blocks|key|575197|text|果冻，11+字节数|type|header-two|depth|inlineStyleRanges|entityRanges|offset|length|data|575198|o-1.’d3ạ/ḞS|code-block|syntax|javascript|575199|在网上试试！或验证所有测试用例.|unstyled|575200|是如何工作的|575201|o-1.’d3ạ/ḞS++Main+link.+Argument:+[a,+b]+(list+of+integers)

+-1.+++++++++Yield+-1.5.
o++++++++++++Take+the+logical+OR+of+a+and+b+with+-1.5.
+++++++++++++This+maps+0+to+-1.5,+and+all+other+integers+to+themselves.
++++’++++++++Decrement+the+results.
+++++d3++++++Divmod;+compute+quotient+and+remainder+of+each+result+divided+by+3.
+++++++ạ/++++Reduce+by+absolute+difference+across+columns.
+++++++++Ḟ+++Floor;+round+all+differences+down+to+the+nearest+integer.
++++++++++S++Sum;+add+the+rounded+absolute+differences.|entityMap|0|LINK|mutability|MUTABLE|url|http://github.com/DennisMitchell/jelly|1|https://github.com/DennisMitchell/jelly/wiki/Code-page|2|http://jelly.tryitonline.net/#code=by0xLuKAmWQz4bqhL-G4nlM&input=&args=WzEsIDRd|3|http://jelly.tryitonline.net/#code=by0xLuKAmWQz4bqhL-G4nlMKOyLDh-KCrEc&input=&args=WzEsIDRdLCBbMSwgOF0sIFszLCA3XSwgWzgsIDJdLCBbNiwgMV0sIFswLCA5XSwgWzQsIDRdLCBbMCwgNF0^0|0|2|0|6|3|1|0|0|0|6|2|7|8|3|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]|$A|16|B|17|1|18]]|C|$]]|$1|D|3|E|5|F|7|19|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|K|7|1A|8|@]|9|@$A|1B|B|1C|1|1D]|$A|1E|B|1F|1|1G]]|C|$]]|$1|L|3|M|5|6|7|1H|8|@]|9|@]|C|$]]|$1|N|3|O|5|F|7|1I|8|@]|9|@]|C|$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]]|10|$5|R|S|T|C|$U|11]]]]

<h1><a href="http://github.com/DennisMitchell/jelly" rel="nofollow">Jelly</a>, 11 <a href="https://github.com/DennisMitchell/jelly/wiki/Code-page" rel="nofollow">bytes</a></h1>

<pre><code>o-1.’d3ạ/ḞS
</code></pre>

<p><a href="http://jelly.tryitonline.net/#code=by0xLuKAmWQz4bqhL-G4nlM&amp;input=&amp;args=WzEsIDRd" rel="nofollow">Try it online!</a> or <a href="http://jelly.tryitonline.net/#code=by0xLuKAmWQz4bqhL-G4nlMKOyLDh-KCrEc&amp;input=&amp;args=WzEsIDRdLCBbMSwgOF0sIFszLCA3XSwgWzgsIDJdLCBbNiwgMV0sIFswLCA5XSwgWzQsIDRdLCBbMCwgNF0" rel="nofollow">verify all test cases</a>.</p>

<h3>How it works</h3>

<pre><code>o-1.’d3ạ/ḞS  Main link. Argument: [a, b] (list of integers)

 -1.         Yield -1.5.
o            Take the logical OR of a and b with -1.5.
             This maps 0 to -1.5, and all other integers to themselves.
    ’        Decrement the results.
     d3      Divmod; compute quotient and remainder of each result divided by 3.
       ạ/    Reduce by absolute difference across columns.
         Ḟ   Floor; round all differences down to the nearest integer.
          S  Sum; add the rounded absolute differences.
</code></pre>


