2019-01-03 04:13

First that I've come up with.

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

EDIT: Sorry, I didn't notice the tag C. But you can use the idea about string formatting, I guess...

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