blocks|key|856175|text|Mathematica，9字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|856176|Fibonacci|code-block|syntax|javascript|856177|是的，这个内置函数支持负数.|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|$]]

<h1>Mathematica, 9 bytes</h1>

<pre><code>Fibonacci
</code></pre>

<p>Yes, this built-in function supports negative numbers.</p>


blocks|key|6282004|text|倍频程，20字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|6282005|+@(n)([1,1;1,0]%5En)(2)|code-block|syntax|javascript|6282006|在网上试试！|unstyled|offset|length|6282007|解释|6282008|这利用了fibonacci序列f(n)可以编写为(这应该是矩阵向量表示法)的事实：|style|CODE|6282009|递归地：|6282010|[f(n%2B1)]++=+[1++1]+*+[f(n)++]
[f(n)++]++++[1++0]+++[f(n-1)]|6282011|明文规定：|6282012|[f(n%2B1)]++=+[1++1]+%5En+*+[1]
[f(n)++]++++[1++0]++++++[0]|6282013|这意味着，该矩阵的右上角项到n的幂值是我们正在寻找的值f(n)。显然，我们也可以反演这个矩阵，因为它有满秩，并且这个关系仍然描述相同的递推关系。这意味着它也适用于负输入。|entityMap|0|LINK|mutability|MUTABLE|url|https://tio.run/nexus/octave#@@@gkaepEW2oY2htqGMQGwfkGGn@T8wr1jDQ/A8A^0|0|0|0|6|0|0|0|F|4|0|0|0|0|0|E|1|R|4^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|1A|8|@]|9|@$J|1B|K|1C|1|1D]]|A|$]]|$1|L|3|M|5|6|7|1E|8|@]|9|@]|A|$]]|$1|N|3|O|5|I|7|1F|8|@$J|1G|K|1H|P|Q]]|9|@]|A|$]]|$1|R|3|S|5|I|7|1I|8|@]|9|@]|A|$]]|$1|T|3|U|5|D|7|1J|8|@]|9|@]|A|$E|F]]|$1|V|3|W|5|I|7|1K|8|@]|9|@]|A|$]]|$1|X|3|Y|5|D|7|1L|8|@]|9|@]|A|$E|F]]|$1|Z|3|10|5|I|7|1M|8|@$J|1N|K|1O|P|Q]|$J|1P|K|1Q|P|Q]]|9|@]|A|$]]]|11|$12|$5|13|14|15|A|$16|17]]]]

<h1>Octave, 20 bytes</h1>
<pre><code> @(n)([1,1;1,0]^n)(2)
</code></pre>
<p><a href="https://tio.run/nexus/octave#@@@gkaepEW2oY2htqGMQGwfkGGn@T8wr1jDQ/A8A" rel="noreferrer" title="Octave – TIO Nexus">Try it online!</a></p>
<h3>Explanation</h3>
<p>This makes use of the fact that the fibonacci sequence <code>f(n)</code> can be written as (this should be a matrix vector notation):</p>
<p>Recursively:</p>
<pre><code>[f(n+1)]  = [1  1] * [f(n)  ]
[f(n)  ]    [1  0]   [f(n-1)]
</code></pre>
<p>Explicitly:</p>
<pre><code>[f(n+1)]  = [1  1] ^n * [1]
[f(n)  ]    [1  0]      [0]
</code></pre>
<p>This means that the top right entry of this matrix to the power of <code>n</code> is the value <code>f(n)</code> we're looking for. Obviously we can also invert this matrix as it has full rank, and the relationship still describes the same recurrence relation. That means it also works for negative inputs.</p>


blocks|key|6282098|text|最大值，3字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|6282099|+fib|code-block|syntax|javascript|6282100|支持正数和负数。|unstyled|6282101|在极大值在线上试用(粘贴)|offset|length|entityMap|0|LINK|mutability|MUTABLE|url|http://maxima.cesga.es/^0|0|0|0|1|5|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]]|$1|G|3|H|5|I|7|W|8|@]|9|@]|A|$]]|$1|J|3|K|5|I|7|X|8|@]|9|@$L|Y|M|Z|1|10]]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

