ee1e8abd87
* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
41 lines
1.2 KiB
Markdown
41 lines
1.2 KiB
Markdown
---
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id: 5900f3ec1000cf542c50feff
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title: 问题128：六边形瓷砖差异
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challengeType: 5
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videoUrl: ''
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dashedName: problem-128-hexagonal-tile-differences
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---
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# --description--
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编号为1的六边形瓷砖由六个六边形瓷砖环围绕，从“12点钟”开始，并以逆时针方向对瓷砖2至7进行编号。新环以相同的方式添加，下一个环编号为8至19,20至37,38至61，依此类推。下图显示了前三个环。
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通过找到tile n与其六个邻居中的每一个之间的差异，我们将PD（n）定义为那些作为素数的差异的数量。例如，在瓦片8周围顺时针工作，差异是12,29,11,6,1和13.因此PD（8）= 3.以相同的方式，瓦片17周围的差异是1,17,16,1 ，11和10，因此PD（17）= 2.可以证明PD（n）的最大值是3.如果PD（n）= 3的所有瓦片按升序列出以形成一个序列，第10个瓦片将是271.按此顺序找到第2000个瓦片。
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# --hints--
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`euler128()`应该返回14516824220。
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```js
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assert.strictEqual(euler128(), 14516824220);
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```
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# --seed--
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## --seed-contents--
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```js
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function euler128() {
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return true;
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}
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euler128();
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```
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# --solutions--
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```js
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// solution required
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```
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