blocks|key|6407536|text|Mathematica，41字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|6407537|Plot[Re[(-#)%5Ex],{x,-3,3},PlotRange->All]&|code-block|syntax|javascript|6407538|除了数字的字体(我怀疑它是使用Wolfram创建的)外，输出看起来与挑战中显示的完全一样。|unstyled|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|M|8|@]|9|@]|A|$]]]|J|$]]

<h2>Mathematica, 41 bytes</h2>

<pre><code>Plot[Re[(-#)^x],{x,-3,3},PlotRange-&gt;All]&amp;
</code></pre>

<p>Output looks exactly as shown in the challenge except for the font of the numbers (which I suspect was created with Wolfram Alpha).</p>


blocks|key|6407754|text|R，29字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|6407755|curve(Re((0i-scan())%5Ex),-3,3)|code-block|syntax|javascript|6407756|n是通过stdin提供的。N=1的结果：|unstyled|offset|length|style|CODE|6407757|📷|atomic|6407758|对于n=2：|6407759|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://i.stack.imgur.com/EJyeC.png|imageAlt|1|https://i.stack.imgur.com/RFBiL.png^0|0|0|0|1|0|0|1|0|0|0|0|1|1^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|15|8|@$J|16|K|17|L|M]]|9|@]|A|$]]|$1|N|3|O|5|P|7|18|8|@]|9|@$J|19|K|1A|1|1B]]|A|$]]|$1|Q|3|R|5|I|7|1C|8|@]|9|@]|A|$]]|$1|S|3|O|5|P|7|1D|8|@]|9|@$J|1E|K|1F|1|1G]]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z|10|-4]]|11|$5|V|W|X|A|$Y|12|10|-4]]]]

<h2>R, 29 bytes</h2>
<pre><code>curve(Re((0i-scan())^x),-3,3)
</code></pre>
<p><code>n</code> is provided through stdin.
Result for n=1:
<a href="https://i.stack.imgur.com/EJyeC.png" rel="noreferrer"><img src="https://i.stack.imgur.com/EJyeC.png" alt="enter image description here" /></a></p>
<p>And for n=2:</p>
<p><a href="https://i.stack.imgur.com/RFBiL.png" rel="noreferrer"><img src="https://i.stack.imgur.com/RFBiL.png" alt="enter image description here" /></a></p>


blocks|key|980314|text|R，30字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|980315|plot(Re((0i-n)%5Eseq(-3,3,.05)))|code-block|syntax|javascript|980316|n+=+1|offset|length|style|CODE|980317|📷|atomic|980318|n+=+2|980319|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://i.stack.imgur.com/XVqu3.png|imageAlt|1|https://i.stack.imgur.com/w38G9.png^0|0|0|0|1|0|0|1|0|0|0|1|0|0|1|1^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|14|8|@$I|15|J|16|K|L]]|9|@]|A|$]]|$1|M|3|N|5|O|7|17|8|@]|9|@$I|18|J|19|1|1A]]|A|$]]|$1|P|3|Q|5|6|7|1B|8|@$I|1C|J|1D|K|L]]|9|@]|A|$]]|$1|R|3|N|5|O|7|1E|8|@]|9|@$I|1F|J|1G|1|1H]]|A|$]]]|S|$T|$5|U|V|W|A|$X|Y|Z|-4]]|10|$5|U|V|W|A|$X|11|Z|-4]]]]

<h1>R, 30 bytes</h1>

<pre><code>plot(Re((0i-n)^seq(-3,3,.05)))
</code></pre>

<h3><code>n</code> = 1</h3>

<p><a href="https://i.stack.imgur.com/XVqu3.png" rel="noreferrer"><img src="https://i.stack.imgur.com/XVqu3.png" alt="enter image description here"></a></p>

<h3><code>n</code> = 2</h3>

<p><a href="https://i.stack.imgur.com/w38G9.png" rel="noreferrer"><img src="https://i.stack.imgur.com/w38G9.png" alt="enter image description here"></a></p>


blocks|key|980458|text|木星笔记本和Python3；53字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|980459|%25pylab
def+f(n):x=arange(121)/20-3%2B0j;plot(x,(-n)**x)|code-block|syntax|javascript|980460|多亏了@Jonathan，节省了三个字节。|unstyled|980461|📷|atomic|980462|entityMap|0|LINK|mutability|MUTABLE|url|https://jupyter.org/|1|IMAGE|IMMUTABLE|imageUrl|https://i.stack.imgur.com/ixRfC.png|imageAlt|2|https://i.stack.imgur.com/b0zV3.png^0|0|5|0|0|0|0|0|1|1|0|0|1|2^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]|$1|D|3|E|5|F|7|18|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|19|8|@]|9|@]|C|$]]|$1|L|3|M|5|N|7|1A|8|@]|9|@$A|1B|B|1C|1|1D]]|C|$]]|$1|O|3|M|5|N|7|1E|8|@]|9|@$A|1F|B|1G|1|1H]]|C|$]]]|P|$Q|$5|R|S|T|C|$U|V]]|W|$5|X|S|Y|C|$Z|10|11|-4]]|12|$5|X|S|Y|C|$Z|13|11|-4]]]]

