如何将一个数除3而不使用*、/、+、-、%等运算符?
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如何将一个数除3而不使用*、/、+、-、%等运算符?
号码可以有签名,也可以没有签名。
当前回答
如果你提醒自己标准的学校除法方法,用二进制来做,你会发现在3的情况下,你只是在有限的一组值中除法和减法(在这种情况下,从0到5)。这些可以用switch语句处理,以摆脱算术运算符。
static unsigned lamediv3(unsigned n)
{
unsigned result = 0, remainder = 0, mask = 0x80000000;
// Go through all bits of n from MSB to LSB.
for (int i = 0; i < 32; i++, mask >>= 1)
{
result <<= 1;
// Shift in the next bit of n into remainder.
remainder = remainder << 1 | !!(n & mask);
// Divide remainder by 3, update result and remainer.
// If remainder is less than 3, it remains intact.
switch (remainder)
{
case 3:
result |= 1;
remainder = 0;
break;
case 4:
result |= 1;
remainder = 1;
break;
case 5:
result |= 1;
remainder = 2;
break;
}
}
return result;
}
#include <cstdio>
int main()
{
// Verify for all possible values of a 32-bit unsigned integer.
unsigned i = 0;
do
{
unsigned d = lamediv3(i);
if (i / 3 != d)
{
printf("failed for %u: %u != %u\n", i, d, i / 3);
return 1;
}
}
while (++i != 0);
}
其他回答
一般来说,解决这个问题的方法是:
log(pow(exp(numerator),pow(deliminator,-1)))
使用黑客的喜悦魔术数字计算器
int divideByThree(int num)
{
return (fma(num, 1431655766, 0) >> 32);
}
其中fma是在math.h头文件中定义的标准库函数。
第一:
x/3 = (x/4) / (1-1/4)
然后求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))
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)
}
虽然它使用了+,但有人已经实现了按位操作的add。
#include <stdio.h>
typedef struct { char a,b,c; } Triple;
unsigned long div3(Triple *v, char *r) {
if ((long)v <= 2)
return (unsigned long)r;
return div3(&v[-1], &r[1]);
}
int main() {
unsigned long v = 21;
int r = div3((Triple*)v, 0);
printf("%ld / 3 = %d\n", v, r);
return 0;
}
很好bc:
$ num=1337; printf "scale=5;${num}\x2F3;\n" | bc
445.66666