blocks|key|575241|text|果冻，13字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|575242|%2B2$.¹?€d3ạ/SḞ|code-block|syntax|javascript|575243|在网上试试！|unstyled|offset|length|575244|我的柏斯回答港。|575245|测试套件。|entityMap|0|LINK|mutability|MUTABLE|url|http://jelly.tryitonline.net/#code=KzIkLsK5P-KCrGQz4bqhL1PhuJ4&input=&args=WzAsM10|1|https://codegolf.stackexchange.com/a/80359/48934|2|http://jelly.tryitonline.net/#code=KzIkLsK5P-KCrGQz4bqhL1PhuJ4Kw4figqw&input=&args=W1sxLCA0XSwgWzEsIDhdLCBbMywgN10sIFs4LCAyXSwgWzYsIDFdLCBbMCwgOV0sIFs0LCA0XSwgWzAsIDRdXQ^0|0|0|0|6|0|0|0|6|1|0|0|5|2^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|12|8|@]|9|@$J|13|K|14|1|15]]|A|$]]|$1|L|3|M|5|I|7|16|8|@]|9|@$J|17|K|18|1|19]]|A|$]]|$1|N|3|O|5|I|7|1A|8|@]|9|@$J|1B|K|1C|1|1D]]|A|$]]]|P|$Q|$5|R|S|T|A|$U|V]]|W|$5|R|S|T|A|$U|X]]|Y|$5|R|S|T|A|$U|Z]]]]

<h1>Jelly, 13 bytes</h1>
<pre><code>+2$.¹?€d3ạ/SḞ
</code></pre>
<p><a href="http://jelly.tryitonline.net/#code=KzIkLsK5P-KCrGQz4bqhL1PhuJ4&amp;input=&amp;args=WzAsM10" rel="nofollow noreferrer">Try it online!</a></p>
<p>Port of <a href="https://codegolf.stackexchange.com/a/80359/48934">my Pyth answer</a>.</p>
<p><a href="http://jelly.tryitonline.net/#code=KzIkLsK5P-KCrGQz4bqhL1PhuJ4Kw4figqw&amp;input=&amp;args=W1sxLCA0XSwgWzEsIDhdLCBbMywgN10sIFs4LCAyXSwgWzYsIDFdLCBbMCwgOV0sIFs0LCA0XSwgWzAsIDRdXQ" rel="nofollow noreferrer">Test suite.</a></p>


blocks|key|6002828|text|JavaScript+(ES6)，59|type|header-two|depth|inlineStyleRanges|entityRanges|data|6002829|不幸的是，javascript中没有divmod。事实上，没有整数div，也没有mod：%25并不完全是mod。但就这一次而言，带有负数的%25的奇怪行为是有用的。|unstyled|offset|length|style|CODE|6002830|(a,b,A=Math.abs)=>A(~(--a/3)-~(--b/3))%2BA(a%253-b%253)-(a*b%253<0)|code-block|syntax|javascript|6002831|测试|6002832|f=(a,b,A=Math.abs)=>A(~(--a/3)-~(--b/3))%2BA(a%253-b%253)-(a*b%253<0)

for(i=0;i<10;console.log(r),i%2B%2B)
++for(r='',j=0;j<10;j%2B%2B)
++++r%2B=[i,j,f(i,j)]%2B'+'|entityMap^0|0|I|6|X|3|14|3|18|1|1E|3|1V|1|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@$E|U|F|V|G|H]|$E|W|F|X|G|H]|$E|Y|F|Z|G|H]|$E|10|F|11|G|H]|$E|12|F|13|G|H]|$E|14|F|15|G|H]]|9|@]|A|$]]|$1|I|3|J|5|K|7|16|8|@]|9|@]|A|$L|M]]|$1|N|3|O|5|6|7|17|8|@]|9|@]|A|$]]|$1|P|3|Q|5|K|7|18|8|@]|9|@]|A|$L|M]]]|R|$]]

<h1>JavaScript (ES6), 59</h1>

<p>Unfortunately, no <code>divmod</code> in javascript. In fact, no integer <code>div</code>, and no <code>mod</code> either: the <code>%</code> is not exactly <code>mod</code>. But for once, the weird behaviour of <code>%</code> with negative numbers is useful.</p>

<pre><code>(a,b,A=Math.abs)=&gt;A(~(--a/3)-~(--b/3))+A(a%3-b%3)-(a*b%3&lt;0)
</code></pre>