<h1><a href="https://jupyter.org/" rel="nofollow noreferrer">Jupyter notebook</a> and Python 3; 53 bytes</h1>

<pre><code>%pylab
def f(n):x=arange(121)/20-3+0j;plot(x,(-n)**x)
</code></pre>

<p>Three bytes saved thanks to @Jonathan Allan.</p>

<p><a href="https://i.stack.imgur.com/ixRfC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ixRfC.png" alt="n=1"></a>
<a href="https://i.stack.imgur.com/b0zV3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/b0zV3.png" alt="n=2"></a></p>


blocks|key|980685|text|封装的PostScript；232个字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|980686|%25!PS-Adobe-3.0+EPSF-3.0
%25%25BoundingBox:+0+0+500+500
%25%25EndComments
/n+1+def+.02+setlinewidth+/f{dup+dup+n+exch+exp+exch+180+mul+cos+mul+3+div}def
250+250+translate+80+80+scale+newpath+-3+f+moveto+-3+.05+3{f+lineto}for+stroke
%25%25EOF|code-block|syntax|javascript|980687|既然这是矢量图像本身..。|unstyled|980688|📷|atomic|offset|length|980689|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://i.stack.imgur.com/us3SR.png|imageAlt|1|https://i.stack.imgur.com/fYp2i.png^0|0|0|0|0|1|0|0|0|1|1^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|10|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|11|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|12|8|@]|9|@$M|13|N|14|1|15]]|A|$]]|$1|O|3|K|5|L|7|16|8|@]|9|@$M|17|N|18|1|19]]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V|W|-4]]|X|$5|R|S|T|A|$U|Y|W|-4]]]]

<h1>Encapsulated PostScript; 232 bytes</h1>

<pre><code>%!PS-Adobe-3.0 EPSF-3.0
%%BoundingBox: 0 0 500 500
%%EndComments
/n 1 def .02 setlinewidth /f{dup dup n exch exp exch 180 mul cos mul 3 div}def
250 250 translate 80 80 scale newpath -3 f moveto -3 .05 3{f lineto}for stroke
%%EOF
</code></pre>

<p>Now since this is a vector image itself... </p>

<p><a href="https://i.stack.imgur.com/us3SR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/us3SR.png" alt="enter image description here"></a></p>

<p><a href="https://i.stack.imgur.com/fYp2i.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fYp2i.png" alt="enter image description here"></a></p>


blocks|key|6408060|text|TikZ+%2B+PGFPlots，175个字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6408061|\documentclass{standalone}\usepackage{tikz,pgfplots}\begin{document}\typein[\n]{}\tikz{\begin{axis}\addplot[domain=-3:3,samples=120]{\n%5Ex*cos(180*x)};\end{axis}}\end{document}|code-block|syntax|javascript|6408062|编译与，例如，latexmk+-cd+-f+-pdf+in.tex的一个pdf输出。在编译期间，将提示用户输入n。|unstyled|style|CODE|6408063|N=1和n=2的样本输出(转换为png)：|6408064|📷|atomic|6408065|entityMap|0|LINK|mutability|MUTABLE|url|https://www.ctan.org/pkg/pgf?lang=en|1|https://www.ctan.org/pkg/pgfplots?lang=en|2|IMAGE|IMMUTABLE|imageUrl|https://i.imgur.com/ZLbrawl.png|imageAlt|3|https://i.imgur.com/LAlpZwY.png^0|0|4|0|7|8|1|0|0|7|Q|1J|1|0|0|0|1|2|0|0|1|3^^$0|@$1|2|3|4|5|6|7|1A|8|@]|9|@$A|1B|B|1C|1|1D]|$A|1E|B|1F|1|1G]]|C|$]]|$1|D|3|E|5|F|7|1H|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1I|8|@$A|1J|B|1K|L|M]|$A|1L|B|1M|L|M]]|9|@]|C|$]]|$1|N|3|O|5|K|7|1N|8|@]|9|@]|C|$]]|$1|P|3|Q|5|R|7|1O|8|@]|9|@$A|1P|B|1Q|1|1R]]|C|$]]|$1|S|3|Q|5|R|7|1S|8|@]|9|@$A|1T|B|1U|1|1V]]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]|10|$5|V|W|X|C|$Y|11]]|12|$5|13|W|14|C|$15|16|17|-4]]|18|$5|13|W|14|C|$15|19|17|-4]]]]

