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。
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
好吧,我想我们都同意这不是一个现实世界的问题。为了好玩,这里是如何用Ada和多线程来做这件事:
with Ada.Text_IO;
procedure Divide_By_3 is
protected type Divisor_Type is
entry Poke;
entry Finish;
private
entry Release;
entry Stop_Emptying;
Emptying : Boolean := False;
end Divisor_Type;
protected type Collector_Type is
entry Poke;
entry Finish;
private
Emptying : Boolean := False;
end Collector_Type;
task type Input is
end Input;
task type Output is
end Output;
protected body Divisor_Type is
entry Poke when not Emptying and Stop_Emptying'Count = 0 is
begin
requeue Release;
end Poke;
entry Release when Release'Count >= 3 or Emptying is
New_Output : access Output;
begin
if not Emptying then
New_Output := new Output;
Emptying := True;
requeue Stop_Emptying;
end if;
end Release;
entry Stop_Emptying when Release'Count = 0 is
begin
Emptying := False;
end Stop_Emptying;
entry Finish when Poke'Count = 0 and Release'Count < 3 is
begin
Emptying := True;
requeue Stop_Emptying;
end Finish;
end Divisor_Type;
protected body Collector_Type is
entry Poke when Emptying is
begin
null;
end Poke;
entry Finish when True is
begin
Ada.Text_IO.Put_Line (Poke'Count'Img);
Emptying := True;
end Finish;
end Collector_Type;
Collector : Collector_Type;
Divisor : Divisor_Type;
task body Input is
begin
Divisor.Poke;
end Input;
task body Output is
begin
Collector.Poke;
end Output;
Cur_Input : access Input;
-- Input value:
Number : Integer := 18;
begin
for I in 1 .. Number loop
Cur_Input := new Input;
end loop;
Divisor.Finish;
Collector.Finish;
end Divide_By_3;
其他回答
这是Python中的,基本上,字符串比较和一个状态机。
def divide_by_3(input):
to_do = {}
enque_index = 0
zero_to_9 = (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
leave_over = 0
for left_over in (0, 1, 2):
for digit in zero_to_9:
# left_over, digit => enque, leave_over
to_do[(left_over, digit)] = (zero_to_9[enque_index], leave_over)
if leave_over == 0:
leave_over = 1
elif leave_over == 1:
leave_over = 2
elif leave_over == 2 and enque_index != 9:
leave_over = 0
enque_index = (1, 2, 3, 4, 5, 6, 7, 8, 9)[enque_index]
answer_q = []
left_over = 0
digits = list(str(input))
if digits[0] == "-":
answer_q.append("-")
digits = digits[1:]
for digit in digits:
enque, left_over = to_do[(left_over, int(digit))]
if enque or len(answer_q):
answer_q.append(enque)
answer = 0
if len(answer_q):
answer = int("".join([str(a) for a in answer_q]))
return answer
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。
你可以使用(依赖于平台)内联程序集,例如,对于x86:(也适用于负数)
#include <stdio.h>
int main() {
int dividend = -42, divisor = 5, quotient, remainder;
__asm__ ( "cdq; idivl %%ebx;"
: "=a" (quotient), "=d" (remainder)
: "a" (dividend), "b" (divisor)
: );
printf("%i / %i = %i, remainder: %i\n", dividend, divisor, quotient, remainder);
return 0;
}
要将一个数除以3,而不使用乘法、除法、余数、减法或加法操作,在汇编编程语言中,惟一可用的指令是LEA(地址有效负载)、SHL(向左移动)和SHR(向右移动)。
在这个解决方案中,我没有使用与运算符+ - * /%相关的操作
我假设有输出数字在定点格式(16位整数部分和16位小数部分)和输入数字的类型是短int;但是,我已经近似输出的数量,因为我只能信任整数部分,因此我返回一个短int类型的值。
65536/6是固定点值,相当于1/3浮点数,等于21845。
21845 = 16384 + 4096 + 1024 + 256 + 64 + 16 + 4 + 1.
因此,要用1/3(21845)来做乘法,我使用指令LEA和SHL。
short int DivideBy3( short int num )
//In : eax= 16 Bit short int input number (N)
//Out: eax= N/3 (32 Bit fixed point output number
// (Bit31-Bit16: integer part, Bit15-Bit0: digits after comma)
{
__asm
{
movsx eax, num // Get first argument
// 65536 / 3 = 21845 = 16384 + 4096 + 1024 + 256 + 64 + 16 + 4 + 1
lea edx,[4*eax+eax] // EDX= EAX * 5
shl eax,4
lea edx,[eax+edx] // EDX= EDX + EAX * 16
shl eax,2
lea edx,[eax+edx] // EDX= EDX + EAX * 64
shl eax,2
lea edx,[eax+edx] // EDX= EDX + EAX * 256
shl eax,2
lea edx,[eax+edx] // EDX= EDX + EAX * 1024
shl eax,2
lea edx,[eax+edx] // EDX= EDX + EAX * 4096
shl eax,2
lea edx,[eax+edx+08000h] // EDX= EDX + EAX * 16384
shr edx,010h
movsx eax,dx
}
// Return with result in EAX
}
它也适用于负数;结果具有正数的最小近似值(逗号后的最后一位数字为-1)。
如果您不打算使用运算符+ - * /%来执行除3的操作,但可以使用与它们相关的操作,我建议另一种解决方案。
int DivideBy3Bis( short int num )
//In : eax= 16 Bit short int input number (N)
//Out: eax= N/3 (32 Bit fixed point output number
// (Bit31-Bit16: integer part, Bit15-Bit0: digits after comma)
{
__asm
{
movsx eax, num // Get first argument
mov edx,21845
imul edx
}
// Return with result in EAX
}
使用itoa转换为以3为基数的字符串。去掉最后一个小调,转换回10进制。
// Note: itoa is non-standard but actual implementations
// 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) // Remove sign if positive
str[0] = ' ';
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
}