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Created January 10, 2012 12:09
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ScalaでもHaskellとほぼ同じ方法でできるよ!
scala> import com.github.okomok.ken._
import com.github.okomok.ken._
scala> def solve(n:Int) = List.filter(List.isInfixOf("AAG") _ )( List.replicateM(n)("ATCG") )
solve: (n: Int)com.github.okomok.ken.List[com.github.okomok.ken.List[Char]]
scala> solve(5)
res0: com.github.okomok.ken.List[com.github.okomok.ken.List[Char]] = List("AAAAG","AAAGA","AAAGT","AAAGC","AAAGG","AAGAA","AAGAT","AAGAC","AAGAG","AAGTA","AAGTT","AAGTC","AAGTG","AAGCA","AAGCT","AAGCC","AAGCG","AAGGA","AAGGT","AAGGC","AAGGG","ATAAG","ACAAG","AGAAG","TAAAG","TAAGA","TAAGT","TAAGC","TAAGG","TTAAG","TCAAG","TGAAG","CAAAG","CAAGA","CAAGT","CAAGC","CAAGG","CTAAG","CCAAG","CGAAG","GAAAG","GAAGA","GAAGT","GAAGC","GAAGG","GTAAG","GCAAG","GGAAG")
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