blocks|key|968559|text|不是，41字节，这家伙|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968560|NOOOOOO?NOOOOOOOOOOO!nnOOOOO
NOOOOOOOO?no|code-block|syntax|javascript|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/ValyrioCode/NO|1|https://codegolf.stackexchange.com/a/115253/47066^0|0|2|0|8|3|1|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@$A|S|B|T|1|U]|$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]]|I|$J|$5|K|L|M|C|$N|O]]|P|$5|K|L|M|C|$N|Q]]]]

<h1><a href="https://github.com/ValyrioCode/NO" rel="nofollow noreferrer">NO</a>, 41 bytes, <a href="https://codegolf.stackexchange.com/a/115253/47066">This Guy</a></h1>
<pre><code>NOOOOOO?NOOOOOOOOOOO!nnOOOOO
NOOOOOOOO?no
</code></pre>


blocks|key|6396438|text|JavaScript+(ES7)，尼尔|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396439|_26_=>_26_**6.25**.5|code-block|syntax|javascript|6396440|当然，最难的是弄清楚如何处理这些额外的角色。(也不要在错误的线程中发布此解决方案，就像我一开始不小心做的那样。.)|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115127^0|H|2|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|X|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

<h1>JavaScript (ES7), <a href="https://codegolf.stackexchange.com/a/115127">Neil</a></h1>

<pre><code>_26_=&gt;_26_**6.25**.5
</code></pre>

<p>The hard part, of course, was figuring out what to do with all the extra characters.  (And also not posting this solution in the wrong thread, like I accidentally did at first.  Oopsie...)</p>


blocks|key|967294|text|Python，15字节，X码器先生|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|967295|lambda+x:x**2.5|code-block|syntax|javascript|967296|很简单。只需将x提升到2.5第四大电源即可。|unstyled|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115064/41805^0|C|5|0|0|0|7|1|B|3^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@$A|10|B|11|L|M]|$A|12|B|13|L|M]]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

<h1>Python, 15 bytes, <a href="https://codegolf.stackexchange.com/a/115064/41805">Mr. Xcoder</a></h1>
<pre><code>lambda x:x**2.5
</code></pre>
<p>Pretty simple. Just takes <code>x</code> and raises it to the <code>2.5</code>th power.</p>


blocks|key|967434|text|OCaml，13字节，苏琪|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|967435|fun+f->f**2.5|code-block|syntax|javascript|967436|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115068/55508|1|https://tio.run/nexus/ocaml#@19QlJlXEp@Wk59YoqGRVpqnkKZrl6alZaRnqqlgqaf5/z8A^0|B|2|0|0|0|0|6|1^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<h2>OCaml, 13 bytes, <a href="https://codegolf.stackexchange.com/a/115068/55508">shooqie</a></h2>

<pre><code>fun f-&gt;f**2.5
</code></pre>

<p><a href="https://tio.run/nexus/ocaml#@19QlJlXEp@Wk59YoqGRVpqnkKZrl6alZaRnqqlgqaf5/z8A" rel="nofollow noreferrer" title="OCaml – TIO Nexus">Try it online!</a></p>


blocks|key|6397847|text|Python3，44个字节，凯尔·格利恩|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6397848|那些*s非常误导人。很聪明！|unstyled|style|CODE|6397849|lambda+i:i**(lambda+o,r:o/r)(*map(ord,'i*'))|code-block|syntax|javascript|6397850|由于字符集非常有限，如果除了对参数进行简单的重命名或重新排序之外，还有其他有效的解决方案，我会感到非常惊讶。|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115855/15858^0|E|6|0|0|2|1|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@$A|11|B|12|G|H]]|9|@]|C|$]]|$1|I|3|J|5|K|7|13|8|@]|9|@]|C|$L|M]]|$1|N|3|O|5|F|7|14|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

<h1>Python 3, 44  bytes, <a href="https://codegolf.stackexchange.com/a/115855/15858">Kyle Gullion</a></h1>

<p>Those <code>*</code>s were quite misleading. Very clever!</p>



<pre class="lang-python prettyprint-override"><code>lambda i:i**(lambda o,r:o/r)(*map(ord,'i*'))
</code></pre>

<p>Due to the quite limited character set I would be very surprised if there were any other valid solutions beyond trivial renaming or reordering of arguments.</p>


blocks|key|967386|text|C%2B%2B+(gcc)，100个字节，X码器先生|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|967387|#include<math.h>
#include"iostream"
using+namespace+std;int+main(){float+n;cin>>n;cout<<pow(n,2.5);}|code-block|syntax|javascript|967388|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://gcc.gnu.org/|1|https://codegolf.stackexchange.com/a/115067/66323|2|https://tio.run/nexus/cpp-gcc#NcgxCsMwDADAPa8w6ZJAmyGQIdj4L0JRa0EshVimQ@mzO7tdOh1cu7DgXjcKGSxNKXb/6FmLnQS572pheTiBTOUAJFds8yzmMrAM4@u@K5gTjywx/tBqIRz6HOQ6T8vo362tH9EbAib6Ag^0|0|9|0|H|5|1|0|0|0|6|2^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]|$A|10|B|11|1|12]]|C|$]]|$1|D|3|E|5|F|7|13|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]|U|$5|N|O|P|C|$Q|V]]]]

<h1><a href="https://gcc.gnu.org/" rel="nofollow noreferrer">C++ (gcc)</a>, 100 bytes, <a href="https://codegolf.stackexchange.com/a/115067/66323">Mr. Xcoder</a></h1>



<pre class="lang-cpp prettyprint-override"><code>#include&lt;math.h&gt;
#include"iostream"
using namespace std;int main(){float n;cin&gt;&gt;n;cout&lt;&lt;pow(n,2.5);}
</code></pre>

<p><a href="https://tio.run/nexus/cpp-gcc#NcgxCsMwDADAPa8w6ZJAmyGQIdj4L0JRa0EshVimQ@mzO7tdOh1cu7DgXjcKGSxNKXb/6FmLnQS572pheTiBTOUAJFds8yzmMrAM4@u@K5gTjywx/tBqIRz6HOQ6T8vo362tH9EbAib6Ag" rel="nofollow noreferrer" title="C++ (gcc) – TIO Nexus">Try it online!</a></p>


blocks|key|967537|text|Haskell，狮子座|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|967538|x=exp.(2.5*).log|code-block|syntax|javascript|967539|一个名为x的无点函数。用法：x+4+->+32.0|unstyled|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115077/34531^0|8|3|0|0|0|4|1|E|3|L|4^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@$A|10|B|11|L|M]|$A|12|B|13|L|M]|$A|14|B|15|L|M]]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

<h2>Haskell, <a href="https://codegolf.stackexchange.com/a/115077/34531">Leo</a></h2>

<pre><code>x=exp.(2.5*).log
</code></pre>

<p>A pointfree function named <code>x</code>. Usage: <code>x 4</code> -> <code>32.0</code></p>


blocks|key|6396357|text|通知7，科弗斯_192|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396358|酷，一个Inform7条目。)我只需要试一试。|unstyled|6396359|我很确定这就是我们想要的解决方案：|6396360|R+is+a+room.

To+f+(n+-+number):+say+"[n+*+n+*+real+square+root+of+n]".|code-block|syntax|javascript|6396361|请注意，由于使用了real+square+root+of函数，此解决方案只有在使用Glulx后端编译时才能工作。|style|CODE|6396362|顺便说一句，双引号和方括号实际上是不必要的；只要say+n+*+n+*+real+square+root+of+n就行了。命令末尾的句点也可以省略；或者我们可以保留第一个句点，而不是换行符。我们可以修剪掉的代码的其他部分包括文章"a“前面的"room”，以及括号前和冒号后面的空格。幸运的是，由于我们有一个备用的括号，所以我们总是可以使用它们来注释掉所有这些额外的字符。)这也是一个有效的解决方案：|6396363|R+is+room.To+f(n+-+number):say+n+*+n+*+real+square+root+of+n[
"+a+.+"
]|6396364|要以交互方式测试此解决方案，可以方便地将下面的测试工具添加到代码中：|6396365|Effing+is+an+action+applying+to+one+number.
Understand+"f+[number]"+as+effing.
Carry+out+effing:+f+the+number+understood.|6396366|在编译和运行程序之后，您可以在f+4.+f+6.+f+9.+f+25提示符下输入例如>，并接收如下输出：|6396367|Welcome
An+Interactive+Fiction
Release+1+/+Serial+number+170404+/+Inform+7+build+6L38+(I6/v6.33+lib+6/12N)+SD

