p或~p [英] p or ~p
问题描述
我的逻辑有点模糊。在一个法律上不可接受的环境中,一位女士给我签了4封信。由于新近,最后我可以识别为''e''。
我今天早上无法验证。另外两个是{''r'',''a''}
或{'''',''r}。订单很重要。有没有人在c中模糊逻辑看到了单词
问题的处理? EC
My logic is a little fuzzy. In a legally-inadmissable setting a woman
signed 4 letters to me. Because of recency, the last I can identify as ''e''.
One I could not verify this morning. The other two were either {''r'', ''a''}
or {''a'', ''r}. Order matters. Has anyone seen a treatment of the word
problem with fuzzy logic in c? EC
推荐答案
Elijah Cardon发布:
Elijah Cardon posted:
我的逻辑有点模糊。
My logic is a little fuzzy.
你的语言也是如此。
So is your language.
在法律上不可接受的情况下,女人< br $>
给我签了4封信。
In a legally-inadmissable setting a woman
signed 4 letters to me.
具体。
(1)字母表中的字母
(2)一个人写给另一个人的一封信
Be specific.
(1) A letter of the alphabet
(2) A letter which a person writes to another person
由于新近,最后我可以识别为''e''。
Because of recency, the last I can identify as ''e''.
没有任何意义,但我隐约知道你想要兑现
表示最近的一封信更多比之前的
更难忘。
Makes no sense whatsoever, but I vaguely get the idea that you''re trying to
convey that the most recent letter is more memorable than the previous
ones.
今天早上我无法验证。
One I could not verify this morning.
其中一个是什么?这些字母?
One of what? The letters?
其他两个是{''r'',''a''}或{''a''',' 'R''}。
The other two were either {''r'', ''a''} or {''a'', ''r''}.
哦,所以你在谈论字母表的字母?
Oh, so you''re talking about letters of the alphabet?
订单很重要。有没有人在c中看到过用
模糊逻辑处理单词问题?
Order matters. Has anyone seen a treatment of the word problem with
fuzzy logic in c?
四个字母。其中一个是e。在''r'旁边有一个''a''。
编码应该不难:
int ContainsE (char const * p)
{
for(;;)
{
switch(* p ++)
{
case''e'':返回1;
案例0:返回0;
}
}
}
int HasAbesideR(char const * p)
{
for(;;)
{
switch(* p ++)
{
case''a'':if(''r''== p [1])返回1;
case''r'':if(''a''= = p [1])返回1;
案例0:返回0;
}
}
}
#include< string.h>
int FitsCriteria(char const * p)
{
返回4 == strlen(p)&&包含E(p)&& HasAbesideR(p);
}
现在你所要做的就是打开一个字典文本文件并运行每个单词
通过它。
-
Frederick Gotham
Four letters. One of them is ''e''. There''s an ''a'' directly beside an ''r''.
Shouldn''t be hard to code:
int ContainsE(char const *p)
{
for(;;)
{
switch(*p++)
{
case ''e'': return 1;
case 0: return 0;
}
}
}
int HasAbesideR(char const *p)
{
for(;;)
{
switch(*p++)
{
case ''a'': if(''r'' == p[1]) return 1;
case ''r'': if(''a'' == p[1]) return 1;
case 0: return 0;
}
}
}
#include <string.h>
int FitsCriteria(char const *p)
{
return 4==strlen(p) && ContainsE(p) && HasAbesideR(p);
}
Now all you have to do is open a dictionary text file and run every word
through it.
--
Frederick Gotham
Frederick Gotham发布:
Frederick Gotham posted:
switch(* p ++)
{
case''a'':if(''r ''== p [1])返回1;
switch(*p++)
{
case ''a'': if(''r'' == p[1]) return 1;
Wups,我需要一个休息那里。
Wups, I need a "break" there.
case''r'':if(''a''== p [1])返回1;
case ''r'': if(''a'' == p[1]) return 1;
在这里。
-
Frederick Gotham >
我可以提高效率:
int FitsCriteria(char const * p)
{
char const * const pover = p + 4;
int contains_e = 0,has_a_beside_r = 0;
do
{
开关(* p ++)
{
case''e'':
{
if(has_a_beside_r)返回1;
contains_e = 1 ;
}
休息;
案例''a'':
{
if(''r''== p [1])
{
if(contains_e)返回1;
has_a_beside_r = 1;
}
}
休息;
case''r'':
{
if(''a'= = p [1])
{
如果(contains_e)返回1;
has_a_beside_r = 1;
}
}
休息;
案例0:返回0;
}
}
while(pover!= p);
返回0;
}
-
Frederick Gotham
May as well boost the efficiency while I''m at it:
int FitsCriteria(char const *p)
{
char const *const pover = p + 4;
int contains_e = 0, has_a_beside_r = 0;
do
{
switch(*p++)
{
case ''e'':
{
if(has_a_beside_r) return 1;
contains_e = 1;
}
break;
case ''a'':
{
if(''r'' == p[1])
{
if(contains_e) return 1;
has_a_beside_r = 1;
}
}
break;
case ''r'':
{
if(''a'' == p[1])
{
if(contains_e) return 1;
has_a_beside_r = 1;
}
}
break;
case 0: return 0;
}
}
while(pover != p);
return 0;
}
--
Frederick Gotham
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