<p><strong>Test</strong></p>

<p><div class="snippet" data-lang="js" data-hide="false" data-console="true">
<div class="snippet-code">
<pre class="snippet-code-js lang-js prettyprint-override"><code>f=(a,b,A=Math.abs)=&gt;A(~(--a/3)-~(--b/3))+A(a%3-b%3)-(a*b%3&lt;0)

for(i=0;i&lt;10;console.log(r),i++)
  for(r='',j=0;j&lt;10;j++)
    r+=[i,j,f(i,j)]+' '</code></pre>
</div>
</div>
</p>


blocks|key|575278|text|Python2，61个字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|575279|lambda+x,y:abs(x/-3-y/-3)%2Babs((x+or+1.5)%25-3-(y+or+1.5)%25-3)//1|code-block|syntax|javascript|575280|在艾德龙上进行测试。|unstyled|offset|length|575281|是如何工作的|575282|垂直距离整数除法总是舍入Python中最接近的整数，得到10种可能的红利除以-3的结果如下。0+1+2+3+4+5+6+7+8+9+0+-1+-1+-1+-2+-2+-2+-3+-3+-3，因此，我们可以作为abs(x/-3-y/-3)计算x和y之间的垂直距离.|unordered-list-item|style|CODE|575283|对于列的水平距离，我们可以避免将0作为特例，将其替换为1.5，从而将其“放在”第一列和第二列之间。模(%25)在Python中总是有除数符号，对于10种可能的分红模-3，得到以下结果。0+1+2+3+4+5+6+7+8+9+-1+5+-1+0+-1+0-1+0-1+0-1+0-1+0-1+0-1+0-1，通过舍入(//1)+abs((x+or+1.5)%25-3-(y+or+1.5)%25-3)的结果，我们可以计算x与y的水平差.|entityMap|0|LINK|mutability|MUTABLE|url|http://ideone.com/C7nrCE^0|0|0|1|3|0|0|0|2W|E|0|1F|1|4C|3|4H|W^^$0|@$1|2|3|4|5|6|7|11|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|12|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|13|8|@]|9|@$J|14|K|15|1|16]]|A|$]]|$1|L|3|M|5|6|7|17|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|18|8|@$J|19|K|1A|Q|R]]|9|@]|A|$]]|$1|S|3|T|5|P|7|1B|8|@$J|1C|K|1D|Q|R]|$J|1E|K|1F|Q|R]|$J|1G|K|1H|Q|R]]|9|@]|A|$]]]|U|$V|$5|W|X|Y|A|$Z|10]]]]

<h1>Python 2, 61 bytes</h1>



<pre class="lang-python prettyprint-override"><code>lambda x,y:abs(x/-3-y/-3)+abs((x or 1.5)%-3-(y or 1.5)%-3)//1
</code></pre>

<p>Test it on <a href="http://ideone.com/C7nrCE" rel="nofollow">Ideone</a>.</p>

<h3>How it works</h3>

<ul>
<li><p><strong>Vertical distance</strong></p>

<p>Integer division always round <em>down</em> to the nearest integer in Python, yielding the following results for the 10 possible dividends divided by <strong>-3</strong>.</p>

<pre class="lang-python prettyprint-override"><code> 0  1  2  3  4  5  6  7  8  9
 0 -1 -1 -1 -2 -2 -2 -3 -3 -3
</code></pre>

<p>Thus, we can compute the vertical distance between <strong>x</strong> and <strong>y</strong> as <code>abs(x/-3-y/-3)</code>.</p></li>
<li><p><strong>Horizontal distance</strong></p>

<p>For columns, we can avoid treating <strong>0</strong> as a special case by replacing it with <strong>1.5</strong>, thus placing it "between" the first and second column.</p>

<p>Modulus (<code>%</code>) always has the sign of the divisor in Python, yielding the following results for the 10 possible dividend modulo <strong>-3</strong>.</p>

<pre class="lang-python prettyprint-override"><code>   0  1  2  3  4  5  6  7  8  9
-1.5 -2 -1  0 -2 -1  0 -2 -1  0
</code></pre>

