Oliver Eyton-Williams ee1e8abd87

feat(curriculum): restore seed + solution to Chinese (#40683 )

* 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>

2021-01-12 19:31:00 -07:00

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id, title, challengeType, videoUrl, dashedName

id	title	challengeType	videoUrl	dashedName
5900f3901000cf542c50fea3	问题36：双基回文	5		problem-36-double-base-palindromes

--description--

十进制数，585 = 10010010012（二进制），在两个碱基中都是回文。找到所有数字的总和，小于n，而1000 <= n <= 1000000，它们在基数10和基数2中是回文的。（请注意，任一基数中的回文数可能不包括前导零。）

--hints--

doubleBasePalindromes(1000)应该返回1772。

assert(doubleBasePalindromes(1000) == 1772);

doubleBasePalindromes(50000)应该返回105795。

assert(doubleBasePalindromes(50000) == 105795);

doubleBasePalindromes(500000)应该返回286602。

assert(doubleBasePalindromes(500000) == 286602);

doubleBasePalindromes(1000000)应该返回872187。

assert(doubleBasePalindromes(1000000) == 872187);

--seed--

--seed-contents--

function doubleBasePalindromes(n) {

  return n;
}

doubleBasePalindromes(1000000);

--solutions--

// solution required

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