我正在尝试用Haskell解决这个问题,但是时间限制已经超过了。我运用了我所有的Haskell和数学技巧来优化它,但都徒劳无功。请问有人能进一步建议我如何优化这段代码吗?序列F_3 + F_7 + F_11 .... + F_(4n+3) = F_2n*F_(2n+1)。我使用了O(log n)方法来计算斐波那契数。
import Data.List
import Data.Maybe
import qualified Data.ByteString.Lazy.Char8 as BS
matmul :: [Integer] -> [Integer] -> Integer -> [Integer]
matmul [a,b,c] [d,e,f] m = [x,y,z] where
y = (a*e + b*f) `mod` m
z = (b*e + c*f) `mod` m
x = y + z
powM ::[Integer] -> Integer -> Integer -> [Integer]
powM a n m | n == 1 = a
| n == 2 = matmul a a m
| even n = powM ( matmul a a m ) ( div n 2 ) m
| otherwise = matmul a ( powM ( matmul a a m ) ( div n 2 ) m ) m
readInt :: BS.ByteString -> Integer
readInt = fst.fromJust.BS.readInteger
solve::Integer -> BS.ByteString
solve n = BS.pack.show $ mod ( c*d ) 1000000007 where
[c,d,_] = powM [1,1,0] ( 2*n ) 1000000007
--([_,a,_]:_) = powM [[1,2,1],[0,5,3],[0,3,2]] n 1000000007
-- f_3+f_7+f_11+f_15 = f_2n*f_(2n+1)
main = BS.interact $ BS.unlines. map ( solve.readInt ) . tail . BS.lines
Word64
或Int64
代替Integer
可能会产生巨大的差异。 - Daniel Wagnermatmul
中的列表模式中添加 bang 模式,这可能是 thunk 建立的地方。 - John L