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>
This commit is contained in:
@ -1,31 +1,31 @@
|
||||
---
|
||||
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`<`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`<`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);
|
||||
|
@ -1,31 +1,31 @@
|
||||
---
|
||||
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`<`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`<`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?
|
||||
求 `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);
|
||||
|
@ -1,31 +1,31 @@
|
||||
---
|
||||
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`<`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`<`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 和 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?
|
||||
将 `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);
|
||||
|
Reference in New Issue
Block a user