<h1><a href="https://www.ctan.org/pkg/pgf?lang=en" rel="nofollow noreferrer">TikZ</a> + <a href="https://www.ctan.org/pkg/pgfplots?lang=en" rel="nofollow noreferrer">PGFPlots</a>, 175 bytes</h1>

<pre><code>\documentclass{standalone}\usepackage{tikz,pgfplots}\begin{document}\typein[\n]{}\tikz{\begin{axis}\addplot[domain=-3:3,samples=120]{\n^x*cos(180*x)};\end{axis}}\end{document}
</code></pre>

<p>Compile with, <em>e.g.</em>, <code>latexmk -cd -f -pdf in.tex</code> for a pdf output. During compilation, the user is prompted for <code>n</code>.</p>

<p>Sample outputs (converted to png) for n = 1 and n = 2:</p>

<p><img src="https://i.imgur.com/ZLbrawl.png" alt="n = 1">
<img src="https://i.imgur.com/LAlpZwY.png" alt="n = 2"></p>


blocks|key|6407972|text|Math.JS绘图器，20+Bytes|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6407973|r(n)=f(x)=re((-n)%5Ex)|code-block|syntax|javascript|6407974|侥幸的是，这个绘图实用程序是TC+(大多数情况下，无限循环只会使它崩溃)，而本质上，它的主要输出是图形。|unstyled|6407975|是如何工作的|6407976|r(n)=分配一个函数r，该函数将参数n接受给表达式f(x)=re((-n)%5Ex)。re((-n)%5Ex)几乎是字母的挑战描述。但是这会将函数f(x)分配给这个函数，图形绘制程序将其隐式地输出为一个线状图。|style|CODE|6407977|如何测试它|6407978|您可以使用这站点，在其中输入该函数，然后用r(input)调用它。|6407979|输出|6407980|📷|atomic|entityMap|0|LINK|mutability|MUTABLE|url|http://a-ta.co/grapher|1|2|IMAGE|IMMUTABLE|imageUrl|https://i.stack.imgur.com/gnKgV.png|imageAlt^0|0|A|0|0|0|0|0|0|5|B|1|J|1|Q|F|16|A|1Y|4|0|0|L|8|5|1|1|0|0|0|1|2^^$0|@$1|2|3|4|5|6|7|1E|8|@]|9|@$A|1F|B|1G|1|1H]]|C|$]]|$1|D|3|E|5|F|7|1I|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1J|8|@]|9|@]|C|$]]|$1|L|3|M|5|6|7|1K|8|@]|9|@]|C|$]]|$1|N|3|O|5|K|7|1L|8|@$A|1M|B|1N|P|Q]|$A|1O|B|1P|P|Q]|$A|1Q|B|1R|P|Q]|$A|1S|B|1T|P|Q]|$A|1U|B|1V|P|Q]|$A|1W|B|1X|P|Q]]|9|@]|C|$]]|$1|R|3|S|5|6|7|1Y|8|@]|9|@]|C|$]]|$1|T|3|U|5|K|7|1Z|8|@$A|20|B|21|P|Q]]|9|@$A|22|B|23|1|24]]|C|$]]|$1|V|3|W|5|6|7|25|8|@]|9|@]|C|$]]|$1|X|3|Y|5|Z|7|26|8|@]|9|@$A|27|B|28|1|29]]|C|$]]]|10|$11|$5|12|13|14|C|$15|16]]|17|$5|12|13|14|C|$15|16]]|18|$5|19|13|1A|C|$1B|1C|1D|-4]]]]

