blocks|key|1950996|text|如果您有一个整数的字符串表示，您可以像这样编写isIncre函数(ord将字符转换为整数，string只是一个字符列表)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1950997|isIncre+(x:y:xs)+=+ord+x+<=+ord+y+&&+isIncre+(y:xs)
isIncre+_+=+True|code-block|syntax|javascript|1950998|如果编写不带ordered函数，处理任何有序类型，然后在调用它时将其与"map+ordered“组合在一起，可能会更好。然后，实现就是：|1950999|isIncre+(x:y:xs)+=+x+<=+y+&&+isIncre+(y:xs)
isIncre+_+=+True|1951000|如果x是一个整数，则可以这样调用|1951001|isIncre+(map+ord+(show+x))|1951002|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

If you have a string representation of an integer number, you could write the isIncre function like this (ord converts a character to an integer and string is just a list of chars):

<pre><code>isIncre (x:y:xs) = ord x &lt;= ord y &amp;&amp; isIncre (y:xs)
isIncre _ = True
</code></pre>

It could be even nicer to write the isIncre function without ord, working on any ordered type, then combine it with "map ord" when you call it instead. The implementation would then be just:

<pre><code>isIncre (x:y:xs) = x &lt;= y &amp;&amp; isIncre (y:xs)
isIncre _ = True
</code></pre>

That could be called like this, if x is an integer number

<pre><code>isIncre (map ord (show x))
</code></pre>

blocks|key|1951030|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1951031|数字0.99不能以2为基数精确表示。因此，您可能希望避免在此赋值中使用浮点数。相反，要查看数字<=+x中是否有99%25是保镖，请测试|ordered-list-item|offset|length|style|CODE|1951032|1951033|100+*+(x+-保镖x)+==+x|1951034|这是可行的，因为它(在数学上)与(x+-+bouncers+x)+==+x+/+100相同，如果(x+-+bouncers+x)+(非跳跃数字的数量)是x的1%25，则是真的。注意，因此不需要定义nonBouncers.|1951035|Also，。定义bouncers的另一种方法是|unordered-list-item|1951036|1951037|bouncers+x=+length+$+filter+1..x|1951038|However，isBouncy你应该重新考虑你的设计。当前，对于您尝试的每个x，您正在重新计算最高可达x的弹跳数。因此，大量的工作被一遍又一遍地完成。相反，您可能想要做的是生成一个元组序列(x,+n)，其中n是反弹数<=+x的数量。注意到，如果存在n反弹数<=+x，则存在n或n+%2B+1反弹数<=+x+%2B+1。|BOLD|1951039|1951040|更具体地说，要计算(x+%2B+1,+n')，只需要(x,+n)和isbouncy+(x+%2B+1).的输出|1951041|entityMap^0|0|1B|4|0|0|0|G|R|1C|G|24|1|2O|B|0|8|8|0|0|0|0|7|13|1|1G|1|2N|6|2W|1|31|4|3H|1|3L|4|3T|1|3V|5|43|8|0|0|9|B|O|6|V|G|0^^$0|@$1|2|3|-4|4|5|6|10|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|11|7|@$D|12|E|13|F|G]]|8|@]|9|$]]|$1|H|3|-4|4|5|6|14|7|@]|8|@]|9|$]]|$1|I|3|J|4|5|6|15|7|@]|8|@]|9|$]]|$1|K|3|L|4|5|6|16|7|@$D|17|E|18|F|G]|$D|19|E|1A|F|G]|$D|1B|E|1C|F|G]|$D|1D|E|1E|F|G]]|8|@]|9|$]]|$1|M|3|N|4|O|6|1F|7|@$D|1G|E|1H|F|G]]|8|@]|9|$]]|$1|P|3|-4|4|5|6|1I|7|@]|8|@]|9|$]]|$1|Q|3|R|4|5|6|1J|7|@]|8|@]|9|$]]|$1|S|3|T|4|O|6|1K|7|@$D|1L|E|1M|F|U]|$D|1N|E|1O|F|G]|$D|1P|E|1Q|F|G]|$D|1R|E|1S|F|G]|$D|1T|E|1U|F|G]|$D|1V|E|1W|F|G]|$D|1X|E|1Y|F|G]|$D|1Z|E|20|F|G]|$D|21|E|22|F|G]|$D|23|E|24|F|G]|$D|25|E|26|F|G]]|8|@]|9|$]]|$1|V|3|-4|4|5|6|27|7|@]|8|@]|9|$]]|$1|W|3|X|4|5|6|28|7|@$D|29|E|2A|F|G]|$D|2B|E|2C|F|G]|$D|2D|E|2E|F|G]]|8|@]|9|$]]|$1|Y|3|-4|4|5|6|2F|7|@]|8|@]|9|$]]]|Z|$]]

