translated some Chinese curriculum files (#40531)

* translate task #1 - #4

* add translation of zhang-suen-thinning and markov algorithm

Co-authored-by: S1ngS1ng <liuxing0514@gmail.com>
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miyaliu666
2021-01-09 02:59:18 +08:00
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---
id: 5900f3b31000cf542c50fec6
title: 'Problem 71: Ordered fractions'
title: '关卡 71有序分数'
challengeType: 5
forumTopicId: 302184
---
# --description--
Consider the fraction, `n`/`d`, where `n` and `d` are positive integers. If `n`&lt;`d` and HCF(`n`,`d`)=1, it is called a reduced proper fraction.
考虑形如 `n`/`d` 的分数,其中 `n` `d` 均为正整数。如果 `n`&lt;`d`,且最大公约数 HCF(`n`,`d`)=1,则该分数被称为最简真分数。
If we list the set of reduced proper fractions for `d` ≤ 8 in ascending order of size, we get:
如果我们将 `d` ≤ 8 的最简真分数构成的集合按大小升序列出,将得到:
<div style='text-align: center;'>1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, <strong>2/5</strong>, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8</div>
It can be seen that 2/5 is the fraction immediately to the left of 3/7.
可以看出 2/5 是 3/7 的直接左邻的分数。
By listing the set of reduced proper fractions for `d` ≤ 1,000,000 in ascending order of size, find the numerator of the fraction immediately to the left of 3/7.
将所有 `d` ≤ 1,000,000 的最简真分数按大小升序排列,求此时 3/7 直接左邻的分数的分子。
# --hints--
`orderedFractions()` should return a number.
`orderedFractions()` 应该返回一个数字。
```js
assert(typeof orderedFractions() === 'number');
```
`orderedFractions()` should return 428570.
`orderedFractions()` 应该返回 428570
```js
assert.strictEqual(orderedFractions(), 428570);

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---
id: 5900f3b41000cf542c50fec7
title: 'Problem 72: Counting fractions'
title: '关卡 72分数计数'
challengeType: 5
forumTopicId: 302185
---
# --description--
Consider the fraction, `n`/`d`, where n and d are positive integers. If `n`&lt;`d` and HCF(`n`,`d`)=1, it is called a reduced proper fraction.
考虑形如 `n`/`d` 的分数,其中 n 和 d 均为正整数。如果 `n`&lt;`d`,且最大公约数 HCF(`n`,`d`)=1,则该分数被称为最简真分数。
If we list the set of reduced proper fractions for `d` ≤ 8 in ascending order of size, we get:
如果我们将 `d` ≤ 8 的最简真分数构成的集合按大小升序列出,将得到:
<div style='text-align: center;'>1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8</div>
It can be seen that there are 21 elements in this set.
可以看出该集合中共有 21 个元素。
How many elements would be contained in the set of reduced proper fractions for `d` ≤ 1,000,000?
`d` ≤ 1,000,000 的最简真分数构成的集合中共有多少个元素。
# --hints--
`countingFractions()` should return a number.
`countingFractions()` 应该返回一个数字。
```js
assert(typeof countingFractions() === 'number');
```
`countingFractions()` should return 303963552391.
`countingFractions()` 应该返回 303963552391
```js
assert.strictEqual(countingFractions(), 303963552391);

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---
id: 5900f3b61000cf542c50fec8
title: 'Problem 73: Counting fractions in a range'
title: '关卡 73区间内的分数个数'
challengeType: 5
forumTopicId: 302186
---
# --description--
Consider the fraction, `n`/`d`, where n and d are positive integers. If `n`&lt;`d` and HCF(`n`,`d`)=1, it is called a reduced proper fraction.
考虑形如 `n`/`d` 的分数,其中 n 和 d 均为正整数。如果 `n`&lt;`d`,且其最大公约数 HCF(`n`,`d`)=1,则该分数被称为最简真分数。
If we list the set of reduced proper fractions for `d` ≤ 8 in ascending order of size, we get:
如果我们将 `d` ≤ 8 的最简真分数构成的集合按大小升序列出,将得到:
<div style='text-align: center;'>1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, <strong>3/8</strong>, <strong>2/5</strong>, <strong>3/7</strong>, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8</div>
It can be seen that there are 3 fractions between 1/3 and 1/2.
可以看出在 1/3 1/2 之间有3个分数。
How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for `d` ≤ 12,000?
`d` ≤ 12,000 的最简真分数构成的集合排序后,在 1/3 和 1/2 之间有多少个分数?
# --hints--
`countingFractionsInARange()` should return a number.
`countingFractionsInARange()` 应该返回一个数字。
```js
assert(typeof countingFractionsInARange() === 'number');
```
`countingFractionsInARange()` should return 7295372.
`countingFractionsInARange()` 应该返回 7295372
```js
assert.strictEqual(countingFractionsInARange(), 7295372);