<h1><a href="http://a-ta.co/grapher" rel="nofollow noreferrer">Math.JS Grapher</a>, 20 Bytes</h1>
<pre><code>r(n)=f(x)=re((-n)^x)
</code></pre>
<p>By sheer fluke, this graphing utility is TC (For the most part, Infinite loops just crash it.), and by nature, it's primary output is graphs.</p>
<h2>How it works</h2>
<p><code>r(n)=</code> assigns a function <code>r</code> which takes the argument <code>n</code> to the expression <code>f(x)=re((-n)^x)</code>. <code>re((-n)^x)</code> is pretty much letter for letter the challenge description. But this assigns the function <code>f(x)</code> to this, which the grapher implicitly outputs as a line graph.</p>
<h2>How to test it</h2>
<p>You can use <a href="http://a-ta.co/grapher" rel="nofollow noreferrer">this</a> site, punch that function in there, then call it with <code>r(input)</code>.</p>
<h2>Output</h2>
<p><img src="https://i.stack.imgur.com/gnKgV.png" alt="Output" /></p>


blocks|key|6408428|text|Google+Sheets，65|type|header-two|depth|inlineStyleRanges|entityRanges|data|6408429|输入N在A1中|unordered-list-item|6408430|图形化：|unstyled|6408431|=SPARKLINE(ArrayFormula(IMREAL(IMPOWER(-A1,.05*SEQUENCE(121,1,-60或|offset|length|style|CODE|6408432|=SPARKLINE(ArrayFormula(IMREAL(IMPOWER(-A1,SEQUENCE(121,1,-60)/20|6408433|将图形的行和列调整为500x500。|6408434|我试着使用COS()公式，但我所能做的最好是66个字符具有一个命名范围。我遇到的问题：|6408435|X由序列()生成，然后使用它两次。您自然会考虑缓存它，但是每次缓存到列中时，要再次使用它，您必须指定一个范围，或者使用一个筛选器。命名范围显然是更好的，但它确实花费了一个字符使其公平。|6408436|尝试标量乘一个序列()会迫使我们进入一个ArrayFormula+(或者MMULT()充其量)，我们不能对序列()使用非整数步骤。这意味着使用它的最佳方法是将它交给SPARKLINE公式。|6408437|这也意味着，每次我们使用这个范围时，我们的标量乘/除必须发生。|entityMap^0|0|0|0|0|1T|0|0|1T|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|12|8|@]|9|@]|A|$]]|$1|H|3|I|5|D|7|13|8|@$J|14|K|15|L|M]]|9|@]|A|$]]|$1|N|3|O|5|D|7|16|8|@$J|17|K|18|L|M]]|9|@]|A|$]]|$1|P|3|Q|5|G|7|19|8|@]|9|@]|A|$]]|$1|R|3|S|5|G|7|1A|8|@]|9|@]|A|$]]|$1|T|3|U|5|D|7|1B|8|@]|9|@]|A|$]]|$1|V|3|W|5|D|7|1C|8|@]|9|@]|A|$]]|$1|X|3|Y|5|D|7|1D|8|@]|9|@]|A|$]]]|Z|$]]

<h1>Google Sheets, 65</h1>
<ul>
<li>Input N is in A1</li>
</ul>
<p>Graphing:</p>
<ul>
<li><code>=SPARKLINE(ArrayFormula(IMREAL(IMPOWER(-A1,.05*SEQUENCE(121,1,-60</code>
or</li>
<li><code>=SPARKLINE(ArrayFormula(IMREAL(IMPOWER(-A1,SEQUENCE(121,1,-60)/20</code></li>
</ul>
<p>Resize row and column of graph to 500x500.</p>
<p>I tried to use the COS() formula, but the best I can do is 66 characters with a named range. Problems that I came across:</p>
<ul>
<li>X is generated by the SEQUENCE(), which is then used twice. You would naturally think to cache this, but every time you cache into a column, to use it again, you have to either name a range, or use a FILTER. Named Range is obviously better, but it does cost 1 character to make it fair.</li>
<li>Attempting to scalar multiply a SEQUENCE() forces us into an ArrayFormula (or MMULT() at best), and we cannot use a non-integer step for SEQUENCE(). This means that the most optimal way to use it is to give it to the SPARKLINE formula.</li>
<li>This also means that our scalar multiply/divide has to occur each time we use the range.</li>
</ul>


