679. Freefarea

Let S be the set consisting of the four letters {A,E,F,R}. For n0, let S(n) denote the set of words of length n consisting of letters belonging to S. We designate the words FREE,FARE,AREA,REEF as keywords.

Let f(n) be the number of words in S(n) that contains all four keywords exactly once.

This first happens for n=9, and indeed there is a unique 9 lettered word that contain each of the keywords once: FREEFAREA So, f(9)=1.

You are also given that f(15)=72863.

Find f(30).

679. Freefarea

S 为四元字符集 {A,E,F,R} 对于正整数 n0,记 S(n) 为:只包含 S 中的字符且长度为 n 的字符串的集合。我们将 FREE,FARE,AREA,REEF 四个词指定为 关键词

f(n) 为:S(n) 中,恰好包含四个关键词各一次的字符串的个数。

n=9 时,出现第一个合法的字符串 FREEFAREA。从而 f(9)=1。亦已知 f(15)=72863

f(30)


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