chore(i8n,learn): processed translations

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---
id: 5900f49a1000cf542c50ffac
title: 问题300蛋白质折叠
title: 'Problem 300: Protein folding'
challengeType: 5
videoUrl: ''
forumTopicId: 301954
dashedName: problem-300-protein-folding
---
# --description--
以一种非常简化的形式我们可以将蛋白质视为由疏水H和极性P元素例如HHPPHHHPHHPH
In a very simplified form, we can consider proteins as strings consisting of hydrophobic (H) and polar (P) elements, e.g. HHPPHHHPHHPH.
对于这个问题蛋白质的方向很重要。例如HPP被认为与PPH不同。因此存在2n个由n个元素组成的不同蛋白质。
For this problem, the orientation of a protein is important; e.g. HPP is considered distinct from PPH. Thus, there are 2n distinct proteins consisting of n elements.
当人们自然地遇到这些弦时它们总是以H-H接触点的数量尽可能多的方式折叠因为这在能量上是有利的。 结果H元素倾向于在内部积累而P元素在外部。 天然蛋白质当然会在三个维度上折叠,但我们只会考虑蛋白质在两个维度上的折叠。
When one encounters these strings in nature, they are always folded in such a way that the number of H-H contact points is as large as possible, since this is energetically advantageous. As a result, the H-elements tend to accumulate in the inner part, with the P-elements on the outside. Natural proteins are folded in three dimensions of course, but we will only consider protein folding in two dimensions.
下图显示了我们的示例蛋白质可以折叠的两种可能方式H-H接触点用红色圆点显示
The figure below shows two possible ways that our example protein could be folded (H-H contact points are shown with red dots).
左侧的折叠只有六个H-H接触点因此永远不会自然发生。 另一方面右侧的折叠具有9个H-H接触点这对于此琴弦是最佳的。
The folding on the left has only six H-H contact points, thus it would never occur naturally. On the other hand, the folding on the right has nine H-H contact points, which is optimal for this string.
假设H和P元素同样可能出现在沿着字符串的任何位置那么长度为8的随机蛋白质字符串最优折叠中H-H接触点的平均数量为850/28 = 3.3203125
Assuming that H and P elements are equally likely to occur in any position along the string, the average number of H-H contact points in an optimal folding of a random protein string of length 8 turns out to be 850 / 28=3.3203125.
在最佳折叠长度为15的随机蛋白质串中H-H接触点的平均数量是多少 要获得准确的结果,请使用尽可能多的小数位给出答案。
What is the average number of H-H contact points in an optimal folding of a random protein string of length 15? Give your answer using as many decimal places as necessary for an exact result.
# --hints--
`euler300()`应该返回8.0540771484375
`euler300()` should return 8.0540771484375.
```js
assert.strictEqual(euler300(), 8.0540771484375);