<p>Thus, by rounding down (<code>//1</code>) the result of <code>abs((x or 1.5)%-3-(y or 1.5)%-3)</code>, we can compute the horizontal difference between <strong>x</strong> and <strong>y</strong>.</p></li>
</ul>


blocks|key|575305|text|Julia，56字节|type|header-two|depth|inlineStyleRanges|entityRanges|data|575306|x->sum(abs(-([[divrem(t>0?t%2B2:.5,3)...]for+t=x]...)))÷1|code-block|syntax|javascript|575307|在网上试试！|unstyled|offset|length|entityMap|0|LINK|mutability|MUTABLE|url|http://julia.tryitonline.net/#code=ZiA9IHgtPnN1bShhYnMoLShbW2RpdnJlbSh0PjA_dCsyOi41LDMpLi4uXWZvciB0PXhdLi4uKSkpw7cxCgpmb3IgKHgsIHkpIGluICgoMSwgNCksICgxLCA4KSwgKDMsIDcpLCAoOCwgMiksICg2LCAxKSwgKDAsIDkpLCAoNCwgNCksICgwLCA0KSkKICAgIEBwcmludGYoIiVkLCAlZDogJWRcbiIsIHgsIHksIGYoKHgsIHkpKSkKZW5k&input=^0|0|0|0|6|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|I|7|U|8|@]|9|@$J|V|K|W|1|X]]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

<h1>Julia, 56 bytes</h1>



<pre class="lang-julia prettyprint-override"><code>x-&gt;sum(abs(-([[divrem(t&gt;0?t+2:.5,3)...]for t=x]...)))÷1
</code></pre>

<p><a href="http://julia.tryitonline.net/#code=ZiA9IHgtPnN1bShhYnMoLShbW2RpdnJlbSh0PjA_dCsyOi41LDMpLi4uXWZvciB0PXhdLi4uKSkpw7cxCgpmb3IgKHgsIHkpIGluICgoMSwgNCksICgxLCA4KSwgKDMsIDcpLCAoOCwgMiksICg2LCAxKSwgKDAsIDkpLCAoNCwgNCksICgwLCA0KSkKICAgIEBwcmludGYoIiVkLCAlZDogJWRcbiIsIHgsIHksIGYoKHgsIHkpKSkKZW5k&amp;input=" rel="nofollow">Try it online!</a></p>


<p><em>Credit goes where <a href="http://chat.stackexchange.com/transcript/message/29818745#29818745">credit is due</a></em></p>
<hr />
<p>Given two digits, x, and y, calculate the shortest number of horizontal or vertical jumps to get from x to y on a standard numpad, e.g.</p>
<pre><code>789
456
123
00
</code></pre>
<p>You can safely assume that all inputs will be 0-9, and you do not have to handle invalid inputs. The input <em>may</em> be the same number twice, which has a distance of 0.</p>
<p>IO can be in <a href="http://meta.codegolf.stackexchange.com/questions/2447/default-for-code-golf-input-output-methods">any reasonable format</a>, and standard loopholes are forbidden. Shortest answer in bytes wins!</p>
<h1>Sample IO:</h1>
<pre><code>1, 4: 1
1, 8: 3
3, 7: 4
8, 2: 2
6, 1: 3
0, 9: 4
4, 4: 0
0, 4: 2
</code></pre>


Calculate the Numpad Taxicab

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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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

信贷到哪里去信用到期给定两个数字，x和y，计算在标准数字上从x到y的水平或垂直跳转的最短次数。78945612300您可以安全地假设所有输入都是0-9，并且不需要处理无效的输入。输入可以是同一数字的两倍，它的距离为0。IO可以出现在任何合理的格式中，并且禁止标准漏洞。最短答案(以字节为单位)获胜！示例IO:1, 4: 11, 8: 33, 7: 48, 2: 26, 1: 30, 9: 44, 4

问计算Numpad出租车
EN

示例IO:

回答 5

Code Golf用户

果冻，11 字节数

是如何工作的

Code Golf用户

果冻，13字节

Code Golf用户

JavaScript (ES6)，59

测试

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活动

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问计算Numpad出租车EN

示例IO:

回答 5

Code Golf用户

果冻，11 字节数

是如何工作的

Code Golf用户

果冻，13字节

Code Golf用户

JavaScript (ES6)，59

测试

社区

活动

圈层

关于

腾讯云开发者

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问计算Numpad出租车
EN