<h1>Maxima, 3 bytes</h1>
<pre><code> fib
</code></pre>
<p>supports positive and negative numbers.</p>
<p>Try it (paste) on <a href="http://maxima.cesga.es/" rel="noreferrer">CESGA - Maxima on line</a></p>


blocks|key|856115|text|Python，43字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|856116|g=5**.5/2%2B.5
lambda+n:(g**n-(1-g)**n)/5**.5|code-block|syntax|javascript|856117|黄金分割率g的直接公式。对于f，上面的函数是：|unstyled|offset|length|style|CODE|856118|for+n+in+range(-10,11):print+f(n)+

-55.0
34.0
-21.0
13.0
-8.0
5.0
-3.0
2.0
-1.0
1.0
0.0
1.0
1.0
2.0
3.0
5.0
8.0
13.0
21.0
34.0
55.0|856119|相同长度的alt，只对5的平方根进行混叠：|856120|a=5**.5
lambda+n:((a%2B1)**n-(1-a)**n)/a/2**n|856121|我看不出有什么方法可以使递归函数与这些函数竞争。一次以57个字节为目标的温和的尝试：|856122|f=lambda+n:n<0and(-1)**n*f(-n)or+n>1and+f(n-1)%2Bf(n-2)or+n|856123|作为比较，迭代方法(+Python+2中的60个字节)：|856124|n=input()
a,b=0,1;exec"a,b=b,a%2Bb;"*n%2B"a,b=b-a,a;"*-n
print+a|856125|或者，58个字节：|856126|n=input()
a,b=0,1;exec"a,b=b,a%2Bcmp(n,0)*b;"*abs(n)
print+a|entityMap^0|0|0|5|1|E|1|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|18|8|@$J|19|K|1A|L|M]|$J|1B|K|1C|L|M]]|9|@]|A|$]]|$1|N|3|O|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|P|3|Q|5|I|7|1E|8|@]|9|@]|A|$]]|$1|R|3|S|5|D|7|1F|8|@]|9|@]|A|$E|F]]|$1|T|3|U|5|I|7|1G|8|@]|9|@]|A|$]]|$1|V|3|W|5|D|7|1H|8|@]|9|@]|A|$E|F]]|$1|X|3|Y|5|I|7|1I|8|@]|9|@]|A|$]]|$1|Z|3|10|5|D|7|1J|8|@]|9|@]|A|$E|F]]|$1|11|3|12|5|I|7|1K|8|@]|9|@]|A|$]]|$1|13|3|14|5|D|7|1L|8|@]|9|@]|A|$E|F]]]|15|$]]

<h2>Python, 43 bytes</h2>



<pre class="lang-python prettyprint-override"><code>g=5**.5/2+.5
lambda n:(g**n-(1-g)**n)/5**.5
</code></pre>

<p>A direct formula with the golden ratio <code>g</code>. With <code>f</code> the above function: </p>

<pre class="lang-python prettyprint-override"><code>for n in range(-10,11):print f(n) 

-55.0
34.0
-21.0
13.0
-8.0
5.0
-3.0
2.0
-1.0
1.0
0.0
1.0
1.0
2.0
3.0
5.0
8.0
13.0
21.0
34.0
55.0
</code></pre>

<p>Same length alt, only aliasing the square root of 5:</p>

<pre class="lang-python prettyprint-override"><code>a=5**.5
lambda n:((a+1)**n-(1-a)**n)/a/2**n
</code></pre>

<p>I didn't see a way to make a recursive function that could compete with these. A mildly-golfed attempt for 57 bytes:</p>

<pre class="lang-python prettyprint-override"><code>f=lambda n:n&lt;0and(-1)**n*f(-n)or n&gt;1and f(n-1)+f(n-2)or n
</code></pre>

<p>For comparison, an iterative method (60 bytes in Python 2):</p>

<pre class="lang-python prettyprint-override"><code>n=input()
a,b=0,1;exec"a,b=b,a+b;"*n+"a,b=b-a,a;"*-n
print a
</code></pre>

<p>Or, for 58 bytes:</p>

<pre class="lang-python prettyprint-override"><code>n=input()
a,b=0,1;exec"a,b=b,a+cmp(n,0)*b;"*abs(n)
print a
</code></pre>


blocks|key|856348|text|Jolf，2字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|856349|mL|code-block|syntax|javascript|856350|在这里试试！|unstyled|offset|length|856351|fibonacci内置器，使用phi公式实现。|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|http://conorobrien-foxx.github.io/Jolf/#code=bUw^0|0|0|0|6|0|0|F|3^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|Y|8|@]|9|@$J|Z|K|10|1|11]]|A|$]]|$1|L|3|M|5|I|7|12|8|@$J|13|K|14|N|O]]|9|@]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]]]

