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

1.1 KiB

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id, title, challengeType, videoUrl, dashedName

id	title	challengeType	videoUrl	dashedName
5900f4201000cf542c50ff33	问题180：三个变量函数的有理零点	5		problem-180-rational-zeros-of-a-function-of-three-variables

--description--

对于任何整数n，考虑三个函数f1，n（x，y，z）= xn + 1 + yn + 1 - zn + 1f2，n（x，y，z）=（xy + yz + zx）*（ xn-1 + yn-1-zn-1）f3，n（x，y，z）= xyz *（xn-2 + yn-2-zn-2）及其组合fn（x，y，z）= f1，n（x，y，z）+ f2，n（x，y，z） - f3，n（x，y，z）如果是x，我们将（x，y，z）称为k阶的黄金三元组， y和z都是形式为a / b的有理数，0 <a <b≤k且存在（至少）一个整数n，因此fn（x，y，z）= 0.设s（x ，y，z）= x + y + z。设t = u / v是所有不同s（x，y，z）的所有黄金三元组（x，y，z）的总和。所有s（x，y，z）和t必须在减少形式。找到你+ v。

--hints--

euler180()应该返回285196020571078980。

assert.strictEqual(euler180(), 285196020571078980);

--seed--

--seed-contents--

function euler180() {

  return true;
}

euler180();

--solutions--

// solution required

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