重点在于效率和速度这是练习： 程序应该从键盘（标准输入）监视可能无限的字符串 。如果它检测到序列aaa，则它输出0。如果它检测到序列aba，它输出1。 不检测序列内的序列。程序在检测到输入结束时应该干净地退出。例如： 以下序列aababaaabaaa<输入结束>将产生 以下结果：100 虽然以下序列aaababaaaabbababa<输入结束>将 产生以下结果：0101 任何接受者？
Emphasis is on efficiancy and speed this is the excercise: The program should monitor a possibly infinite stream of characters from the keyboard (standard input). If it detects the sequence "aaa" it outputs a "0". If it detects the sequence "aba" it outputs a "1". DO NOT detect sequences within sequences. The program should exit cleanly when it detects an End Of Input. For example: The following sequence aababaaabaaa<End Of Input> would produce the following result: 100 While the following sequence aaababaaaabbababa<End Of Input> would produce the following result: 0101 Any takers?
推荐答案Andy Green写道： Andy Green wrote: 重点是效率和速度这是练习：程序应该从键盘监视可能无限的字符流（/>）标准输入）。如果它检测到序列aaa，则它输出0。如果它检测到序列aba，它输出1。不检测序列内的序列。程序在检测到输入结束时应该干净地退出。例如： 以下序列aababaaabaaa<输入结束>将产生以下结果：100 以下序列aaababaaaabbababa<输入结束>会产生以下结果：0101 任何接受者？ Emphasis is on efficiancy and speed this is the excercise: The program should monitor a possibly infinite stream of characters from the keyboard (standard input). If it detects the sequence "aaa" it outputs a "0". If it detects the sequence "aba" it outputs a "1". DO NOT detect sequences within sequences. The program should exit cleanly when it detects an End Of Input. For example: The following sequence aababaaabaaa<End Of Input> would produce the following result: 100 While the following sequence aaababaaaabbababa<End Of Input> would produce the following result: 0101 Any takers?
你监控stdin的要求表明你认为 逐个字符的无缓冲输入可用。这个 表明你有一个特定的实现，并使 问题偏离主题。另一方面，如果挑战可以被重写以检查字符串，无论它们来自何处，那么这是一个简单的算法问题，并且是偏离主题的。 也许你可以解释一下你的问题含有什么内容，使它成为关于C的问题。或者你可以做自己的功课。
The requirement that you monitor stdin suggests that you think unbuffered input on a character-by-character basis is available. This suggests that you have a specific implementation in mind and makes the question off-topic. If, on the other hand, the challenge can be rewritten to examine strings, wherever they came from, then this is a simple algorithm question and is off-topic. Perhaps you can explain what content your question has that makes it a question about C. Or you could do your own homework.
Andy Green写道： Andy Green wrote: 重点是效率和速度这是练习：程序应该监视一个可能无限的流来自键盘的字符（标准输入）。如果它检测到序列aaa，则它输出0。如果它检测到序列aba，它输出1。不检测序列内的序列。程序在检测到输入结束时应该干净地退出。例如： 以下序列aababaaabaaa<输入结束>将产生以下结果：100 以下序列aaababaaaabbababa<输入结束>会产生以下结果：0101 任何接受者？ Emphasis is on efficiancy and speed this is the excercise: The program should monitor a possibly infinite stream of characters from the keyboard (standard input). If it detects the sequence "aaa" it outputs a "0". If it detects the sequence "aba" it outputs a "1". DO NOT detect sequences within sequences. The program should exit cleanly when it detects an End Of Input. For example: The following sequence aababaaabaaa<End Of Input> would produce the following result: 100 While the following sequence aaababaaaabbababa<End Of Input> would produce the following result: 0101 Any takers?