blocks|key|6408099|text|SmileBASIC，82字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|6408100|INPUT+N
FOR+I=0TO+399X=I/66.5-3GPSET+I,120-POW(N,X-3*SGN(N-1))*COS(PI()*X)*120NEXT|code-block|syntax|javascript|6408101|图形填充整个屏幕，即使N小于1时。|unstyled|6408102|当N大于1时，可以用n%5E3除以Y在-+1和1之间。我已经在做n%5Ex了，n%5Ex+/+n%5E3可以简化为n%5E(x-3)。然而，当N小于1时，我不得不用n%5E-3除以Y。这相当于n%5E(x%2B3)。|offset|length|style|CODE|6408103|我可以使用n%5E(x-3*sign(n-1))使用-3+if+n>1和%2B3+if+n<1|6408104|图像即将到来|entityMap^0|0|0|0|A|3|U|3|Z|9|1D|7|20|4|2C|7|0|5|H|O|2|U|3|Y|2|14|3|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|W|8|@]|9|@]|A|$]]|$1|J|3|K|5|I|7|X|8|@$L|Y|M|Z|N|O]|$L|10|M|11|N|O]|$L|12|M|13|N|O]|$L|14|M|15|N|O]|$L|16|M|17|N|O]|$L|18|M|19|N|O]]|9|@]|A|$]]|$1|P|3|Q|5|I|7|1A|8|@$L|1B|M|1C|N|O]|$L|1D|M|1E|N|O]|$L|1F|M|1G|N|O]|$L|1H|M|1I|N|O]|$L|1J|M|1K|N|O]]|9|@]|A|$]]|$1|R|3|S|5|I|7|1L|8|@]|9|@]|A|$]]]|T|$]]

<h1>SmileBASIC, 82 bytes</h1>
<pre><code>INPUT N
FOR I=0TO 399X=I/66.5-3GPSET I,120-POW(N,X-3*SGN(N-1))*COS(PI()*X)*120NEXT
</code></pre>
<p>Graph fills the entire screen, even when N is less than 1.</p>
<p>When N is greater than 1, you can scale Y to be between -1 and 1 by dividing it by <code>n^3</code>. I'm already doing <code>n^x</code>, and <code>n^x / n^3</code> can be simplified to <code>n^(x-3)</code>. However, when N is less than 1, I have to divide Y by <code>n^-3</code> instead. This is equivalent to <code>n^(x+3)</code>.</p>
<p>I can use <code>n^(x-3*sign(n-1))</code> to use <code>-3</code> if <code>n&gt;1</code>, and <code>+3</code> if <code>n&lt;1</code></p>
<p>Images coming soon</p>


<h2>Challenge</h2>
<p>Given an input of an integer, <span class="math-container">\$n\$</span> (where <span class="math-container">\$0&lt;n&lt;50\$</span>), output the graph of <span class="math-container">\$y=\mathrm{Re}((-n)^x)\$</span> from <span class="math-container">\$x = -3\$</span> to <span class="math-container">\$x = 3\$</span> inclusive.</p>
<p>Where <span class="math-container">\$\mathrm{Re}(p)\$</span> is the real part of the complex number <span class="math-container">\$p\$</span>.</p>
<p>Note that <span class="math-container">\$\mathrm{Re}((-n)^x) = n^x \cos{(\pi x)}\$</span></p>
<h2>Output</h2>
<p>The output may be in any form you wish (e.g. an image or a window etc.). ASCII art is disallowed.</p>
<p>The graph does not need to have axes (to allow languages without built-in graphing functions to compete).</p>
<p>If an image is output, it each side must be longer than 500 pixels. Similarly, the plot must fill the image as best it can.</p>
<p>The minimum interval between plots is 0.05.</p>
<p>Vector graphics are allowed.</p>
<h2>Examples</h2>
<p>For an input of <code>2</code>:</p>
<p><img src="https://i.stack.imgur.com/Vpryc.gif" alt="" /></p>
<p>For an input of <code>1</code>:</p>
<p><img src="https://i.stack.imgur.com/qpTPE.gif" alt="" /></p>
<hr />
<p><em><strong>You must put your corresponding outputs in your answer (n = 1 and n = 2).</strong></em></p>
<h2>Winning</h2>
<p>The shortest code in bytes wins.</p>


Draw a graph of \$y=(-n)^x\$

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文章&问答评论现已支持表情

挑战给定整数n (其中0<n<50)的输入，将y=\mathrm{Re}((-n)^x)的图形从x = -3输出到x = 3包含。其中\mathrm{Re}(p)是复数p的真实部分。请注意，\mathrm{Re}((-n)^x) = n^x \cos{(\pi x)}输出输出可以是您想要的任何形式(例如，图像或窗口等)。ASCII艺术是不允许的。图形不需要有轴(允许没有内置图形功能的语言进行竞争)

问绘制\$y=(-n)^x\$的图
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回答 9

Code Golf用户

Mathematica，41字节

Code Golf用户

R，29字节

Code Golf用户

R，30字节

`n` = 1

`n` = 2

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问绘制\$y=(-n)^x\$的图EN

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回答 9

Code Golf用户

Mathematica，41字节

Code Golf用户

R，29字节

Code Golf用户

R，30字节

n = 1

n = 2

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问绘制\$y=(-n)^x\$的图
EN

`n` = 1

`n` = 2