R

>f+4.+f+6.+f+9.+f+25
32.0
88.18164
243.0
3125.0
>|6396368|顺便说一句，我刚刚注意到，通知(或者说，更确切地说，Glulx)将f+6的最后小数位圈错了:正确的值更接近88.18163而不是88.18164。幸运的是，我不认为这会影响解决方案的正确性(S)，特别是因为这一挑战指定了“你选择的任何四舍五入机制”。:)|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Inform#Inform_7|1|https://codegolf.stackexchange.com/a/115121^0|0|3|0|4|7|1|0|0|0|0|9|J|0|O|X|0|0|0|0|F|J|16|1|0|0|X|3^^$0|@$1|2|3|4|5|6|7|1E|8|@]|9|@$A|1F|B|1G|1|1H]|$A|1I|B|1J|1|1K]]|C|$]]|$1|D|3|E|5|F|7|1L|8|@]|9|@]|C|$]]|$1|G|3|H|5|F|7|1M|8|@]|9|@]|C|$]]|$1|I|3|J|5|K|7|1N|8|@]|9|@]|C|$L|M]]|$1|N|3|O|5|F|7|1O|8|@$A|1P|B|1Q|P|Q]]|9|@]|C|$]]|$1|R|3|S|5|F|7|1R|8|@$A|1S|B|1T|P|Q]]|9|@]|C|$]]|$1|T|3|U|5|K|7|1U|8|@]|9|@]|C|$L|M]]|$1|V|3|W|5|F|7|1V|8|@]|9|@]|C|$]]|$1|X|3|Y|5|K|7|1W|8|@]|9|@]|C|$L|M]]|$1|Z|3|10|5|F|7|1X|8|@$A|1Y|B|1Z|P|Q]|$A|20|B|21|P|Q]]|9|@]|C|$]]|$1|11|3|12|5|K|7|22|8|@]|9|@]|C|$L|M]]|$1|13|3|14|5|F|7|23|8|@$A|24|B|25|P|Q]]|9|@]|C|$]]]|15|$16|$5|17|18|19|C|$1A|1B]]|1C|$5|17|18|19|C|$1A|1D]]]]

<h1><a href="https://en.wikipedia.org/wiki/Inform#Inform_7" rel="nofollow noreferrer">Inform 7</a>, <a href="https://codegolf.stackexchange.com/a/115121">corvus_192</a></h1>

<p>Cool, an Inform7 entry. :)  I just had to give this one a try.</p>

<p>I'm pretty sure this is the intended solution:</p>

<pre><code>R is a room.

To f (n - number): say "[n * n * real square root of n]".
</code></pre>

<p>Note that this solution only works if compiled with the Glulx back-end, due to the use of the <code>real square root of</code> function.</p>

<hr>

<p>BTW, the double quotes and square brackets are actually unnecessary; just <code>say n * n * real square root of n</code> would work just as well.  The periods at the end of the commands could be omitted, too; or we could keep the first period and get rid of the newlines instead.  Other parts of the code we could trim away include the article "a" before "room" and the spaces before the parentheses and after the colon.  Fortunately, since we've got a spare pair of brackets, we can always use them to comment out all these extra characters. ;) So this is a valid solution, too:</p>

<pre><code>R is room.To f(n - number):say n * n * real square root of n[
" a . "
]
</code></pre>

<hr>

<p>To test this solution interactively, it's convenient to append something like the following test harness to the code:</p>

<pre><code>Effing is an action applying to one number.
Understand "f [number]" as effing.
Carry out effing: f the number understood.
</code></pre>

<p>After compiling and running the program, you can type e.g. <code>f 4. f 6. f 9. f 25</code> at the <code>&gt;</code> prompt and receive something like the following output:</p>

<pre><code>Welcome
An Interactive Fiction
Release 1 / Serial number 170404 / Inform 7 build 6L38 (I6/v6.33 lib 6/12N) SD

R

&gt;f 4. f 6. f 9. f 25
32.0
88.18164
243.0
3125.0
&gt;
</code></pre>

<p>BTW, I just noticed that Inform (or presumably, rather, Glulx) rounds the last decimal place of <code>f 6</code> wrong: the correct value is <em>much</em> closer to 88.18163 than to 88.18164.  Fortunately, I don't think this affects the correctness of the solution(s), especially since the challenge specified "any rounding mechanism of your choice". :)</p>


blocks|key|967637|text|Mathematica，格雷格·马丁|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|967638|f[y_]:=With[{x=
++++#&@@{#(#)#%5E(1/(1%2B1))&@y,#&@@@{1%5E(1),-1}}
},Print[#,".",IntegerString[Round@#2,10,3]]&@@QuotientRemainder[1000x,1000]]|code-block|syntax|javascript|967639|谢谢你让四舍五入的东西完好无损！|unstyled|967640|说明：#(#)#%5E(1/(1%2B1))&@y做的主要工作是把y平方，也就是y(y)和y's平方根y%5E(1/(1%2B1))相乘。#&@@@{1%5E(1),-1}位只是用来使用其他字母的垃圾，#&@@从垃圾中挑选出有用的位。|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115103/66104^0|C|6|0|0|0|0|3|I|T|1|10|4|15|1|1B|B|1P|F|2J|4^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|11|8|@]|9|@]|C|$]]|$1|L|3|M|5|K|7|12|8|@$A|13|B|14|N|O]|$A|15|B|16|N|O]|$A|17|B|18|N|O]|$A|19|B|1A|N|O]|$A|1B|B|1C|N|O]|$A|1D|B|1E|N|O]|$A|1F|B|1G|N|O]]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

<h2>Mathematica, <a href="https://codegolf.stackexchange.com/a/115103/66104" title="name">Greg Martin</a></h2>

<pre><code>f[y_]:=With[{x=
    #&amp;@@{#(#)#^(1/(1+1))&amp;@y,#&amp;@@@{1^(1),-1}}
},Print[#,".",IntegerString[Round@#2,10,3]]&amp;@@QuotientRemainder[1000x,1000]]
</code></pre>

<p>Thanks for leaving the rounding stuff intact!</p>

<p>Explanation: <code>#(#)#^(1/(1+1))&amp;@y</code> does the main work of multiplying <code>y</code> squared, aka <code>y(y)</code>, and <code>y</code>'s square root, <code>y^(1/(1+1))</code>. The <code>#&amp;@@@{1^(1),-1}</code> bit is just junk to use up the other letters, and <code>#&amp;@@</code> picks out the useful bit from the junk.</p>


blocks|key|6397417|text|马蒂尔，12字节，路易斯·门多|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6397418|10'U&%2B:'n/%5EP|code-block|syntax|javascript|6397419|计算10/4+=+2.5，其中4来自字符串长度。用它作为指数。P在这里是没有行动的。|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/lmendo/MATL|1|https://codegolf.stackexchange.com/a/115071/49958^0|0|3|0|9|6|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]|$A|Y|B|Z|1|10]]|C|$]]|$1|D|3|E|5|F|7|11|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|12|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<h1><a href="https://github.com/lmendo/MATL" rel="nofollow noreferrer">MATL</a>, 12 bytes, <a href="https://codegolf.stackexchange.com/a/115071/49958">Luis Mendo</a></h1>

<pre><code>10'U&amp;+:'n/^P
</code></pre>

<p>Calculate 10/4 = 2.5 with 4 coming from string length. Use this as an exponent. P is a no-op here.</p>


blocks|key|967609|text|R，Steadybox|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|967610|y<-function(p)(p%5E(1/2)*p%5E2);|code-block|syntax|javascript|967611|似乎是funny(p1)-tio(%5E*%5E)/pc(2)<p2;的一个比喻|unstyled|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115093/26905^0|0|B|0|0|0|3|S^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@$A|10|B|11|L|M]]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

<h2><a href="https://codegolf.stackexchange.com/a/115093/26905">R, Steadybox</a></h2>
<pre><code>y&lt;-function(p)(p^(1/2)*p^2);
</code></pre>
<p>seems to be an anagram of <code>funny(p1)-tio(^*^)/pc(2)&lt;p2;</code></p>


blocks|key|6396414|text|Python+2，60个字节，安东尼·范姆|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396415|print++++(input()**(5.0/(2*5554448893999/5554448893840))-0)|code-block|syntax|javascript|6396416|基于通过Python2‘S浮点数除法丢弃字符(整数之间的/的默认设置)。|unstyled|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115118/63641^0|F|6|0|0|0|S|1^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@$A|10|B|11|L|M]]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

<h1>Python 2, 60 bytes, <a href="https://codegolf.stackexchange.com/a/115118/63641">Anthony Pham</a></h1>

<pre><code>print    (input()**(5.0/(2*5554448893999/5554448893840))-0)
</code></pre>

