不用"+","-","*","/","%"运算符球一个数的1/3

第一种方法：使用位操作符并实现“+”操作

// 替换加法运算符 
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

第二种方法：

#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; 
} 

第三种方法：

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

增强版：

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

第四种方法：

#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()是标准C语言函数
    printf("%d\n", r.quot); 
    return 0; 
} 

第五种方法：使用内联汇编

#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); 
} 

第六种方法：

// 注意: itoa 不是个标准函数，但它可以实现
// 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++; // 这是++操作符，不是+ 
    floor = r; 
    r++; // 这也不是 
  } 
  return floor; 
} 

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; // 溢出（注意：这里的x和mask都是0）
} 

这个版本更快一些：

unsigned add(char const zero[], unsigned const x, unsigned const y) { 
  // 这是因为：如果foo是char*类型， &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; 
} 

第八种方法：实现乘法

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

第九种方法：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); 
} 

原理：

因为， 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 + ...

第十种方法：

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; 
} 

注：本例是C#实现，因为作者更熟悉C#，但本题更倾向于算法，所以语言并不是太重要吧？（当然只是在不使用语言特性的前提下。）