导读:算法一直是程序员进阶的一道龙门,通常算法都是为了更高效地解决问题而创造的,但也有的只是出于学术性,并不在意其实际意义。这是近日在国外技术问答网站stackoverflow的一个热门问题,不知道你能给出几种解决方法?
问:在不使用*、/、+、-、%操作符的情况下,如何求一个数的1/3?(用C语言实现)
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);
- 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:
-
- int div3(int i) {
- char str[42];
- sprintf(str, "%d", INT_MIN);
- if (i>0) str[0] = ' ';
- itoa(abs(i), &str[1], 3);
- str[strlen(&str[1])] = '\0';
- return strtol(str, NULL, 3);
- }
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++;
- floor = r;
- r++;
- }
- 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;
- }
这个版本更快一些:
- unsigned add(char const zero[], unsigned const x, unsigned const y) {
-
- 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。)