{-# LANGUAGE
  ScopedTypeVariables, BangPatterns, TupleSections, ExplicitForAll,
  LambdaCase, MultiWayIf, Unsafe, RecordWildCards, FlexibleContexts, CPP,
  NoMonomorphismRestriction, GADTs, PatternGuards, MagicHash, UnboxedTuples,
  RankNTypes, EmptyDataDecls, EmptyCase, ViewPatterns, PolyKinds,
  TypeFamilies, OverloadedStrings, FlexibleInstances, UndecidableInstances,
  DefaultSignatures, GeneralizedNewtypeDeriving, StandaloneDeriving,
  DeriveGeneric, DeriveFunctor, DeriveDataTypeable, DeriveFoldable,
  DeriveTraversable, DeriveDataTypeable #-}
{-# OPTIONS_GHC -O2 #-}

import Prelude
import Data.Bits
import Data.List
import Data.Maybe
import Data.Tuple
import Data.Ord
import Data.Int
import Data.Word
import Data.Char
import Data.Ratio
import Data.Function
import Data.STRef
import Data.IORef
import Data.Monoid
import Data.Functor
import Data.Data
import Data.Typeable
import GHC.Generics
import System.IO
import Control.Arrow
import Control.Applicative
import Control.Monad
import Control.Monad.Primitive
import Control.Monad.State.Strict
import Control.Monad.ST
import Control.Monad.ST.Lazy (strictToLazyST, lazyToStrictST)
import qualified Control.Monad.ST.Lazy as STL
-- import Control.Monad.ST.Safe
import Control.DeepSeq
import Data.Coerce
import qualified Data.ByteString.Char8 as BS
import qualified Data.ByteString.Lazy.Char8 as BSL
import qualified Data.ByteString.Builder as BSB
import Data.IntMap (IntMap)
import qualified Data.IntMap.Strict as IMS
import qualified Data.IntMap.Lazy as IML
import Data.IntSet (IntSet)
import qualified Data.IntSet as IS
import qualified Data.Sequence as Seq
import Data.Sequence (Seq)
import qualified Data.Array.IArray as A
import qualified Data.Array.MArray.Safe as A
import qualified Data.Array.MArray as A
import Data.Array (Array)
import Data.Array.Unboxed (UArray)
import Data.Array.IArray (IArray)
import Data.Array.MArray.Safe (MArray)
import Data.Array.IO.Safe (IOArray, IOUArray)
import Data.Array.ST.Safe (STArray, STUArray, runSTArray, runSTUArray)
import qualified Data.Vector as V
import qualified Data.Vector.Mutable as VM
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.Vector.Storable as VS
import qualified Data.Vector.Storable.Mutable as VSM
import qualified Data.Vector.Primitive as VP
import qualified Data.Vector.Primitive.Mutable as VPM
import qualified Data.Vector.Generic as VG
import qualified Data.Vector.Generic.Mutable as VGM
import qualified Data.Attoparsec.ByteString.Char8 as Atto
import Debug.Trace
import GHC.Exts (build, Int(..), Int#,
                 (+#), (*#), (-#), (<#), (>=#), (==#), quotRemInt#,
                 remInt#, uncheckedIShiftRL#, andI#, orI#,
                isTrue#)
import Control.Monad.Trans.Maybe
rIntS1 = subtract 1 <$> rIntS

main :: IO ()
main = do
  [n,m] <- map readInt . words <$> getLine
  grph <- getDirectedGraphForward n m
    $ fromJust . evalStateT (liftA3 (,,) rIntS1 rIntS1 (return ()))
       <$> BS.getLine
  visited <- VUM.replicate n (0::Int)
  next <- VUM.replicate n (-1)
  let dfs :: Int -> IO Int
      dfs !v = do
        visitedV <- VUM.read visited v
        case visitedV of
          2 -> return (-1)
          1 -> return v
          0 -> do
            VUM.unsafeWrite visited v 1
            VU.foldr (\ (!v1,_) cont -> do
                         VUM.unsafeWrite next v v1
                         !res <- dfs v1
                         case res of
                           -1 -> cont
                           -2 -> do VUM.unsafeWrite next v (-1)
                                    return (-2)
                           _ | res == v  -> return (-2)
                             | otherwise -> return res)
              (do VUM.unsafeWrite visited v 2
                  VUM.unsafeWrite next v (-1)
                  return (-1))
              (edgesOut grph v)
  res <- foldr (\ !v cont -> do
                   res <- dfs v
                   if res /= -1 then return True else cont)
         (return False) [0..n-1]
  if not res then print (-1) else do
  !v0 <- fromJust . VU.findIndex (>=0) <$> VU.unsafeFreeze next
  VUM.set visited 0
  let go :: Int -> IO ()
      go !v0 = do
        visit <- VUM.read visited v0
        unless (visit > 0) $ do
        VUM.unsafeWrite visited v0 1
        VU.forM_ (edgesOut grph v0) $ \ !(v1,_) -> do
          !next1 <- VUM.unsafeRead next v1
          when (next1 >= 0) $ do
            !next0 <- VUM.unsafeRead next v0
            VUM.unsafeWrite next v0 v1
            let go2 !next0
                  | next0 == v1 = return ()
                  | otherwise = do
                      !nextnext0 <- VUM.unsafeRead next next0
                      VUM.unsafeWrite next next0 (-1)
                      go2 nextnext0
            go2 next0
        VUM.unsafeRead next v0 >>= go
  go v0
  next_ <- VU.unsafeFreeze next
  let result = VU.map ((+1).fst)
        $ VU.filter ((>=0) . snd) $ VU.indexed next_
  print $ VU.length result
  printVecInLines $ result

