結果

問題 No.447 ゆきこーだーの雨と雪 (2)
ユーザー atuk721atuk721
提出日時 2017-02-11 21:41:31
言語 Haskell
(9.10.1)
結果
AC  
実行時間 702 ms / 2,000 ms
コード長 2,070 bytes
コンパイル時間 2,382 ms
コンパイル使用メモリ 176,048 KB
実行使用メモリ 13,440 KB
最終ジャッジ日時 2024-12-29 16:37:25
合計ジャッジ時間 10,501 ms
ジャッジサーバーID
(参考情報)
judge4 / judge3
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 3
other AC * 25
権限があれば一括ダウンロードができます
コンパイルメッセージ
Loaded package environment from /home/judge/.ghc/x86_64-linux-9.8.2/environments/default
[1 of 2] Compiling Main             ( Main.hs, Main.o )
[2 of 2] Linking a.out

ソースコード

diff #
プレゼンテーションモードにする

import qualified Control.Monad as M
import qualified Data.Char as C
import qualified Data.List as L
type Result = (String, ([Int], Int))
type Answer = (Char, Int)
main :: IO ()
main = do
examN <- readLn :: IO Int
levels <- fmap (map read . words) getLine :: IO [Int]
submitN <- readLn :: IO Int
submits <- fmap lines getContents
let answers = zip (take examN ['A'..]) $ repeat 0
result = proc [] answers levels examN submits
M.forM_ (zip [1..] result) $ \(n, (name, (scores, s))) -> do
putStrLn $ show n ++ " " ++ name ++ " " ++ (unwords $ map show scores) ++ " " ++ show s
getLevel :: [Int] -> Char -> (Int, Int)
getLevel levels c = (index, levels !! index)
where index = C.ord c - 65
proc :: [Result] -> [Answer] -> [Int] -> Int -> [String] -> [Result]
proc rs _ _ _ [] = rs
proc rs as ls examN (z:zs) = proc rs'' as' ls examN zs
where [name, levelC:[]] = words z
(levelIndex, level) = getLevel ls levelC
order = case lookup levelC as of
Just x -> x + 1
Nothing -> error "No level"
score = floor $ 50 * (realToFrac level) + 50 * (realToFrac level) / (0.8 + 0.2 * (realToFrac order))
(i, ss) = case L.findIndex (\(n, _) -> n == name) rs of
Just i -> (i, fst . snd $ rs !! i)
Nothing -> (-1, replicate examN 0)
ss' = take levelIndex ss ++ (score : drop (levelIndex + 1) ss)
as' = take levelIndex as ++ ((levelC, order) : drop (levelIndex + 1) as)
rs' = if i == -1
then rs ++ [(name, (ss', sum ss'))]
else take i rs ++ (name, (ss', sum ss')) : (drop (i + 1) rs)
rs'' = L.sortOn (\(_, (_, s)) -> (-s)) rs'
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