<p>Based on discarding characters through Python 2's float division (the default for <code>/</code> between integers).</p>


blocks|key|967666|text|C，50个字节，戴夫|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|967667|double+b(float+\ufa2d){return+pow(\ufa2d,25e-1);%25>|code-block|syntax|javascript|967668|在网上试试！|unstyled|967669|这需要-lm编译器标记，但我不知道没有它怎么可能解决这个问题。|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115130/66323|1|https://tio.run/nexus/c-gcc#U87MS84pTUlVsCkuScnM18uw@5@SX5qUk6qQpJGWk59YohBTmpZolKJZXZRaUlqUp1CQX64BEdIxMk3VNdS0VrX7n5lXopCbmJmnoalQzcVZUATkp2koqabF5CnpAA0yMtXUtOaq/f8vOS0nMb34v25OLgA^0|8|2|0|0|0|0|6|1|0|3|3^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|13|8|@]|9|@$A|14|B|15|1|16]]|C|$]]|$1|L|3|M|5|K|7|17|8|@$A|18|B|19|N|O]]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

<h1>C, 50 bytes, <a href="https://codegolf.stackexchange.com/a/115130/66323">Dave</a></h1>



<pre class="lang-c prettyprint-override"><code>double b(float \ufa2d){return pow(\ufa2d,25e-1);%&gt;
</code></pre>

<p><a href="https://tio.run/nexus/c-gcc#U87MS84pTUlVsCkuScnM18uw@5@SX5qUk6qQpJGWk59YohBTmpZolKJZXZRaUlqUp1CQX64BEdIxMk3VNdS0VrX7n5lXopCbmJmnoalQzcVZUATkp2koqabF5CnpAA0yMtXUtOaq/f8vOS0nMb34v25OLgA" rel="nofollow noreferrer" title="C (gcc) – TIO Nexus">Try it online!</a></p>

<p>This requires <code>-lm</code> compiler flag, but I don't know how it would be possible to solve this without it.</p>


blocks|key|6396773|text|R，比目鱼|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396774|这是@Flounderer+33字节解决方案中的一个漏洞。|unstyled|6396775|scan()%5E(floor(pi)-1/2)-sin(7*0e1)|code-block|syntax|javascript|6396776|用法：|6396777|>+scan()%5E(floor(pi)-1/2)-sin(7*0e1)
1:+4
2:+
Read+1+item
[1]+32
>+scan()%5E(floor(pi)-1/2)-sin(7*0e1)
1:+6
2:+
Read+1+item
[1]+88.18163
>+scan()%5E(floor(pi)-1/2)-sin(7*0e1)
1:+9
2:+
Read+1+item
[1]+243
>+scan()%5E(floor(pi)-1/2)-sin(7*0e1)
1:+25
2:+
Read+1+item
[1]+3125|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115133/6741^0|2|3|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$]]|$1|G|3|H|5|I|7|11|8|@]|9|@]|C|$J|K]]|$1|L|3|M|5|F|7|12|8|@]|9|@]|C|$]]|$1|N|3|O|5|I|7|13|8|@]|9|@]|C|$J|K]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

<h1>R, <a href="https://codegolf.stackexchange.com/a/115133/6741">Flounderer</a></h1>
<p>This is a crack of the 33 bytes solution of @Flounderer</p>
<pre><code>scan()^(floor(pi)-1/2)-sin(7*0e1)
</code></pre>
<p>Usage:</p>
<pre><code>&gt; scan()^(floor(pi)-1/2)-sin(7*0e1)
1: 4
2: 
Read 1 item
[1] 32
&gt; scan()^(floor(pi)-1/2)-sin(7*0e1)
1: 6
2: 
Read 1 item
[1] 88.18163
&gt; scan()^(floor(pi)-1/2)-sin(7*0e1)
1: 9
2: 
Read 1 item
[1] 243
&gt; scan()^(floor(pi)-1/2)-sin(7*0e1)
1: 25
2: 
Read 1 item
[1] 3125
</code></pre>


blocks|key|968270|text|EXCEL，26位字节巴约克|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968271|=SQRT(A1)*A1%5E2/1/ISNA(IP2)|code-block|syntax|javascript|968272|A1作为输入IP2包含一个#N/A错误的第二个输入，在本例中ISNA(IP2)属于1|unstyled|968273|对于额外的()，我们可以这样做|style|CODE|968274|=SQRT(A1)*A1%5E2/ISNA(PI(1/2))|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/questions/115036/cops-square-times-square-root/115233#115233^0|B|3|0|0|0|0|5|2|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|13|8|@]|9|@]|C|$]]|$1|L|3|M|5|K|7|14|8|@$A|15|B|16|N|O]]|9|@]|C|$]]|$1|P|3|Q|5|F|7|17|8|@]|9|@]|C|$G|H]]]|R|$S|$5|T|U|V|C|$W|X]]]]

<h1>EXCEL, 26 Bytes <a href="https://codegolf.stackexchange.com/questions/115036/cops-square-times-square-root/115233#115233">pajonk</a></h1>

<pre><code>=SQRT(A1)*A1^2/1/ISNA(IP2)
</code></pre>

<p>A1 as input
IP2 contain a second input with a #N/A Error
in this case ISNA(IP2) belongs to 1</p>

<p>For an additional <code>()</code>
we can do this</p>

<pre><code>=SQRT(A1)*A1^2/ISNA(PI(1/2))
</code></pre>


blocks|key|968330|text|RProgN+2，ATaco|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968331|]2%5E\š*|code-block|syntax|javascript|968332|显然，StackExchange需要额外的角色，所以给你。|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115120/63641^0|9|5|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|X|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

<h1>RProgN 2, <a href="https://codegolf.stackexchange.com/a/115120/63641">ATaco</a></h1>
<pre><code>]2^\š*
</code></pre>
<p>Apparently StackExchange needs extra characters, so here you go.</p>


blocks|key|968384|text|HODOR，198岁，这家伙|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968385|Walder
Hodor?!
hodor.
Hodor,+Hodor+Hodor+Hodor+Hodor+Hodor+Hodor+Hodor,+Hodor+Hodor,
hodor,
Hodor,+Hodor+Hodor+Hodor+Hodor+Hodor+Hodor,+Hodor+Hodor,
Hodor,+Hodor+Hodor+Hodor,+hodor!,
HODOR!!
HODOR!!!|code-block|syntax|javascript|968386|解释：|unstyled|968387|Start
read+input+into+accumulator
copy+accumulator+to+storage
Do+math,+option+7(nth+root),+n=2
swap+storage+and+accumulator
Do+math,+option+6(nth+power),+n=2
Do+math,+option+3(times),+storage
output+accumulator+as+a+number
end|968388|注意:我必须对get解释器进行一些修改才能在我的机器上运行(您发布的那个似乎不接受小写h，以及其他一些东西)|968389|而且，我似乎没有足够的代表发表评论，所以如果有人能让这个家伙知道的话，我将非常感激|968390|我认为这修复了错误，代码现在以Walder开始，而不是Wylis，后者增加了额外的字节。|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115262/67716^0|B|3|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|15|8|@]|9|@]|C|$]]|$1|L|3|M|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|N|3|O|5|K|7|17|8|@]|9|@]|C|$]]|$1|P|3|Q|5|K|7|18|8|@]|9|@]|C|$]]|$1|R|3|S|5|K|7|19|8|@]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]]]

<h1>HODOR, 198, <a href="https://codegolf.stackexchange.com/a/115262/67716">This Guy</a></h1>
<pre><code>Walder
Hodor?!
hodor.
Hodor, Hodor Hodor Hodor Hodor Hodor Hodor Hodor, Hodor Hodor,
hodor,
Hodor, Hodor Hodor Hodor Hodor Hodor Hodor, Hodor Hodor,
Hodor, Hodor Hodor Hodor, hodor!,
HODOR!!
HODOR!!!
</code></pre>
<p>Explanation:</p>
<pre><code>Start
read input into accumulator
copy accumulator to storage
Do math, option 7(nth root), n=2
swap storage and accumulator
Do math, option 6(nth power), n=2
Do math, option 3(times), storage
output accumulator as a number
end
</code></pre>
<p>note: I had to make some modifications to the get interpreter to run on my machine (the one you have posted doesn't seem to accept lowercase h, among some other things)</p>
<p>Also, I don't seem to have enough rep to comment, so if someone could let @This Guy know, I would be grateful</p>
<p>I think this fixed the error, the code now starts with Walder instead of Wylis, which adds the extra byte</p>


blocks|key|6397133|text|C#，172个字节，剃须刀|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6397134|最难的部分是弄清楚如何处理所有的剩菜。|unstyled|6397135|using+System;using+S=System.Console;class+PMabddellorttuuv{static+void+Main(){S.Write(Math.Pow(double.Parse(S.ReadLine()),2.5));Func<double>+o;int+q=1,M=q*2,b,e;q*=(q*M);}}|code-block|syntax|javascript|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115069/32628^0|A|3|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$]]|$1|G|3|H|5|I|7|X|8|@]|9|@]|C|$J|K]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

