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

1.2 KiB

Raw Blame History

id, title, challengeType, videoUrl, dashedName

id	title	challengeType	videoUrl	dashedName
5900f3c51000cf542c50fed6	问题88：产品总和数	5		problem-88-product-sum-numbers

--description--

可以写为至少两个自然数{a1，a2，...，ak}的给定集合的和和乘积的自然数N称为乘积和数：N = a1 + a2 + ... + ak = a1×a2×...×ak。例如，6 = 1 + 2 + 3 = 1×2×3。对于给定的大小集合k，我们将使用此属性调用最小的N作为最小乘积和数。尺寸组k = 2,3,4,5和6的最小乘积和数如下。 k = 2：4 = 2×2 = 2 + 2k = 3：6 = 1×2×3 = 1 + 2 + 3k = 4：8 = 1×1×2×4 = 1 + 1 + 2 + 4k = 5：8 = 1×1×2×2×2 = 1 + 1 + 2 + 2 + 2k = 6：12 = 1×1×1×1×2×6 = 1 + 1 + 1 + 1 + 2 +因此，对于2≤k≤6，所有最小乘积和数之和为4 + 6 + 8 + 12 = 30;请注意，8只计算总和一次。实际上，由于2≤k≤12的完整最小乘积和数是{4,6,8,12,15,16}，因此总和为61.所有最小乘积和的总和是多少数字为2≤k≤12000？

--hints--

euler88()应该返回7587457。

assert.strictEqual(euler88(), 7587457);

--seed--

--seed-contents--

function productSumNumbers() {

  return true;
}

productSumNumbers();

--solutions--

// solution required

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