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72 lines
1.3 KiB
Markdown
72 lines
1.3 KiB
Markdown
---
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id: 594810f028c0303b75339acf
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title: 阿克曼函数
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challengeType: 1
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forumTopicId: 302223
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dashedName: ackermann-function
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---
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# --description--
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The Ackermann function is a classic example of a recursive function, notable especially because it is not a primitive recursive function. It grows very quickly in value, as does the size of its call tree.
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阿克曼函数通常定义如下：
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$A(m, n) = \\begin{cases} n+1 & \\mbox{if } m = 0 \\\\ A(m-1, 1) & \\mbox{if } m > 0 \\mbox{ and } n = 0 \\\\ A(m-1, A(m, n-1)) & \\mbox{if } m > 0 \\mbox{ and } n > 0. \\end{cases}$
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它的参数从不是负的，而且总是有终点。
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# --instructions--
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编写一个函数，返回 $A(m, n)$ 的值。 任意精度是首选（因为函数增长如此之快），但不是必需的。
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# --hints--
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`ack` 应该是一个函数。
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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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