<h2>C#, 172 bytes, <a href="https://codegolf.stackexchange.com/a/115069/32628">raznagul</a></h2>
<p>Hardest part was figuring out what to do with all the leftovers.</p>
<pre><code>using System;using S=System.Console;class PMabddellorttuuv{static void Main(){S.Write(Math.Pow(double.Parse(S.ReadLine()),2.5));Func&lt;double&gt; o;int q=1,M=q*2,b,e;q*=(q*M);}}
</code></pre>


blocks|key|968635|text|Python3.6，64个字节，X码器先生|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968636|php38af4r2aoot2srm0itpfpmm0726991i=+++++(lambda+x:x**2.5*1*1/1);|code-block|syntax|javascript|968637|也许不是本意，但有效;)|unstyled|968638|$+python3
Python+3.6.1+(default,+Apr++4+2017,+09:36:47)+
[GCC+4.2.1+Compatible+Apple+LLVM+7.0.2+(clang-700.1.81)]+on+darwin
Type+"help",+"copyright",+"credits"+or+"license"+for+more+information.
>>>+php38af4r2aoot2srm0itpfpmm0726991i=+++++(lambda+x:x**2.5*1*1/1);
>>>+php38af4r2aoot2srm0itpfpmm0726991i(6)
88.18163074019441
>>>+php38af4r2aoot2srm0itpfpmm0726991i(4)
32.0
>>>+php38af4r2aoot2srm0itpfpmm0726991i(25)
3125.0|968639|还没有足够的代表来评论警察的答案，抱歉.如果有人能帮我，我会很感激的，谢谢！|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115339/44706^0|G|5|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|11|8|@]|9|@]|C|$]]|$1|L|3|M|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|N|3|O|5|K|7|13|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]]]

<h2>Python 3.6, 64 bytes, <a href="https://codegolf.stackexchange.com/a/115339/44706">Mr. Xcoder</a></h2>
<pre><code>php38af4r2aoot2srm0itpfpmm0726991i=     (lambda x:x**2.5*1*1/1);
</code></pre>
<p>Maybe not what was intended, but works ;)</p>
<pre><code>$ python3
Python 3.6.1 (default, Apr  4 2017, 09:36:47) 
[GCC 4.2.1 Compatible Apple LLVM 7.0.2 (clang-700.1.81)] on darwin
Type &quot;help&quot;, &quot;copyright&quot;, &quot;credits&quot; or &quot;license&quot; for more information.
&gt;&gt;&gt; php38af4r2aoot2srm0itpfpmm0726991i=     (lambda x:x**2.5*1*1/1);
&gt;&gt;&gt; php38af4r2aoot2srm0itpfpmm0726991i(6)
88.18163074019441
&gt;&gt;&gt; php38af4r2aoot2srm0itpfpmm0726991i(4)
32.0
&gt;&gt;&gt; php38af4r2aoot2srm0itpfpmm0726991i(25)
3125.0
</code></pre>
<p>Not enough rep to comment on the cops' thread's answer yet, sorry... Would appreciate if someone could do it for me, thanks!</p>


blocks|key|968657|text|罗达，28字节，费古斯克|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968658|f(n){push+n%5E0.5,n%5E2%7Cproduct}|code-block|syntax|javascript|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/fergusq/roda/tree/roda-0.12|1|https://codegolf.stackexchange.com/a/115053/49958^0|0|2|0|8|4|1|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@$A|S|B|T|1|U]|$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]]|I|$J|$5|K|L|M|C|$N|O]]|P|$5|K|L|M|C|$N|Q]]]]

<h1><a href="https://github.com/fergusq/roda/tree/roda-0.12" rel="nofollow noreferrer">Röda</a>, 28 bytes, <a href="https://codegolf.stackexchange.com/a/115053/49958">fergusq</a></h1>

<pre><code>f(n){push n^0.5,n^2|product}
</code></pre>


blocks|key|968854|text|Python2.7，Koishore+Roy|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968855|s=e=x=y=input()**0.5
print'%25.3f'%25(y**(5.0))|code-block|syntax|javascript|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115778/64121^0|A|C|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|C|$G|H]]]|I|$J|$5|K|L|M|C|$N|O]]]]

<h1>Python 2.7, <a href="https://codegolf.stackexchange.com/a/115778/64121">Koishore Roy</a></h1>

<pre><code>s=e=x=y=input()**0.5
print'%.3f'%(y**(5.0))
</code></pre>


blocks|key|6397615|text|R，比目鱼|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6397616|这是@Flounderer的31字节解决方案中的一个漏洞：|unstyled|6397617|`[.`=function(`]`)`]`%5E`[`(lh,9)|code-block|syntax|javascript|6397618|好吧，这是个艰难的决定。它创建一个名为`[.`的函数。这一功能的论证被称为`]`，通过使用内置的time-serie+lh的第9个元素，它被提升到了2.5的功率(“一个规则的时间序列，将血液样本中的黄体生成素以10分钟的间隔从一个人类女性身上得到，48个样本)”。在R的一个基本包中用作示例)。lh[9]在此之上，它被其等效的`[`(lh,+9)所取代。将f替换为函数名，将n替换为参数名，从而消除混淆，然后函数变为f=function(n)n%5Elh[9]。|style|CODE|6397619|用法：|6397620|>+`[.`=function(`]`)`]`%5E`[`(lh,9)
>+`[.`(4)
[1]+32
>+`[.`(6)
[1]+88.18163
>+`[.`(9)
[1]+243
>+`[.`(25)
[1]+3125|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115133/6741^0|2|3|0|0|0|0|J|4|11|3|1N|2|43|5|4J|A|4Y|1|57|1|5S|K|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$]]|$1|G|3|H|5|I|7|15|8|@]|9|@]|C|$J|K]]|$1|L|3|M|5|F|7|16|8|@$A|17|B|18|N|O]|$A|19|B|1A|N|O]|$A|1B|B|1C|N|O]|$A|1D|B|1E|N|O]|$A|1F|B|1G|N|O]|$A|1H|B|1I|N|O]|$A|1J|B|1K|N|O]|$A|1L|B|1M|N|O]]|9|@]|C|$]]|$1|P|3|Q|5|F|7|1N|8|@]|9|@]|C|$]]|$1|R|3|S|5|I|7|1O|8|@]|9|@]|C|$J|K]]]|T|$U|$5|V|W|X|C|$Y|Z]]]]

<h1>R, <a href="https://codegolf.stackexchange.com/a/115133/6741">Flounderer</a></h1>
<p>This is a crack of @Flounderer's 31-byte solution:</p>
<pre><code>`[.`=function(`]`)`]`^`[`(lh,9)
</code></pre>
<p>Ok that was a tough one. It creates a function called <code>`[.`</code>. The argument to the function is called <code>`]`</code> which is elevated to power 2.5 by using the 9th element of the built-in time-serie <code>lh</code> (&quot;a regular time series giving the luteinizing hormone in blood samples at 10 mins intervals from a human female, 48 samples.&quot; that is used as example in one of R's base packages). <code>lh[9]</code> is here on top of it replaced by its equivalent <code>`[`(lh, 9)</code>. De-obfuscated by substituting <code>f</code> for the function name and <code>n</code> for the argument name, the function then becomes <code>f=function(n)n^lh[9]</code>.</p>
<p>Usage:</p>
<pre><code>&gt; `[.`=function(`]`)`]`^`[`(lh,9)
&gt; `[.`(4)
[1] 32
&gt; `[.`(6)
[1] 88.18163
&gt; `[.`(9)
[1] 243
&gt; `[.`(25)
[1] 3125
</code></pre>


blocks|key|967410|text|Python2，44个字节，安东尼·范姆|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|967411|print+int(raw_input())**(0%2B000000000000.5*5)|code-block|syntax|javascript|967412|从raw_input获取输入，转换为int并将其提升到power+2.5|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115074/64505^0|E|6|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|X|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

<h1>Python 2, 44 bytes, <a href="https://codegolf.stackexchange.com/a/115074/64505">Anthony Pham</a></h1>