type Vertex = Int
data GraphData d = GraphData {-# UNPACK #-} !(VU.Vector (Vertex,d))
                             {-# UNPACK #-} !(VU.Vector Int)
data MGraphData s d = MGraphData {-# UNPACK #-} !(VUM.MVector s (Vertex,d))
                                 {-# UNPACK #-} !(VU.Vector Int)
data TreeData d = TreeData {-# UNPACK #-} !Vertex
                           {-# UNPACK #-} !(GraphData d)

edgesOut :: (VU.Unbox d)
  => GraphData d -> Vertex -> VU.Vector (Vertex,d)
edgesOut (GraphData connTable degCount) vert
  = VU.slice beg (end-beg) connTable
  where
    !end = degCount VU.! (vert+1)
    !beg = degCount VU.! vert

{-# INLINE unsafeFreezeGraphData #-}
unsafeFreezeGraphData :: (PrimMonad m, VU.Unbox d) =>
  MGraphData (PrimState m) d -> m (GraphData d)
unsafeFreezeGraphData (MGraphData connTable degCount)
  = (`GraphData` degCount) <$!> VU.unsafeFreeze connTable

{-# INLINE getDirectedGraphForward #-}
getDirectedGraphForward :: (PrimMonad m, VU.Unbox d) =>
  Int -> Int -> m (Int,Int,d) -> m (GraphData d)
getDirectedGraphForward !numVert !numEdges getEdge = do
  getDirectedGraphMutForward numVert numEdges getEdge >>= unsafeFreezeGraphData

getDirectedGraphMutForward :: (PrimMonad m, VU.Unbox d)
  => Int -> Int -> (m (Int,Int,d))
  -> m (MGraphData (PrimState m) d)
getDirectedGraphMutForward !numVert !numEdges getEdge = do
  !degCount <- VUM.replicate (numVert+1) 0
  !connTable <- VUM.new numEdges
  write degCount connTable
  MGraphData connTable <$!> VU.unsafeFreeze degCount
  where
    write !degCount !connTable = do
      !temp <- do
        !temp <- VUM.new numEdges
        VU.forM_ (VU.generate numEdges id) $ \ !i -> do
          (!a,!b,!d) <- getEdge
          VUM.unsafeWrite temp i (a,b,d)
          VUM.modify degCount (+1) (a+1)
        VU.unsafeFreeze temp
      VU.foldM'
        (\ !total !i -> do
            !di <- VUM.unsafeRead degCount i
            VUM.unsafeWrite degCount i total
            return $! total+di)
        0 (VU.generate (numVert+1) id)
      VU.forM_ temp $ \(!a,!b,!d) -> do
        !ia <- VUM.read degCount (a+1)
        VUM.unsafeWrite degCount (a+1) $! ia+1
        VUM.unsafeWrite connTable ia $! (b,d)

#define IL(f) {-# INLINE f #-}; f

class ShowAsBuilder a where
  showAsBuilder :: a -> BSB.Builder
  default showAsBuilder :: (Show a) => a -> BSB.Builder
  showAsBuilder = BSB.string8 . show

#define INS(t,f) instance ShowAsBuilder t where \
  { {-# INLINE showAsBuilder #-}; showAsBuilder=f }
INS(Int,BSB.intDec)
INS(Int8,BSB.int8Dec)
INS(Int16,BSB.int16Dec)
INS(Int32,BSB.int32Dec)
INS(Int64,BSB.int64Dec)
INS(Word,BSB.wordDec)
INS(Word8,BSB.word8Dec)
INS(Word16,BSB.word16Dec)
INS(Word32,BSB.word32Dec)
INS(Word64,BSB.word64Dec)
INS(Integer,BSB.integerDec)
INS(Float,BSB.floatDec)
INS(Double,BSB.doubleDec)
-- INS(String,BSB.string8) -- Inconsistent with Show
-- INS(BS.ByteString,BSB.byteString) -- Inconsistent with Show
-- INS(BSL.ByteString,BSB.lazyByteString) -- Inconsisitent with Show
#undef INS