<h1>Jolf, 2 bytes</h1>

<pre><code>mL
</code></pre>

<p><a href="http://conorobrien-foxx.github.io/Jolf/#code=bUw" rel="nofollow noreferrer">Try it here!</a></p>

<p>The fibonacci builtin, implemented using the <code>phi</code> formula.</p>


blocks|key|856401|text|Haskell，51字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|856402|s=0:scanl(%2B)1s;f+z%7Ceven+z,z<0=+-f(-z);f+z=s!!abs+z|code-block|syntax|javascript|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|I|8|@]|9|@]|A|$E|F]]]|G|$]]

<h1>Haskell, 51 bytes</h1>

<pre><code>s=0:scanl(+)1s;f z|even z,z&lt;0= -f(-z);f z=s!!abs z
</code></pre>


blocks|key|6282314|text|利普，88字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6282315|#N::((if(<+N+2)((if(<+N+0)((-(f(%2B+N+2))(f(%2B+N+1))))((%2B+N))))((%2B(f(-+N+2))(f(-+N+1))))))|code-block|syntax|javascript|6282316|我看看那些括号。|unstyled|6282317|在网上试试！|6282318|不是很小真的。目前存在一个解析错误，需要使用(get+N)或(%2B+N)，而不是简单的N。我选了小一点的。然而，我不认为有什么可以做进一步的高尔夫。|style|CODE|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/andrakis/node-lithp|1|https://xkcd.com/297/|2|https://andrakis.github.io/lithp-webide/?code=KAogICAgKGRlZiBmICNOOjooCiAgICAgICAgKGlmICg8IE4gMikgKAogICAgICAgICAgICAoaWYgKDwgTiAwKSAoCiAgICAgICAgICAgICAgICAoLSAoZiAoKyBOIDIpKSAoZiAoKyBOIDEpKSkKICAgICAgICAgICAgKSAoCiAgICAgICAgICAgICAgICAoKyBOKQogICAgICAgICAgICApKQogICAgICAgICkgKAogICAgICAgICAgICAoKyAoZiAoLSBOIDIpKSAoZiAoLSBOIDEpKSkKICAgICAgICApKQogICAgKSkKICAgIChpbXBvcnQgbGlzdHMpCiAgICAoZWFjaCAoc2VxICgtIDkpIDgpICNJIDo6ICgocHJpbnQgIkZvciAiIEkgIjogIiAoZiBJKSkpKQop^0|0|2|0|0|0|5|2|1|0|0|6|2|0|M|7|U|5|16|1^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|17|8|@]|9|@$A|18|B|19|1|1A]]|C|$]]|$1|L|3|M|5|K|7|1B|8|@]|9|@$A|1C|B|1D|1|1E]]|C|$]]|$1|N|3|O|5|K|7|1F|8|@$A|1G|B|1H|P|Q]|$A|1I|B|1J|P|Q]|$A|1K|B|1L|P|Q]]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]|Y|$5|T|U|V|C|$W|Z]]|10|$5|T|U|V|C|$W|11]]]]

<h2><a href="https://github.com/andrakis/node-lithp" rel="nofollow noreferrer">Lithp</a>, 88 bytes</h2>
<pre><code>#N::((if(&lt; N 2)((if(&lt; N 0)((-(f(+ N 2))(f(+ N 1))))((+ N))))((+(f(- N 2))(f(- N 1))))))
</code></pre>
<p>My look at all those <a href="https://xkcd.com/297/" rel="nofollow noreferrer">parentheses</a>.</p>
<p><a href="https://andrakis.github.io/lithp-webide/?code=KAogICAgKGRlZiBmICNOOjooCiAgICAgICAgKGlmICg8IE4gMikgKAogICAgICAgICAgICAoaWYgKDwgTiAwKSAoCiAgICAgICAgICAgICAgICAoLSAoZiAoKyBOIDIpKSAoZiAoKyBOIDEpKSkKICAgICAgICAgICAgKSAoCiAgICAgICAgICAgICAgICAoKyBOKQogICAgICAgICAgICApKQogICAgICAgICkgKAogICAgICAgICAgICAoKyAoZiAoLSBOIDIpKSAoZiAoLSBOIDEpKSkKICAgICAgICApKQogICAgKSkKICAgIChpbXBvcnQgbGlzdHMpCiAgICAoZWFjaCAoc2VxICgtIDkpIDgpICNJIDo6ICgocHJpbnQgIkZvciAiIEkgIjogIiAoZiBJKSkpKQop" rel="nofollow noreferrer">Try it online!</a></p>
<p>Not very small really. There's currently a parsing bug that requires one to use <code>(get N)</code> or <code>(+ N)</code> instead of simply <code>N</code>. I chose the smaller one. However I don't think there's anything that can be done further to golf this.</p>


