[算法]不使用*、/、+、-、%操作符求一個數的1/3

問：在不使用*、/、+、-、%操作符的情況下，如何求一個數的1/3？（用C語言實現）

第一種方法：使用位操作符並實現“+”操作

// 替換加法運算符
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;
}

原理：n = 4 * a + b; n / 3 = a + (a + b) / 3; 然後 sum += a, n = a + b 並迭代; 當 a == 0 (n < 4)時，sum += floor(n / 3); i.e. 1, if n == 3, else 0

第二種方法：

#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;
}

第三種方法：

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

增強版：

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

第四種方法：

#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()是標準C語言函數
    printf("%d\n", r.quot);
    return 0;
}

第五種方法：使用內聯彙編

#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);
}

第六種方法：

// 注意: itoa 不是個標準函數，但它可以實現
// 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
}

第七種方法：

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++; // 這是++操作符，不是+
    floor = r;
    r++; // 這也不是
  }
  return floor;
}

替換掉上面算法的++運算符：

unsigned inc(unsigned x) {
  for (unsigned mask = 1; mask; mask <<= 1) {
    if (mask & x)
      x &= ~mask;
    else
      return x & mask;
  }
  return 0; // 溢出（注意：這裏的x和mask都是0）
}

這個版本更快一些：

unsigned add(char const zero[], unsigned const x, unsigned const y) {
  // 這是因爲：如果foo是char*類型， &foo[bar] == foo+bar
  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;
}

第八種方法：實現乘法

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

第九種方法：極限

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);
}

原理：

因爲， 1/3 = 1/4 + 1/16 + 1/64 + ...

所以，

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 + ...

第十種方法：

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;
}