除以7而不使用除法运算符 [英] division by 7 without using division operator
问题描述
上个月，我出现在接受EA体育采访时，他们问了这个问题。
如果不使用数字，你如何将数字除以7分部操作员？
我做了一个减法并保留了一个计数器，保留了一个标签
减去了多少次。
后来，EA体育人告诉我，当然可以通过使用位操作符来更好地使用
技术。
自7有一个二进制表示111，我的猜测是左移
3位的操作应该给出答案，但是我无法得到
工作。
有何意见？
< kr ******** ***@gmail.com在留言中写道
新闻：11 ********************** @ p10g2000cwp.googlegr psps.com。 ..
上个月，我出现在EA体育的采访中，他们问了这个问题。 n。
如果不使用除法运算符，你如何将数字除以7？
我做了减法并保留了一个计数器保留了一个标签
减去了多少次。
后来，EA运动家伙告诉我当然有可能更好
使用位运算符的
技术。
最明显的方法是移动，比较和有条件地减去和b / b掩盖一个1的商数（非常像必要的汇编语言
在没有原生除法指令的处理器上。然而，这是非常低效的。
。这涉及一些操作，但可能不是面试官正在寻找的那些
。
您可能会发布另一篇文章并交叉发布到sci.math同样。可能
来自那些人的一些好答案。
由于7有二进制表示111，我的猜测是左移
3位的操作应该给出答案，但是我无法得到
的工作。
我不确定你要去的方向是什么，除了死胡同。
我不是意识到你所建议的方向的一般方法
将与师合作。有一些很好的近似值（例如，
1/7 = 1/8 + 1/56，大约等于1/8 + 1/64），但我不是
意识到你建议的方向的确切方法。
-
David T. Ashley（dt * @ e3ft.com）
http://www.e3ft.com （咨询主页页面）
http://www.dtashley.com （个人主页）
http://gpl.e3ft.com （GPL出版物和项目）
kr ********** *@gmail.com 写道：
上个月我出现在接受EA体育采访的时候他们问了我这个问题。
如果不使用divi，你如何将数字除以7 sion运营商？
除以并截断为积分结果？
或
见如果一个数字可被7整除而没有余数？
kr *********** @ gmail.com 写道：
上个月我出现在接受EA sports和他们问了这个问题。
如果不使用除法运算符，你如何将数字除以7？
我做了一个减法并保留了一个计数器，保留了一个标签
减去了多少次。
后来，EA运动员告诉我那当然可以通过使用位运算符来更好地使用
技术。
由于7具有二进制表示111，我的猜测是左移
3位的操作应该给出答案，但我无法得到
工作。
有何评论？
（数量> 3）+ 1似乎有用吗？
Last month I appeared for an interview with EA sports and they asked
me this question.
How would you divide a number by 7 without using division operator ?
I did by doing a subtraction and keeping a counter that kept a tab on
how many times I subtracted.
Later, the EA sport guy told me that of course there are can be better
technique by using bit operator.
Since 7 has a binary representation 111, my guess is that a left shift
operation of 3 bits should give the answer, but I couldn''t get it to
work.
Any comments ?
解决方案<kr***********@gmail.comwrote in message
news:11**********************@p10g2000cwp.googlegr oups.com...Last month I appeared for an interview with EA sports and they asked
me this question.
How would you divide a number by 7 without using division operator ?
I did by doing a subtraction and keeping a counter that kept a tab on
how many times I subtracted.
Later, the EA sport guy told me that of course there are can be better
technique by using bit operator.The most obvious method is to shift, compare, and conditionally subtract and
mask a 1 into the quotient (very much like the necessary assembly-language
on a processor without a native division instruction). This is, however,
very inefficient. This involves some bit operations, but may not be what
the interviewer was looking for.
You might make another post and cross-post to sci.math as well. There may
be some good answers coming from those folks.
Since 7 has a binary representation 111, my guess is that a left shift
operation of 3 bits should give the answer, but I couldn''t get it to
work.I''m not sure that the direction you''re going is anything except a dead end.
I''m not aware of a general method in the direction you''ve suggested that
will work with division. There are some good approximations (for example,
1/7 = 1/8 + 1/56 which is approximately equal to 1/8 + 1/64), but I''m not
aware of an exact method in the direction you''re suggesting.
--
David T. Ashley (dt*@e3ft.com)
http://www.e3ft.com (Consulting Home Page)
http://www.dtashley.com (Personal Home Page)
http://gpl.e3ft.com (GPL Publications and Projects)
kr***********@gmail.com wrote:
Last month I appeared for an interview with EA sports and they asked
me this question.
How would you divide a number by 7 without using division operator ?Divide by and truncate to integral result?
or
See if a number is divisible by 7 with no remainder?
kr***********@gmail.com wrote:Last month I appeared for an interview with EA sports and they asked
me this question.
How would you divide a number by 7 without using division operator ?
I did by doing a subtraction and keeping a counter that kept a tab on
how many times I subtracted.
Later, the EA sport guy told me that of course there are can be better
technique by using bit operator.
Since 7 has a binary representation 111, my guess is that a left shift
operation of 3 bits should give the answer, but I couldn''t get it to
work.
Any comments ?
(num >3) + 1 seems to work?
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