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>
62 lines
1.3 KiB
Markdown
62 lines
1.3 KiB
Markdown
---
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id: 594810f028c0303b75339acf
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title: 阿克曼功能
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challengeType: 5
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videoUrl: ''
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dashedName: ackermann-function
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---
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# --description--
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<p> Ackermann函数是递归函数的典型示例，尤其值得注意的是它不是原始递归函数。它的值增长非常快，其调用树的大小也是如此。 </p><p> Ackermann函数通常定义如下： </p> $$ A（m，n）= \\ begin {cases} n + 1＆\\ mbox {if} m = 0 \\\\ A（m-1,1）＆\\ mbox {if} m> 0 \\ mbox {和} n = 0 \\\\ A（m-1，A（m，n-1））＆\\ mbox {if} m> 0 \\ mbox {和} n> 0. \\ end {cases} $$ <p>它的论点永远不会消极，它总是终止。编写一个返回$ A（m，n）$的值的函数。任意精度是首选（因为函数增长如此之快），但不是必需的。 </p>
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# --hints--
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`ack`是一个功能。
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```js
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assert(typeof ack === 'function');
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```
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`ack(0, 0)`应该返回1。
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```js
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assert(ack(0, 0) === 1);
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```
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`ack(1, 1)`应该返回3。
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```js
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assert(ack(1, 1) === 3);
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```
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`ack(2, 5)`应该返回13。
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```js
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assert(ack(2, 5) === 13);
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```
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`ack(3, 3)`应该返回61。
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```js
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assert(ack(3, 3) === 61);
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```
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# --seed--
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## --seed-contents--
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```js
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function ack(m, n) {
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}
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```
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# --solutions--
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```js
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function ack(m, n) {
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return m === 0 ? n + 1 : ack(m - 1, n === 0 ? 1 : ack(m, n - 1));
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}
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```
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