chore(i8n,learn): processed translations

curriculum/challenges/chinese/10-coding-interview-prep/rosetta-code/ackermann-function.md

@@ -1,42 +1,52 @@
 ---
 id: 594810f028c0303b75339acf
-title: 阿克曼功能
+title: Ackermann function
 challengeType: 5
-videoUrl: ''
+forumTopicId: 302223
 dashedName: ackermann-function
 ---
 
 # --description--
 
-<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）$的值的函数。任意精度是首选（因为函数增长如此之快），但不是必需的。 </p>
+The Ackermann function is a classic example of a recursive function, notable especially because it is not a primitive recursive function. It grows very quickly in value, as does the size of its call tree.
+
+The Ackermann function is usually defined as follows:
+
+$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}$
+
+Its arguments are never negative and it always terminates.
+
+# --instructions--
+
+Write a function which returns the value of $A(m, n)$. Arbitrary precision is preferred (since the function grows so quickly), but not required.
 
 # --hints--
 
-`ack`是一个功能。
+`ack` should be a function.
 
 ```js
 assert(typeof ack === 'function');
 ```
 
-`ack(0, 0)`应该返回1。
+`ack(0, 0)` should return 1.
 
 ```js
 assert(ack(0, 0) === 1);
 ```
 
-`ack(1, 1)`应该返回3。
+`ack(1, 1)` should return 3.
 
 ```js
 assert(ack(1, 1) === 3);
 ```
 
-`ack(2, 5)`应该返回13。
+`ack(2, 5)` should return 13.
 
 ```js
 assert(ack(2, 5) === 13);
 ```
 
-`ack(3, 3)`应该返回61。
+`ack(3, 3)` should return 61.
 
 ```js
 assert(ack(3, 3) === 61);