<ol>
<li>The number 0.99 cannot be represented exactly in base 2. Hence you may want to avoid the use of floating point numbers for this assignment. Instead, to see whether exactly 99% of the numbers <code>&lt;= x</code> are bouncers, test whether

<pre><code>100 * (x - bouncers x) == x
</code></pre>

This works because it is (mathematically) the same as <code>(x - bouncers x) == x / 100</code>, which is true if <code>(x - bouncers x)</code> (the number of non-bouncy numbers) is 1% of <code>x</code>. Observe that there is thus no need to define <code>nonBouncers</code>.</li>
<li>Also, another way to define <code>bouncers</code> is

<pre><code>bouncers x = length $ filter isBouncy [1..x]
</code></pre></li>
<li>However, you should reconsider your design. Currently you are recalculating the number of bouncy numbers up to <code>x</code>, for every <code>x</code> that you try. So a lot of work is being done over and over. What you may instead want to do, is generate a sequence of tuples <code>(x, n)</code>, where <code>n</code> is the number of bouncy numbers <code>&lt;= x</code>. Observe that if there are <code>n</code> bouncy numbers <code>&lt;= x</code>, then there are either <code>n</code> or <code>n + 1</code> bouncy number <code>&lt;= x + 1</code>.

More specifically, to calculate <code>(x + 1, n')</code>, all you need is <code>(x, n)</code> and the output of <code>isbouncy (x + 1)</code>.</li>
</ol>

blocks|key|543502|text|如果你有intetger的字符串表示，我会使用非常好的isIncre函数版本。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|543503|isIncre+::+(Ord+a)+=>+[a]+->+Bool
isIncre+list+=+and+$+zipWith+(<=)+list+(tail+list)|code-block|syntax|javascript|543504|如果不是，就用show编写它。|543505|isIncreNum+::+Integer+->+Bool
isIncreNum+=+isIncre+.+show|543506|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I would use really nice functional version of isIncre if you have string representation of intetger.

<pre><code>isIncre :: (Ord a) =&gt; [a] -&gt; Bool
isIncre list = and $ zipWith (&lt;=) list (tail list)
</code></pre>

If not, just compose it with show.

<pre><code>isIncreNum :: Integer -&gt; Bool
isIncreNum = isIncre . show
</code></pre>

I am doing <a href="http://projecteuler.net/index.php?section=problems&amp;id=112" rel="nofollow noreferrer">problem 112</a> on Project Euler and came up with the following to test the example case (I'll change the number in <code>answer</code> to 0.99 to get the real answer):

<pre><code>isIncre x | x == 99 = False
 | otherwise = isIncre' x
 where
 isIncre' x = ???

isDecre x = isIncre (read $ reverse $ show x :: Int)
isBouncy x = (isIncre x == False) &amp;&amp; (isDecre x == False)

bouncers x = length [n|n&lt;-[1..x],isBouncy n]
nonBouncers x = length [n|n&lt;-[1..x],(isBouncy n) == False] 

answer = head [x|x&lt;-[1..],((bouncers x) / (nonBouncers x)) == 0.5]
</code></pre>

But what I don't know how to do is define a function <code>isIncre'</code> which tests to see if the digits in a number are greater than or equal to the one on their left. I know it needs to be done recursively but how? 

On a side note, I know I can only use <code>/</code> on two floating point numbers but how can I make the output of <code>bouncers</code> to be floating point number instead of an integer?

Edit:

Thanks for the help, but it didn't like the <code>=</code> when I changed <code>isIncre</code> to:

<pre><code>isIncre x | x &lt;= 99 = False
 | otherwise = isIncre' (mshow x)
 where
 isIncre' (x:y:xs) = (x &lt;= y) &amp;&amp; (isIncre' (y:xs))
 isIncre' _ = True
</code></pre>

How to recursively compare the digits in a number in Haskell

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在对Project Euler执行，并提出了以下内容来测试示例用例(我将answer中的数字更改为0.99以获得真正的答案)：isIncre  x | x == 99 = False           | otherwise = isIncre' x                where           ...

问如何在Haskell中递归比较数字中的数字
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