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
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
并不是所有的答案都是面试官想听到的:
我的回答:
“我绝不会那样做,谁会为这种愚蠢的事情付出代价呢?”没有人 会有一个优势,它不是更快,它只是愚蠢。 教授设计师必须知道这一点,但这必须适用于所有数字,而不仅仅是除以3。”
为什么我们不直接用在大学里学过的定义呢?结果可能效率低,但很清楚,因为乘法只是递归的减法,减法是加法,那么加法可以通过递归的异或/和逻辑端口组合来执行。
#include <stdio.h>
int add(int a, int b){
int rc;
int carry;
rc = a ^ b;
carry = (a & b) << 1;
if (rc & carry)
return add(rc, carry);
else
return rc ^ carry;
}
int sub(int a, int b){
return add(a, add(~b, 1));
}
int div( int D, int Q )
{
/* lets do only positive and then
* add the sign at the end
* inversion needs to be performed only for +Q/-D or -Q/+D
*/
int result=0;
int sign=0;
if( D < 0 ) {
D=sub(0,D);
if( Q<0 )
Q=sub(0,Q);
else
sign=1;
} else {
if( Q<0 ) {
Q=sub(0,Q);
sign=1;
}
}
while(D>=Q) {
D = sub( D, Q );
result++;
}
/*
* Apply sign
*/
if( sign )
result = sub(0,result);
return result;
}
int main( int argc, char ** argv )
{
printf( "2 plus 3=%d\n", add(2,3) );
printf( "22 div 3=%d\n", div(22,3) );
printf( "-22 div 3=%d\n", div(-22,3) );
printf( "-22 div -3=%d\n", div(-22,-3) );
printf( "22 div 03=%d\n", div(22,-3) );
return 0;
}
有人说……首先让它工作。注意,该算法应该适用于负Q…
下面的脚本生成了一个C程序,可以在不使用运算符* / + - %的情况下解决这个问题:
#!/usr/bin/env python3
print('''#include <stdint.h>
#include <stdio.h>
const int32_t div_by_3(const int32_t input)
{
''')
for i in range(-2**31, 2**31):
print(' if(input == %d) return %d;' % (i, i / 3))
print(r'''
return 42; // impossible
}
int main()
{
const int32_t number = 8;
printf("%d / 3 = %d\n", number, div_by_3(number));
}
''')
Yet another solution. This should handle all ints (including negative ints) except the min value of an int, which would need to be handled as a hard coded exception. This basically does division by subtraction but only using bit operators (shifts, xor, & and complement). For faster speed, it subtracts 3 * (decreasing powers of 2). In c#, it executes around 444 of these DivideBy3 calls per millisecond (2.2 seconds for 1,000,000 divides), so not horrendously slow, but no where near as fast as a simple x/3. By comparison, Coodey's nice solution is about 5 times faster than this one.
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;
}
这是c#,因为这是我手边的东西,但与c的区别应该很小。
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;