C语言标准对速度没什么可说的/> 和效率，但我觉得这个节目很难敲打：b $ b $ main（无效）{ 返回0; } 您可能想知道为什么这个程序按要求工作，并且您的教授可能会想到同样的事情。答案在于 仔细阅读家庭作业陈述： - 只有* * aaa"或aba或aba是检测到。这个程序既没有检测到（也没有检测到，b / b 是否需要检测任何东西的b / b ）因此可以解除任何责任 的任何责任 - 明确禁止该程序检测序列中的序列。 输入字符的流是一个序列（注意在两个例子中使用 这个术语），所以程序是 禁止检测其输入中的任何内容。 - 该程序需要监控。它的输入， 但是动词监视器不是由C 语言标准定义，也不是由家庭作业定义。 因此我们可以自由地定义它 我们找到最方便的。第六个定义 由 www.yourdictionary for 及物动词监视器是指导，并且 上面的程序指示输入 高效位桶。 - 实际上，该程序不需要做任何事情 与输入有关。家庭作业 说程序应该......。 （强调我的），和 应该仅仅是一个提示。要求总是用应表示正确表达。 （和禁止 ，不得），如 C语言标准第4节所述。 我希望这可以帮助你获得你应得的成绩。 - Er ********* @ sun
The C language Standard has nothing to say about speed and "efficiancy," but I think this program will be hard to beat: int main(void) { return 0; } You may be wondering why this program works as required, and your professor may wonder the same thing. The answer lies in a close reading of the homework statement: - Output is only required *if* "aaa" or "aba" is detected. This program detects neither (nowhere does the homework assignment state a requirement that anything be detected), and is thus relieved of any responsibility to produce output. - The program is expressly forbidden to detect "sequences within sequences." The stream of input characters is a sequence (note the use of the term in the two examples), so the program is forbidden to detect anything in its input. - The program is required to "monitor" its input, but the verb "monitor" is not defined by the C language Standard, nor by the homework assignment. Therefore we are free to define it in the manner we find most convenient. The sixth definition given by www.yourdictionary for the transitive verb "monitor" is "to direct," and the program above "directs" the input to the highly efficient bit bucket. - Actually, the program is not required to do anything at all with or to the input. The homework assignment says "The program SHOULD ..." (emphasis mine), and "should" is merely a hint. A requirement is always properly expressed with "shall" (and a prohibition with "shall not"), as described in Section 4 of the C language Standard. I hope this helps you get the grade you deserve. -- Er*********@sun
Andy Green写道： Andy Green wrote: 重点在于效率和速度这是练习：程序应该从键盘监视可能无限的字符流（标准输入）。如果它检测到序列aaa，则它输出0。如果它检测到序列aba，它输出1。不检测序列内的序列。程序在检测到输入结束时应该干净地退出。例如： 以下序列aababaaabaaa<输入结束>将产生以下结果：100 以下序列aaababaaaabbababa<输入结束>会产生以下结果：0101 任何接受者？ Emphasis is on efficiancy and speed this is the excercise: The program should monitor a possibly infinite stream of characters from the keyboard (standard input). If it detects the sequence "aaa" it outputs a "0". If it detects the sequence "aba" it outputs a "1". DO NOT detect sequences within sequences. The program should exit cleanly when it detects an End Of Input. For example: The following sequence aababaaabaaa<End Of Input> would produce the following result: 100 While the following sequence aaababaaaabbababa<End Of Input> would produce the following result: 0101 Any takers?
这在lexing和解析下更多。 IMO，自然的 行动方案是为这项任务创建一种语言。 类似于： zero_sequence :: = aaa one_sequance :: = aba 通过lex或flex等程序运行，生成 a解析表。然后写一个程序来支持解析 表。保留这些作品，因为你需要它们来做你的下一个家庭作业。 或者一个简单的状态图可以帮助你： b + --- + + ------> | 4 | - + | + --- + | | | + --- + a + --- + a + --- + | O - > | 1 | ---> | 2 | ---> | 3 | | + --- + + --- + + --- + | ^ | | | | vv | + --------------------------- + 状态1：保持这种状态，直到检测到a为止。 状态2：如果检测到a，则转换到状态3。 /> 如果检测到b，则转换到状态4. 否则转换到状态1. 状态3：如果检测到a，则打印0。 在所有情况下转换到状态1. 状态4：如果是检测到'a''，打印''1'。 在所有情况下转换到状态1. 如果在任何情况下检测到EOF状态，该程序应该退出（即状态5）。 现在，编码。 - Thomas Matthews C ++新闻组欢迎辞： www.slack/~shiva/welcome.txt C ++常见问题： www.parashift/c++-faq-lite C常见问题： http：//www.eskimo。 com /~scs / c-faq / top.html altp.lang.learn.c-c ++ faq： www.raos.demon.uk/acllc-c++/faq.html 其他网站： www.josuttis - C ++ STL图书馆书
This falls more under lexing and parsing. IMO, the natural course of action is to create a language for this assignment. Something like: zero_sequence ::= a a a one_sequance ::= a b a Run this through a program such as lex or flex to generate a parse table. Then write a program to support the parse table. Keep these pieces, as you will need them to do your next homework assignments. Or perhaps a simple state diagram would help you: b +---+ +------> | 4 | --+ | +---+ | | | +---+ a +---+ a +---+ | O -> | 1 | ---> | 2 | ---> | 3 | | +---+ +---+ +---+ | ^ | | | | v v | +---------------------------+ State 1: stay in this state until an ''a'' is detected. State 2: if an ''a'' is detected, transition to state 3. if a ''b'' is detected, transition to state 4. Otherwise transition to state 1. State 3: if an ''a'' is detected, print ''0''. Transition in all cases to state 1. State 4: if an ''a'' is detected, print ''1''. Transition in all cases to state 1. If an EOF is detected in any state, the program should exit (i.e. state 5). Now, code it up. -- Thomas Matthews C++ newsgroup welcome message: www.slack/~shiva/welcome.txt C++ Faq: www.parashift/c++-faq-lite C Faq: www.eskimo/~scs/c-faq/top.html altp.lang.learn.c-c++ faq: www.raos.demon.uk/acllc-c++/faq.html Other sites: www.josuttis -- C++ STL Library book
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