ee1e8abd87
* 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>
81 lines
2.1 KiB
Markdown
81 lines
2.1 KiB
Markdown
---
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id: 595608ff8bcd7a50bd490181
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title: 冰雹序列
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challengeType: 5
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videoUrl: ''
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dashedName: hailstone-sequence
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---
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# --description--
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<p> Hailstone数字序列可以从起始正整数n生成： </p>如果n为1，则序列结束。如果n是偶数，那么序列的下一个n <code>= n/2</code>如果n是奇数，那么序列的下一个n <code>= (3 \* n) + 1</code> <p> （未经证实的） <a href='https://en.wikipedia.org/wiki/Collatz conjecture' title='wp：Collatz猜想'>Collatz猜想</a>是任何起始编号的冰雹序列总是终止。 </p><p>冰雹序列也称为冰雹数（因为这些值通常受到多个下降和上升，如云中的冰雹），或者作为Collatz序列。 </p>任务：创建例程以生成数字的hailstone序列。使用例程表明，对于27号的冰雹序列具有开始与112个元件<code>27, 82, 41, 124</code> ，结束时用<code>8, 4, 2, 1</code>与显示具有最长冰雹序列的数目少于100,000一起序列的长度。 （但不要显示实际的序列！）参见： <a href='http://xkcd.com/710' title='链接：http：//xkcd.com/710'>xkcd</a> （幽默）。
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# --hints--
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`hailstoneSequence`是一个函数。
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```js
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assert(typeof hailstoneSequence === 'function');
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```
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`[[27,82,41,124,8,4,2,1], [351, 77031]]` `hailstoneSequence()`应返回`[[27,82,41,124,8,4,2,1], [351, 77031]]`
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```js
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assert.deepEqual(hailstoneSequence(), res);
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```
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# --seed--
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## --after-user-code--
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```js
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const res = [[27, 82, 41, 124, 8, 4, 2, 1], [351, 77031]];
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```
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## --seed-contents--
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```js
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function hailstoneSequence() {
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const res = [];
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return res;
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}
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```
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# --solutions--
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```js
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function hailstoneSequence () {
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const res = [];
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function hailstone(n) {
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const seq = [n];
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while (n > 1) {
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n = n % 2 ? 3 * n + 1 : n / 2;
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seq.push(n);
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}
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return seq;
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}
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const h = hailstone(27);
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const hLen = h.length;
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res.push([...h.slice(0, 4), ...h.slice(hLen - 4, hLen)]);
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let n = 0;
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let max = 0;
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for (let i = 100000; --i;) {
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const seq = hailstone(i);
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const sLen = seq.length;
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if (sLen > max) {
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n = i;
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max = sLen;
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}
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}
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res.push([max, n]);
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return res;
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}
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```
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