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>
64 lines
1.1 KiB
Markdown
64 lines
1.1 KiB
Markdown
---
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id: 5900f3971000cf542c50feaa
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title: 问题43：子串可分性
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challengeType: 5
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videoUrl: ''
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dashedName: problem-43-sub-string-divisibility
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---
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# --description--
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数字1406357289是0到9的全数字，因为它是由0到9的每个数字以某种顺序组成的，但是它也具有相当有趣的子字符串可除性。
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令d1为第一位数，d2为第二位数，依此类推。 这样，我们注意以下几点：
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d2d3d4 = 406可被2整除
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d3d4d5 = 063被3整除
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d4d5d6 = 635可被5整除
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d5d6d7 = 357可被7整除
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d6d7d8 = 572被11整除
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d7d8d9 = 728被13整除
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d8d9d10 = 289被17整除
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使用此属性查找所有0到9个泛数字的数字。
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# --hints--
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`substringDivisibility()`应该返回[1430952867, 1460357289, 1406357289, 4130952867, 4160357289, 4106357289]。
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```js
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assert.deepEqual(substringDivisibility(), [
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1430952867,
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1460357289,
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1406357289,
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4130952867,
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4160357289,
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4106357289
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]);
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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 substringDivisibility() {
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return [];
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}
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substringDivisibility();
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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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