<pre><code>print int(raw_input())**(0+000000000000.5*5)
</code></pre>

<p>Takes input from raw_input, converts to int and raises to power 2.5</p>


blocks|key|6396287|text|JavaScript，fəˈnɛtɪk|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396288|n=>n**("ggggg".length*2**(-"g".length))//+""((((((()))))))***,-...;;=====>Seeeeegggghhhhhhhhhilllnnnnnnorrrsstttttttttttu{}|code-block|syntax|javascript|6396289|获取5/2到5乘以2的负第一次幂，其中5和1是从字符串长度接收的。从某种意义上说，通过注释掉多余的字符，就可以轻松地走出困境。|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/questions/115036/cops-square-times-square-root/115058#115058^0|B|8|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|X|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

<h1>JavaScript, <a href="https://codegolf.stackexchange.com/questions/115036/cops-square-times-square-root/115058#115058"><strong>fəˈnɛtɪk</strong></a></h1>
<pre class="lang-js prettyprint-override"><code>n=&gt;n**(&quot;ggggg&quot;.length*2**(-&quot;g&quot;.length))// &quot;&quot;((((((()))))))***,-...;;=====&gt;Seeeeegggghhhhhhhhhilllnnnnnnorrrsstttttttttttu{}
</code></pre>
<p>Gets 5/2 through 5 times 2 to the negative first power, where 5 and 1 were received from the length of strings. Took the easy way out in a sense by commenting out the extraneous characters.</p>


blocks|key|6396511|text|C#，112个字节，扬·伊凡|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396512|using+System;class+P{static+void+Main(){var+b=Math.Pow(double.Parse(Console.ReadLine()),2.5);Console.Write(b);}}|code-block|syntax|javascript|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115174/47066^0|A|4|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|C|$G|H]]]|I|$J|$5|K|L|M|C|$N|O]]]]

<h2>C#, 112 bytes, <a href="https://codegolf.stackexchange.com/a/115174/47066">Jan Ivan</a></h2>
<pre><code>using System;class P{static void Main(){var b=Math.Pow(double.Parse(Console.ReadLine()),2.5);Console.Write(b);}}
</code></pre>


blocks|key|6396544|text|火球，8字节，奥克斯|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396545|♥²♥1Z/%5E*|code-block|syntax|javascript|6396546|解释：|unstyled|6396547|♥²♥1Z/%5E*
♥²+++++++Push+first+input+squared.
++♥++++++Push+first+input+again.
+++1Z/+++Push+1/2
++++++%5E++First+input+to+the+1/2th
+++++++*+Multiply+square+and+root|6396548|不确定它是否有效。我的笔记本电脑上目前没有java。:(|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/okx-code/Fireball|1|https://codegolf.stackexchange.com/a/115196/56341^0|0|2|0|7|3|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]|$A|12|B|13|1|14]]|C|$]]|$1|D|3|E|5|F|7|15|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|16|8|@]|9|@]|C|$]]|$1|L|3|M|5|F|7|17|8|@]|9|@]|C|$G|H]]|$1|N|3|O|5|K|7|18|8|@]|9|@]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]]]

<h2><a href="https://github.com/okx-code/Fireball" rel="nofollow noreferrer">Fireball</a>, 8 bytes, <a href="https://codegolf.stackexchange.com/a/115196/56341">Okx</a></h2>
<pre><code>♥²♥1Z/^*
</code></pre>
<p>Explanation:</p>
<pre><code>♥²♥1Z/^*
♥²       Push first input squared.
  ♥      Push first input again.
   1Z/   Push 1/2
      ^  First input to the 1/2th
       * Multiply square and root
</code></pre>
<p>Not sure if it works. I have currently no java on my laptop. :(</p>


blocks|key|6396570|text|05AB1E，47字节，奥克斯|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396571|).2555BFHIJJKKKPQRS``„cg…ghi…lsw…x}T…Áöž«‚¹n¹t*|code-block|syntax|javascript|6396572|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Adriandmen/05AB1E|1|https://codegolf.stackexchange.com/a/115194/47066|2|https://tio.run/nexus/05ab1e#@6@pZ2Rqaurk5uHp5eXt7R0QGBSckPCoYV5y@qOGZekZmUAyp7gcSFbUhgDJw42Htx3dd2j1o4ZZh3bmHdpZovX/vxkA^0|0|6|0|C|3|1|0|0|0|6|2^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]|$A|10|B|11|1|12]]|C|$]]|$1|D|3|E|5|F|7|13|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]|U|$5|N|O|P|C|$Q|V]]]]

<h1><a href="https://github.com/Adriandmen/05AB1E" rel="nofollow noreferrer">05AB1E</a>, 47 bytes, <a href="https://codegolf.stackexchange.com/a/115194/47066">Okx</a></h1>

<pre><code>).2555BFHIJJKKKPQRS``„cg…ghi…lsw…x}T…Áöž«‚¹n¹t*
</code></pre>

<p><a href="https://tio.run/nexus/05ab1e#@6@pZ2Rqaurk5uHp5eXt7R0QGBSckPCoYV5y@qOGZekZmUAyp7gcSFbUhgDJw42Htx3dd2j1o4ZZh3bmHdpZovX/vxkA" rel="nofollow noreferrer" title="05AB1E – TIO Nexus">Try it online!</a></p>


blocks|key|967888|text|CJam，8字节，罗曼·格拉夫|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|967889|我的第一个CJam回答。|unstyled|967890|967891|CnR很适合测试新语言:)|967892|ldYYW#%2B#|code-block|syntax|javascript|967893|在网上试试！|967894|解释|967895|ld+++++++++%25+double(input)
+++++++#+++%25++++++++++++++%5E
++Y++++++++%25+++++++++++++++(2
++++++%2B++++%25+++++++++++++++++%2B
+++Y+++++++%25++++++++++++++++++2
+++++#+++++%25+++++++++++++++++++%5E
++++W++++++%25++++++++++++++++++++-1)|entityMap|0|LINK|mutability|MUTABLE|url|https://sourceforge.net/p/cjam|1|https://codegolf.stackexchange.com/a/115195/47066|2|https://tio.run/nexus/cjam#@5@TEhkZrqyt/P@/GQA^0|0|4|0|9|6|1|0|0|0|0|0|0|6|2|0|0^^$0|@$1|2|3|4|5|6|7|15|8|@]|9|@$A|16|B|17|1|18]|$A|19|B|1A|1|1B]]|C|$]]|$1|D|3|E|5|F|7|1C|8|@]|9|@]|C|$]]|$1|G|3|-4|5|F|7|1D|8|@]|9|@]|C|$]]|$1|H|3|I|5|F|7|1E|8|@]|9|@]|C|$]]|$1|J|3|K|5|L|7|1F|8|@]|9|@]|C|$M|N]]|$1|O|3|P|5|F|7|1G|8|@]|9|@$A|1H|B|1I|1|1J]]|C|$]]|$1|Q|3|R|5|6|7|1K|8|@]|9|@]|C|$]]|$1|S|3|T|5|L|7|1L|8|@]|9|@]|C|$M|N]]]|U|$V|$5|W|X|Y|C|$Z|10]]|11|$5|W|X|Y|C|$Z|12]]|13|$5|W|X|Y|C|$Z|14]]]]

<h1><a href="https://sourceforge.net/p/cjam" rel="nofollow noreferrer">CJam</a>, 8 bytes, <a href="https://codegolf.stackexchange.com/a/115195/47066">Roman Gräf</a></h1>

<p>My First ever CJam answer.<br>
CnR are great for testing new languages :)</p>

<pre><code>ldYYW#+#
</code></pre>

<p><a href="https://tio.run/nexus/cjam#@5@TEhkZrqyt/P@/GQA" rel="nofollow noreferrer" title="CJam – TIO Nexus">Try it online!</a></p>

<p><strong>Explanation</strong></p>

<pre><code>ld         % double(input)
       #   %              ^
  Y        %               (2
      +    %                 +
   Y       %                  2
     #     %                   ^
    W      %                    -1)
</code></pre>


blocks|key|968086|text|R，稳定箱|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968087|a222=function(s)(s**0.5)*s**2**1|code-block|syntax|javascript|968088|用法：|unstyled|968089|>+a222=function(s)(s**0.5)*s**2**1
>+a222(4)
[1]+32
>+a222(6)
[1]+88.18163
>+a222(9)
[1]+243
>+a222(25)
[1]+3125|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115224/6741^0|2|3|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@]|9|@]|C|$]]|$1|L|3|M|5|F|7|10|8|@]|9|@]|C|$G|H]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

<h1>R, <a href="https://codegolf.stackexchange.com/a/115224/6741">steadybox</a></h1>

