Question

我正在尝试在Haskell中解决这个问题problem但是时间限制超过了。我应用了我所有的Haskell和数学技能来优化这一点，但都是徒劳的。有人可以建议我如何进一步优化此代码。序列F_3 + F_7 + F_11 .... + F_（4n + 3）= F_2n * F_（2n + 1）。我使用O（log n）方法来计算Fibonacci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

Answer 1

您的解决方案似乎足够快，但似乎您的主要功能不会在每个新行后打印答案。实际上它需要额外的换行才能得到最后的答案，这可能是你超时的原因！这是一个在输入后直接打印每个答案的版本。

import Data.List
import Data.Maybe
import Control.Monad
import qualified Data.ByteString.Lazy.Char8 as B
import qualified Data.ByteString.Char8 as BC
import qualified Text.Show.ByteString 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 

solve :: Integer -> Integer
solve n = mod ( c*d ) 1000000007 
  where 
  [c,d,_] = powM [1,1,0] ( 2*n ) 1000000007

readInteger :: B.ByteString -> Integer
readInteger  = fst . fromJust . B.readInteger

readInt :: B.ByteString -> Int
readInt = fst . fromJust . B.readInt

get :: IO B.ByteString
get = liftM (B.fromChunks . (:[])) BC.getLine

main :: IO ()
main = do
  n <- liftM readInt get
  replicateM_ n ( liftM readInteger get >>= B.putStrLn . BS.show . solve )

SPOJ问题Flibonakki时限超过

1 个答案: