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

1.3 KiB

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id, title, challengeType, videoUrl, dashedName

id	title	challengeType	videoUrl	dashedName
5900f41c1000cf542c50ff2e	问题175：涉及不同方式的数量的分数数字可以表示为2的幂的总和	5		problem-175-fractions-involving-the-number-of-different-ways-a-number-can-be-expressed-as-a-sum-of-powers-of-2

--description--

将f（0）= 1和f（n）定义为将n作为2的幂之和进行写入的方式的数量，其中没有功率发生超过两次。

例如，f（10）= 5因为有五种不同的表达方式10:10 = 8 + 2 = 8 + 1 + 1 = 4 + 4 + 2 = 4 + 2 + 2 + 1 + 1 = 4 + 4 + 1 + 1

可以证明，对于每个分数p / q（p> 0，q> 0），存在至少一个整数n，使得f（n）/ f（n-1）= p / q。例如，f（n）/ f（n-1）= 13/17的最小n是241. 241的二进制扩展是11110001.从最高有效位到最低有效位读取这个二进制数有4个1，3个零和1个。我们将字符串4,3,1称为缩短的二进制扩展241.找到最小n的缩短二进制扩展，其中f（n）/ f（n-1）= 123456789/987654321。以逗号分隔的整数给出答案，没有任何空格。

--hints--

euler175()应该返回1,13717420,8。

assert.strictEqual(euler175(), 1, 13717420, 8);

--seed--

--seed-contents--

function euler175() {

  return true;
}

euler175();

--solutions--

// solution required

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