blocks|key|6282659|text|果冻，2字节|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|6282660|ÆḞ|code-block|syntax|javascript|6282661|在网上试试！|unstyled|6282662|无聊的内置答案，果冻的Fibonacci内置处理负输入很好|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/DennisMitchell/jelly|1|https://tio.run/##y0rNyan8//9w28Md8/7rWhZZKlgfbld51LRGwf0/AA^0|0|2|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><a href="https://github.com/DennisMitchell/jelly" rel="nofollow noreferrer">Jelly</a>, 2 bytes</h1>
<pre><code>ÆḞ
</code></pre>
<p><a href="https://tio.run/##y0rNyan8//9w28Md8/7rWhZZKlgfbld51LRGwf0/AA" rel="nofollow noreferrer" title="Jelly – Try It Online">Try it online!</a></p>
<p>Boring builtin answer, Jelly’s Fibonacci builtin handles negative inputs just fine</p>


<p>You probably all know the fibonacci sequence:</p>

<pre><code>fibonacci(n)=fibonacci(n-1)+fibonacci(n-2)
fibonacci(0)=0
fibonacci(1)=1
</code></pre>

<p>Your task is as simple as it could be:</p>

<ul>
<li>Given integer <code>N</code> compute <code>fibonacci(n)</code></li>
</ul>

<p>but here is the twist:</p>

<ul>
<li>Also do negative <code>N</code></li>
</ul>

<p>Wait. What?</p>

<pre><code>fibonacci(1)=fibonacci(0)+fibonacci(-1)
</code></pre>

<p>so</p>

<pre><code>fibonacci(-1)=1
</code></pre>

<p>and</p>

<pre><code>fibonacci(-2)=fibonacci(0)-fibonacci(1)=-1
</code></pre>

<p>and so on...</p>

<ul>
<li>This is a <a href="/questions/tagged/code-golf" class="post-tag" title="show questions tagged &#39;code-golf&#39;" rel="tag">code-golf</a> so shortest programm in bytes win.</li>
<li>You may submit a function or a full programm</li>
<li>N is in [-100,100]</li>
</ul>

<p>Testcase(s) in CSV:</p>

<pre><code>-9;-8;-7;-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6;7;8
34;-21;13;-8;5;-3;2;-1;1;0;1;1;2;3;5;8;13;21
</code></pre>

<p>Hint:</p>

<blockquote class="spoiler">
  <p> n&lt;0 and n&amp;1==0:<br/>
 <pre><code>fibonacci(n)=fibonacci(abs(n))*-1</code></pre></p>
</blockquote>


Negative Fibonacci Numbers

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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你们可能都知道斐波纳契序列：fibonacci(n)=fibonacci(n-1)+fibonacci(n-2)fibonacci(0)=0fibonacci(1)=1您的任务尽可能简单：给定整数N计算fibonacci(n)但这是一个转折：同时做负N等。什么？fibonacci(1)=fibonacci(0)+fibonacci(-1)所以fibonacci(-1)=1和fibonacci(-2

问负斐波那契数
EN

回答 8

Code Golf用户

Mathematica，9字节

Code Golf用户

倍频程，20字节

解释

Code Golf用户

最大值，3字节

社区

活动

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问负斐波那契数EN

回答 8

Code Golf用户

Mathematica，9字节

Code Golf用户

倍频程，20字节

解释

Code Golf用户

最大值，3字节

社区

活动

圈层

关于

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问负斐波那契数
EN