<pre><code>a222=function(s)(s**0.5)*s**2**1
</code></pre>

<p>Usage:</p>

<pre><code>&gt; a222=function(s)(s**0.5)*s**2**1
&gt; a222(4)
[1] 32
&gt; a222(6)
[1] 88.18163
&gt; a222(9)
[1] 243
&gt; a222(25)
[1] 3125
</code></pre>


blocks|key|968190|text|05AB1E，22字节，P.+Knops|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968191|n¹t*qA9¥="'?:@->%25#[{!.|code-block|syntax|javascript|968192|在网上试试！|unstyled|968193|解释|968194|n++++++#+square+of+input
+++*+++#+times
+¹t++++#+square+root+of+input
++++q++#+end+program|968195|其余的操作永远不会被执行。|968196|968197|我们也可以在没有q的情况下完成这一任务，方法是在计算之后让?退出，例如用'=转义相等号。|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Adriandmen/05AB1E|1|https://codegolf.stackexchange.com/a/115240/47066|2|https://tio.run/nexus/05ab1e#@593aGeJVqGj5aGltkrq9lYOunaqytHVinr//5sBAA^0|0|6|0|C|8|1|0|0|0|6|2|0|0|0|0|0|8|1|T|1|10|2^^$0|@$1|2|3|4|5|6|7|17|8|@]|9|@$A|18|B|19|1|1A]|$A|1B|B|1C|1|1D]]|C|$]]|$1|D|3|E|5|F|7|1E|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1F|8|@]|9|@$A|1G|B|1H|1|1I]]|C|$]]|$1|L|3|M|5|6|7|1J|8|@]|9|@]|C|$]]|$1|N|3|O|5|F|7|1K|8|@]|9|@]|C|$G|H]]|$1|P|3|Q|5|K|7|1L|8|@]|9|@]|C|$]]|$1|R|3|-4|5|K|7|1M|8|@]|9|@]|C|$]]|$1|S|3|T|5|K|7|1N|8|@$A|1O|B|1P|U|V]|$A|1Q|B|1R|U|V]|$A|1S|B|1T|U|V]]|9|@]|C|$]]]|W|$X|$5|Y|Z|10|C|$11|12]]|13|$5|Y|Z|10|C|$11|14]]|15|$5|Y|Z|10|C|$11|16]]]]

<h1><a href="https://github.com/Adriandmen/05AB1E" rel="nofollow noreferrer">05AB1E</a>, 22 bytes, <a href="https://codegolf.stackexchange.com/a/115240/47066">P. Knops</a></h1>

<pre><code>n¹t*qA9¥="'?:@-&gt;%#[{!.
</code></pre>

<p><a href="https://tio.run/nexus/05ab1e#@593aGeJVqGj5aGltkrq9lYOunaqytHVinr//5sBAA" rel="nofollow noreferrer" title="05AB1E – TIO Nexus">Try it online!</a></p>

<p><strong>Explanation</strong></p>

<pre><code>n      # square of input
   *   # times
 ¹t    # square root of input
    q  # end program
</code></pre>

<p>The rest of the operations never get executed.<br>
We could have done it without the <code>q</code> as well by having <code>?</code> after the calculation and escaping the equality sign for example with <code>'=</code>.</p>


blocks|key|968434|text|霍多尔，171个字节，wwj|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968435|Walder
Hodor+
Hodor+
Hodor+
Hodor+
Hodor
Hodor,+Hodor+Hodor+Hodor+Hodor,+Hodor+Hodor,
hodor.
Hodor.
Hodor?!
Hodor,+Hodor+Hodor+Hodor+Hodor+Hodor+Hodor,+hodor!,
HODOR!!
HODOR!!!|code-block|syntax|javascript|968436|解释|968437|增量累加器到5|unordered-list-item|968438|累加器除以2|968439|保存累加器值的副本|968440|将累加器设置为0|968441|读输入|968442|将输入提高到存储副本(2.5)的值|968443|输出为数字|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/ValyrioCode/Hodor/|1|https://codegolf.stackexchange.com/a/115293/47066^0|0|3|0|B|3|1|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@$A|19|B|1A|1|1B]|$A|1C|B|1D|1|1E]]|C|$]]|$1|D|3|E|5|F|7|1F|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1G|8|@]|9|@]|C|$]]|$1|K|3|L|5|M|7|1H|8|@]|9|@]|C|$]]|$1|N|3|O|5|M|7|1I|8|@]|9|@]|C|$]]|$1|P|3|Q|5|M|7|1J|8|@]|9|@]|C|$]]|$1|R|3|S|5|M|7|1K|8|@]|9|@]|C|$]]|$1|T|3|U|5|M|7|1L|8|@]|9|@]|C|$]]|$1|V|3|W|5|M|7|1M|8|@]|9|@]|C|$]]|$1|X|3|Y|5|M|7|1N|8|@]|9|@]|C|$]]]|Z|$10|$5|11|12|13|C|$14|15]]|16|$5|11|12|13|C|$14|17]]]]

<h1><a href="https://github.com/ValyrioCode/Hodor/" rel="nofollow noreferrer">HODOR</a>, 171 bytes, <a href="https://codegolf.stackexchange.com/a/115293/47066">wwj</a></h1>
<pre><code>Walder
Hodor 
Hodor 
Hodor 
Hodor 
Hodor
Hodor, Hodor Hodor Hodor Hodor, Hodor Hodor,
hodor.
Hodor.
Hodor?!
Hodor, Hodor Hodor Hodor Hodor Hodor Hodor, hodor!,
HODOR!!
HODOR!!!
</code></pre>
<p><strong>Explanation</strong></p>
<ul>
<li>Increment accumulator to 5</li>
<li>Divide accumulator by 2</li>
<li>Save a copy of accumulator value</li>
<li>Set accumulator to 0</li>
<li>Read input</li>
<li>Raise input to the value of the stored copy (2.5)</li>
<li>Output as number</li>
</ul>


blocks|key|6397191|text|JavaScript，39字节，SLuck49|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6397192|x=>(l=Math).cbrt(l.exp(l.log(x)*7.5))|code-block|syntax|javascript|6397193|或者类似的|unstyled|6397194|x=>(l=Math).exp(l.log(l.cbrt(x))*7.5)|6397195|测试：|6397196|alert((

x=>(l=Math).cbrt(l.exp(l.log(x)*7.5))

)(prompt()));|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115252/12474^0|G|7|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|13|8|@]|9|@]|C|$]]|$1|L|3|M|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|N|3|O|5|K|7|15|8|@]|9|@]|C|$]]|$1|P|3|Q|5|F|7|16|8|@]|9|@]|C|$G|H]]]|R|$S|$5|T|U|V|C|$W|X]]]]

<h1>JavaScript, 39 bytes, <a href="https://codegolf.stackexchange.com/a/115252/12474">SLuck49</a></h1>

<pre><code>x=&gt;(l=Math).cbrt(l.exp(l.log(x)*7.5))
</code></pre>

<p>or similarly</p>

<pre><code>x=&gt;(l=Math).exp(l.log(l.cbrt(x))*7.5)
</code></pre>

<p>Test:</p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>alert((

x=&gt;(l=Math).cbrt(l.exp(l.log(x)*7.5))

)(prompt()));</code></pre>
</div>
</div>
</p>


blocks|key|968694|text|C，115个字节，user4867444|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968695|适用于0的新解决方案。|unstyled|968696|o(double+n){double+l=1;;for(double+o=0.;o<=8*8*2e3&&(long)o%2Bn*((long)2*3>>1);o%2B%2B)l=(l%2Bn/l)/2;printf("%25.3f",n*n*l);}|code-block|syntax|javascript|968697|不适用于0的旧解决方案：|968698|o(double+n){double+l=n;;for(double+o=1.0;o<=8*8*2e%2B3&&(long)o*((long)2*3>>1);o%2B%2B)l=(l%2Bn/l)/2;printf("%25.3f",n*n*l);}|968699|用牛顿法(对于平方根)，一个非常大的固定次数，似乎为在所需范围内的数字提供了足够的精度。我在前两个for循环子句中浪费了一些字符。|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115287/49958^0|9|B|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$]]|$1|G|3|H|5|I|7|13|8|@]|9|@]|C|$J|K]]|$1|L|3|M|5|F|7|14|8|@]|9|@]|C|$]]|$1|N|3|O|5|I|7|15|8|@]|9|@]|C|$J|K]]|$1|P|3|Q|5|F|7|16|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

