Algorithm SPOJ问题FLIBONAKI时限超过
我试图在Haskell解决这个问题，但时间限制超过了。我用我所有的Haskell和数学技巧来优化这个，但都是徒劳的。有人可以建议我如何进一步优化这段代码吗。顺序F_3+F_7+F_11….+F_4n+3）=F_2n*F_2n+1。我用O（logn）来计算斐波那契数Algorithm SPOJ问题FLIBONAKI时限超过,algorithm,haskell,Algorithm,Haskell,我试图在Haskell解决这个问题，但时间限制超过了。我用我所有的Haskell和数学技巧来优化这个，但都是徒劳的。有人可以建议我如何进一步优化这段代码吗。顺序F_3+F_7+F_11….+F_4n+3）=F_2n*F_2n+1。我用O（logn）来计算斐波那契数 import Data.List import Data.Maybe import qualified Data.ByteString.Lazy.Char8 as BS matmul :: [Integer] -> [Inte
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
您的求解速度似乎足够快，但您的主函数似乎没有在每一行之后打印答案。事实上，它需要一个额外的换行符才能得到最后的答案，所以这可能是你超时的原因！这是一个直接在输入后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 )
导入数据。列表
导入数据，也许吧
进口管制
将限定数据.ByteString.Lazy.Char8作为B导入
将符合条件的Data.ByteString.Char8作为BC导入
将限定的Text.Show.ByteString作为BS导入
matmul:：[Integer]->[Integer]->Integer->[Integer]
matmul[a，b，c][d，e，f]m=[x，y，z]其中
y=（a*e+b*f）`mod`m
z=（b*e+c*f）`mod`m
x=y+z
powM:：[Integer]->Integer->Integer->[Integer]
功率a n m | n==1=a
|n==2=matmul a m
|偶数n=功率（matmul a m）（第2部分）m
|否则=matmul a（功率（matmul a m）（分区n 2）m）m
求解：：整数->整数
求解n=mod（c*d）100000007
哪里
[c，d，_]=powM[1,1,0]（2*n）100000007
readInteger:：B.ByteString->Integer
readInteger=fst。从刚才开始。B.readInteger
readInt:：B.ByteString->Int
readInt=fst。从刚才开始。B.readInt
get:：IO B.ByteString
get=liftM（B.fromChunks.（:[]）BC.getLine
main:：IO（）
main=do
n>=B.putStrLn。BS.show。解决）
matmul