chore(i8n,learn): processed translations

curriculum/challenges/chinese/10-coding-interview-prep/project-euler/problem-230-fibonacci-words.md

@@ -1,36 +1,36 @@
 ---
 id: 5900f4531000cf542c50ff65
-title: 问题230：斐波纳契语
+title: 'Problem 230: Fibonacci Words'
 challengeType: 5
-videoUrl: ''
+forumTopicId: 301874
 dashedName: problem-230-fibonacci-words
 ---
 
 # --description--
 
-对于任何两个数字串A和B，我们将FA，B定义为序列（A，B，AB，BAB，ABBAB，...），其中每个术语是前两个术语的串联。
+For any two strings of digits, A and B, we define FA,B to be the sequence (A,B,AB,BAB,ABBAB,...) in which each term is the concatenation of the previous two.
 
-此外，我们将DA，B（n）定义为FA的第一项中的第n个数字，B包含至少n个数字。
+Further, we define DA,B(n) to be the nth digit in the first term of FA,B that contains at least n digits.
 
-例：
+Example:
 
-设A = 1415926535，B = 8979323846。我们希望找到DA，B（35）。
+Let A=1415926535, B=8979323846. We wish to find DA,B(35), say.
 
-FA，B的前几个术语是：1415926535 8979323846 14159265358979323846 897932384614159265358979323846 14159265358979323846897932384614159265358979323846
+The first few terms of FA,B are: 1415926535 8979323846 14159265358979323846 897932384614159265358979323846 14159265358979323846897932384614159265358979323846
 
-然后DA，B（35）是第五项中的第35位，即9。
+Then DA,B(35) is the 35th digit in the fifth term, which is 9.
 
-现在我们使用A小数点后面的前100位数字：14159265358979323846264338327950288419716939937510 58209749445923078164062862089986280348253421170679
+Now we use for A the first 100 digits of π behind the decimal point: 14159265358979323846264338327950288419716939937510 58209749445923078164062862089986280348253421170679
 
-对于B下一百个数字：
+and for B the next hundred digits:
 
-82148086513282306647093844609550582231725359408128 48111745028410270193852110555964462294895493038196。
+82148086513282306647093844609550582231725359408128 48111745028410270193852110555964462294895493038196 .
 
-求Σn= 0,1，...，17 10n×DA，B（（127 + 19n）×7n）。
+Find ∑n = 0,1,...,17 10n× DA,B((127+19n)×7n) .
 
 # --hints--
 
-`euler230()`应返回850481152593119200。
+`euler230()` should return 850481152593119200.
 
 ```js
 assert.strictEqual(euler230(), 850481152593119200);