<h1>C, 115 bytes, <a href="https://codegolf.stackexchange.com/a/115287/49958">user4867444</a></h1>
<p>New solution that works for 0.</p>
<pre><code>o(double n){double l=1;;for(double o=0.;o&lt;=8*8*2e3&amp;&amp;(long)o+n*((long)2*3&gt;&gt;1);o++)l=(l+n/l)/2;printf(&quot;%.3f&quot;,n*n*l);}
</code></pre>
<p>Old solution that doesn't work for 0:</p>
<pre><code>o(double n){double l=n;;for(double o=1.0;o&lt;=8*8*2e+3&amp;&amp;(long)o*((long)2*3&gt;&gt;1);o++)l=(l+n/l)/2;printf(&quot;%.3f&quot;,n*n*l);}
</code></pre>
<p>Applying Newton's method (for square root) a very large fixed number of times seems to give adequate precision for numbers in the required range. I did some character wasting in the first two for loop clauses.</p>


blocks|key|968772|text|05AB1E，15字节，user4867444|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968773|DMžDFD¹s/%2B;}01P|code-block|syntax|javascript|968774|在网上试试！|unstyled|968775|解释|968776|D+++++++++++++++++#+duplicate+input
+M++++++++++++++++#+push+the+largest+value+on+the+stack+(the+input)
++žDF+++++++++++++#+4096+times+do:
+++++D++++++++++++#+current+value
+++++++++%2B++++++++#+added+to
++++++¹s/+++++++++#+input+divided+by+current+value
++++++++++;+++++++#+divide+sum+by+2
+++++++++++}++++++#+end+loop
++++++++++++01++++#+push+1
++++++++++++++P+++#+product+of+the+stack|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/Adriandmen/05AB1E|1|https://codegolf.stackexchange.com/a/115613/47066|2|https://tio.run/nexus/05ab1e#@@/ie3Sfi5vLoZ3F@trWtQaGAf//mxsaGwIA^0|0|6|0|C|B|1|0|0|0|6|2|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]|$A|14|B|15|1|16]]|C|$]]|$1|D|3|E|5|F|7|17|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|18|8|@]|9|@$A|19|B|1A|1|1B]]|C|$]]|$1|L|3|M|5|6|7|1C|8|@]|9|@]|C|$]]|$1|N|3|O|5|F|7|1D|8|@]|9|@]|C|$G|H]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|R|S|T|C|$U|X]]|Y|$5|R|S|T|C|$U|Z]]]]

<h1><a href="https://github.com/Adriandmen/05AB1E" rel="nofollow noreferrer">05AB1E</a>, 15 bytes, <a href="https://codegolf.stackexchange.com/a/115613/47066">user4867444</a></h1>

<pre><code>DMžDFD¹s/+;}01P
</code></pre>

<p><a href="https://tio.run/nexus/05ab1e#@@/ie3Sfi5vLoZ3F@trWtQaGAf//mxsaGwIA" rel="nofollow noreferrer" title="05AB1E – TIO Nexus">Try it online!</a></p>

<p><strong>Explanation</strong></p>

<pre><code>D                 # duplicate input
 M                # push the largest value on the stack (the input)
  žDF             # 4096 times do:
     D            # current value
         +        # added to
      ¹s/         # input divided by current value
          ;       # divide sum by 2
           }      # end loop
            01    # push 1
              P   # product of the stack
</code></pre>


blocks|key|969239|text|JavaScript，SIGSEGV|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|969240|x=>x**(25*10**-1)|code-block|syntax|javascript|969241|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://nodejs.org/|1|https://codegolf.stackexchange.com/a/116225/47066|2|https://tio.run/nexus/javascript-node#@59mW2FrV6GlpWFkqmVooKWla6j5Pzk/rzg/J1UvJz9dI03DTFPzPwA^0|0|A|0|B|7|1|0|0|0|6|2^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]|$A|10|B|11|1|12]]|C|$]]|$1|D|3|E|5|F|7|13|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]|U|$5|N|O|P|C|$Q|V]]]]

<h1><a href="https://nodejs.org" rel="nofollow noreferrer">JavaScript</a>, <a href="https://codegolf.stackexchange.com/a/116225/47066">SIGSEGV</a></h1>



<pre class="lang-javascript prettyprint-override"><code>x=&gt;x**(25*10**-1)
</code></pre>

<p><a href="https://tio.run/nexus/javascript-node#@59mW2FrV6GlpWFkqmVooKWla6j5Pzk/rzg/J1UvJz9dI03DTFPzPwA" rel="nofollow noreferrer" title="JavaScript (Node.js) – TIO Nexus">Try it online!</a></p>


blocks|key|967454|text|Python+-+SparklePony|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|967455|import+math
f=lambda+x:x*x*math.sqrt(x)|code-block|syntax|javascript|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/questions/115036/cops-square-times-square-root/115080#115080^0|9|B|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|C|$G|H]]]|I|$J|$5|K|L|M|C|$N|O]]]]

<h1>Python - <a href="https://codegolf.stackexchange.com/questions/115036/cops-square-times-square-root/115080#115080">SparklePony</a></h1>
<pre><code>import math
f=lambda x:x*x*math.sqrt(x)
</code></pre>


blocks|key|6396196|text|Scala，科弗斯_192|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396197|d=>math.pow(d,5/2d)|code-block|syntax|javascript|6396198|在这里试试！|unstyled|6396199|简单lambda表达式|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115097/65425|1|https://codepad.remoteinterview.io/UJENSUDGIW^0|6|7|0|0|0|0|6|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|11|8|@]|9|@$A|12|B|13|1|14]]|C|$]]|$1|L|3|M|5|K|7|15|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

<h1>Scala, <a href="https://codegolf.stackexchange.com/a/115097/65425">corvus_192</a></h1>

<pre><code>d=&gt;math.pow(d,5/2d)
</code></pre>

<p><a href="https://codepad.remoteinterview.io/UJENSUDGIW" rel="nofollow noreferrer">Try it here!</a></p>

<p>Simple lambda expression</p>


blocks|key|6396263|text|05AB1E+-+P.+Knops|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396264|t¹n*|code-block|syntax|javascript|6396265|相当容易:)|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/questions/115036/cops-square-times-square-root/115105#115105^0|9|8|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|X|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

<h1>05AB1E - <a href="https://codegolf.stackexchange.com/questions/115036/cops-square-times-square-root/115105#115105">P. Knops</a></h1>
<pre><code>t¹n*
</code></pre>
<p>Was quite easy :)</p>


blocks|key|967585|text|C，稳定盒|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|967586|float+q(float+b){return(b-b>0.*4*1/1)-sqrtf(b)*pow(b,2)*-1;}|code-block|syntax|javascript|967587|不知道原版做了什么；我不得不抛弃大量的随机运算符和数字，但是一旦可用的单词清楚了，就不难拼凑出总的想法了。|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115102/8927^0|2|3|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|X|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

<h1>C, <a href="https://codegolf.stackexchange.com/a/115102/8927">Steadybox</a></h1>

<pre class="lang-c prettyprint-override"><code>float q(float b){return(b-b&gt;0.*4*1/1)-sqrtf(b)*pow(b,2)*-1;}
</code></pre>

<p>Not sure what the original did; I had to discard a lot of random operators and numbers, but once the available words were clear it wasn't too hard to piece together the general idea.</p>


blocks|key|6396390|text|R，比目鱼|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396391|scan()**mean(1:4.5)|code-block|syntax|javascript|6396392|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115133/61405|1|https://tio.run/nexus/r#@19QlJlXolGcnJinoamllZsKpA2tTPRMNTX//zfhAgA^0|2|3|0|0|0|0|6|1^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|Z|8|@]|9|@$A|10|B|11|1|12]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]]]

<h1>R, <a href="https://codegolf.stackexchange.com/a/115133/61405">Flounderer</a></h1>

<pre><code>scan()**mean(1:4.5)
</code></pre>

<p><a href="https://tio.run/nexus/r#@19QlJlXolGcnJinoamllZsKpA2tTPRMNTX//zfhAgA" rel="nofollow noreferrer">Try it online!</a></p>


blocks|key|6396470|text|Python3.6，C..。|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396471|f=lambda+x:x**2.5or'1*77*8%2B8/5/(((aafothipie.xml)))'|code-block|syntax|javascript|6396472|在网上试试！|unstyled|6396473|将所有不需要的字符填充到or之后的字符串中。因为x**2.5是非零且如此真实，所以在or之后的部分不会被计算。在这里，任何语法上有效的表达式都可以。|style|CODE|6396474|Python将2.5和or作为单独的标记在2.5or中进行解析，尽管代码的语法高亮符并不识别这一点。|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115145/20260|1|https://tio.run/nexus/python3#@59mm5OYm5SSqFBhVaGlZaRnml@kbqhlbq5loW2hb6qvoaGRmJiWX5KRWZCZqleRm6Opqan@v6AoM69EI03DTFPzPwA^0|A|4|0|0|0|0|6|1|0|C|2|O|6|16|2|0|7|3|B|2|L|5^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|F|7|14|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|15|8|@]|9|@$A|16|B|17|1|18]]|C|$]]|$1|L|3|M|5|K|7|19|8|@$A|1A|B|1B|N|O]|$A|1C|B|1D|N|O]|$A|1E|B|1F|N|O]]|9|@]|C|$]]|$1|P|3|Q|5|K|7|1G|8|@$A|1H|B|1I|N|O]|$A|1J|B|1K|N|O]|$A|1L|B|1M|N|O]]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|T|U|V|C|$W|Z]]]]

<h2>Python 3.6, <a href="https://codegolf.stackexchange.com/a/115145/20260">c..</a></h2>



<pre class="lang-python prettyprint-override"><code>f=lambda x:x**2.5or'1*77*8+8/5/(((aafothipie.xml)))'
</code></pre>

<p><a href="https://tio.run/nexus/python3#@59mm5OYm5SSqFBhVaGlZaRnml@kbqhlbq5loW2hb6qvoaGRmJiWX5KRWZCZqleRm6Opqan@v6AoM69EI03DTFPzPwA" rel="nofollow noreferrer" title="Python 3 – TIO Nexus">Try it online!</a></p>

<p>Stuffs all the unneeded characters into a string after the <code>or</code>. Since <code>x**2.5</code> is nonzero and so truthy, the part after the <code>or</code> isn't evaluated. Any syntactically valid expression would be OK here. </p>

<p>Python parses <code>2.5</code> and <code>or</code> as separate tokens in <code>2.5or</code>, though the syntax highlighter for the code doesn't recognize this.</p>


blocks|key|6396492|text|Ruby，G+B|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6396493|->a{eval''<<97<<42<<42<<50<<46<<53}|code-block|syntax|javascript|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115166/45268^0|5|3|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]]|C|$]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|C|$G|H]]]|I|$J|$5|K|L|M|C|$N|O]]]]

blocks|key|968834|text|C#，135个字节，扬·伊凡|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|968835|这很有趣，我特别喜欢using+static的使用。|unstyled|style|CODE|968836|using+System;using+static+System.Math;class+P{static+void+Main(){Console.Write(Pow(long.Parse(Console.ReadLine()),1/Round(PI-E,1)));}}|code-block|syntax|javascript|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/115492/32628^0|A|4|0|0|A|C|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@$A|V|B|W|1|X]]|C|$]]|$1|D|3|E|5|F|7|Y|8|@$A|Z|B|10|G|H]]|9|@]|C|$]]|$1|I|3|J|5|K|7|11|8|@]|9|@]|C|$L|M]]]|N|$O|$5|P|Q|R|C|$S|T]]]]

<h2>C#, 135 bytes, <a href="https://codegolf.stackexchange.com/a/115492/32628">Jan Ivan</a></h2>
<p>This was a lot of fun, I particularly liked the use of <code>using static</code>.</p>
<pre><code>using System;using static System.Math;class P{static void Main(){Console.Write(Pow(long.Parse(Console.ReadLine()),1/Round(PI-E,1)));}}
</code></pre>


blocks|key|969263|text|JavaScript，SIGSEGV|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|969264|x=>x**(3.5-%2B!%2B[]);|code-block|syntax|javascript|969265|在网上试试！|unstyled|entityMap|0|LINK|mutability|MUTABLE|url|https://nodejs.org/|1|https://codegolf.stackexchange.com/a/116225/47066|2|https://tio.run/nexus/javascript-node#@59mW2FrV6GlpWGsZ6qrragdHatp/T85P684PydVLyc/XSNNw0xT8z8A^0|0|A|0|B|7|1|0|0|0|6|2^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]|$A|10|B|11|1|12]]|C|$]]|$1|D|3|E|5|F|7|13|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]|S|$5|N|O|P|C|$Q|T]]|U|$5|N|O|P|C|$Q|V]]]]

<h1><a href="https://nodejs.org" rel="nofollow noreferrer">JavaScript</a>, <a href="https://codegolf.stackexchange.com/a/116225/47066">SIGSEGV</a></h1>



<pre class="lang-javascript prettyprint-override"><code>x=&gt;x**(3.5-+!+[]);
</code></pre>

<p><a href="https://tio.run/nexus/javascript-node#@59mW2FrV6GlpWGsZ6qrragdHatp/T85P684PydVLyc/XSNNw0xT8z8A" rel="nofollow noreferrer" title="JavaScript (Node.js) – TIO Nexus">Try it online!</a></p>


blocks|key|969409|text|Vyxal，裂纹[医][气]素答|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|969410|5%7C2/en[(`ĖĖ:|code-block|syntax|javascript|969411|在网上试试！|unstyled|969412|很简单。我在淋浴时想出了解决办法，这是很容易的。|entityMap|0|LINK|mutability|MUTABLE|url|https://codegolf.stackexchange.com/a/237660/78850|1|https://lyxal.pythonanywhere.com/?flags=&code=5%257C2%252Fen%255B%2528%2560%25C4%2596%25C4%2596%253A&inputs=6&header=&footer=^0|8|8|0|0|0|0|6|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]]|C|$]]|$1|D|3|E|5|F|7|10|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|11|8|@]|9|@$A|12|B|13|1|14]]|C|$]]|$1|L|3|M|5|K|7|15|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

<h1>Vyxal, cracks <a href="https://codegolf.stackexchange.com/a/237660/78850">emanresuA's answer</a></h1>
<pre><code>5|2/en[(`ĖĖ:
</code></pre>
<p><a href="https://lyxal.pythonanywhere.com?flags=&amp;code=5%7C2%2Fen%5B%28%60%C4%96%C4%96%3A&amp;inputs=6&amp;header=&amp;footer=" rel="nofollow noreferrer">Try it Online!</a></p>
<p>Easy. I came up with the solution in the shower that's how easy it is.</p>


<p><strong>Task:</strong> Crack the scrambled code for multiplying the square root of an integer <em><strong>n</strong></em> by the square of it!</p>
<p>You must post a comment in the <strong><a href="https://codegolf.stackexchange.com/questions/115036/cops-square-times-square-root">cops' thread</a></strong> with a link to your working source, mentioning clearly that you have <em><strong>Cracked it</strong></em>. In your answer's title, you must include the link to the original answer.</p>
<h2>Rules:</h2>
<ul>
<li>You may only change the order of the characters in the original source.</li>
<li>Safe answers cannot be cracked anymore.</li>
<li>The other rules mentioned in the cops' thread</li>
<li><em><strong>Please edit the answer you crack</strong></em></li>
</ul>
<hr />
<h2>WINNER: Emigna - 10 submissons (had some trouble counting)</h2>
<p><em>Honorable mentions: Notjagan, Plannapus, TEHTMI</em></p>


Robbers - square times square root

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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功能1上新10个字符

功能2描述100个字符功能2描述100个字符功能2描述100个字符功能2描述100个字符功能2描述100个字符功能2描述100个字符功能2描述100个字符功能2描述100个字符功能2描述100个字符。

功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符功能2上新100个字符。

功能5描述100个字符功能5描述100个字符功能5描述100个字符功能5描述100个字符功能5描述100个字符功能5描述100个字符

功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符功能5上新100个字符

功能4上新

文章&问答评论现已支持表情

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社区富文本编辑器全新改版！诚邀体验～ 

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社区富文本&Markdown编辑器全新改版上线，欢迎大家体验!

诚挚邀请您参与本次调研，分享您的真实使用感受与建议。您的反馈至关重要，感谢您的支持与参与！

社区新版编辑器体验调研

任务:破解加扰码，将整数n的平方根乘以它的平方！您必须在条子线中发布一条带有指向工作源的链接的评论，清楚地指出您已经破解了它。在你的答案标题中，你必须包括到原始答案的链接。规则：您只能更改原始源中字符的顺序。安全的答案不能再破解了。警察线上提到的其他规则请编辑你破解的答案获胜者:Emigna-10个子弟(数得有点困难)荣誉奖: Notjagan，Plannapus，TEHTMI

问强盗-平方根
EN

规则：

获胜者:Emigna-10个子弟(数得有点困难)

回答 45

Code Golf用户

不是，41字节，这家伙

Code Golf用户

JavaScript (ES7)，尼尔

Code Golf用户

Python，15字节，X码器先生

社区

活动

圈层

关于

腾讯云开发者

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问强盗-平方根EN

规则：

获胜者:Emigna-10个子弟(数得有点困难)

回答 45

Code Golf用户

不是，41字节，这家伙

Code Golf用户

JavaScript (ES7)，尼尔

Code Golf用户

Python，15字节，X码器先生

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问强盗-平方根
EN