-- Inconsistent with show
instance (ShowAsBuilder a, VG.Vector v a) => ShowAsBuilder (v a) where
  showAsBuilder = v2BSpcSep

{-# INLINE putBuilder #-}
putBuilder = BSB.hPutBuilder stdout

printVecInLines, printVecInSpcSepLn ::
  (VG.Vector v a, ShowAsBuilder a) => v a -> IO ()
{-# INLINE printVecInLines #-}
printVecInLines = putBuilder . v2BLines
{-# INLINE printVecInSpcSepLn #-}
printVecInSpcSepLn = putBuilder . v2BSpcSepLn

v2BSpcSepLn, v2BSpcSep, v2BConcat, v2BLines ::
  (VG.Vector v a, ShowAsBuilder a)
  => v a -> BSB.Builder
{-# INLINE v2BSpcSepLn #-}
v2BSpcSepLn = v2BSpcSepLnWith showAsBuilder
{-# INLINE v2BSpcSep #-}
v2BSpcSep = v2BSpcSepWith showAsBuilder
{-# INLINE v2BConcat #-}
v2BConcat = v2BConcatWith showAsBuilder
{-# INLINE v2BLines #-}
v2BLines = v2BLinesWith showAsBuilder

v2BSpcSepLnWith, v2BSpcSepWith, v2BConcatWith, v2BLinesWith ::
  (VG.Vector v a)
  => (a -> BSB.Builder) -- ^ show function
  -> v a -> BSB.Builder
{-# INLINE v2BSpcSepLnWith #-}
v2BSpcSepLnWith = v2BSpcSepPostfWith "\n"
{-# INLINE v2BSpcSepWith #-}
v2BSpcSepWith = v2BSpcSepPostfWith ""
{-# INLINE v2BConcatWith #-}
v2BConcatWith showFct = VG.foldr ((<>) . showFct) mempty
{-# INLINE v2BLinesWith #-}
v2BLinesWith showFct
  = VG.foldr (\ a -> (showFct a <>) . (BSB.char7 '\n' <>)) mempty

v2BSpcSepPostf :: (VG.Vector v a, ShowAsBuilder a)
  => BS.ByteString -- ^ postfix
  -> v a -> BSB.Builder
{-# INLINE v2BSpcSepPostf #-}
v2BSpcSepPostf = (`v2BSpcSepPostfWith` showAsBuilder)

v2BSpcSepPostfWith :: (VG.Vector v a)
  => BS.ByteString -- ^ postfix
  -> (a -> BSB.Builder) -- ^ show function
  -> v a -> BSB.Builder
{-# INLINE v2BSpcSepPostfWith #-}
v2BSpcSepPostfWith = vecToBuilder "" " "

{-# INLINE vecToBuilder #-}
vecToBuilder :: (VG.Vector v a)
  => BS.ByteString -- ^ prefix
  -> BS.ByteString -- ^ separator
  -> BS.ByteString -- ^ postfix
  -> (a -> BSB.Builder) -- ^ show function
  -> v a -> BSB.Builder
vecToBuilder !prefix !separator !postfix
  = vecToBuilder_ (BSB.byteString prefix)
                  (BSB.byteString separator)
                  (BSB.byteString postfix)

{-# INLINE vecToBuilder_ #-}
vecToBuilder_ :: (VG.Vector v a)
  => BSB.Builder -- ^ prefix
  -> BSB.Builder -- ^ separator
  -> BSB.Builder -- ^ postfix
  -> (a -> BSB.Builder) -- ^ show function
  -> v a -> BSB.Builder
vecToBuilder_ !prefix !separator !postfix showFct vec
  = prefix <> VG.foldr
    (\ a rest !prefx -> prefx <> (showFct a <> rest separator))
    (const postfix) vec mempty

IL(evalVals) :: [a] -> [a]
evalVals xs = build $ \c n -> foldr (c $!) n xs
IL(forceVals) :: (NFData a) => [a] -> [a]
forceVals xs = build $ \c n -> foldr (c $!!) n xs

getVecGLn :: (VG.Vector v a) =>
           Int -> StateT BS.ByteString Maybe a -> IO (v a)
getVecGLn n s = VG.unfoldrN n (runStateT s) <$> BS.getLine
{-# INLINE getVecGLn #-}
getVecGRest :: (VG.Vector v a) =>
            Int -> StateT BSL.ByteString Maybe a -> IO (v a)
getVecGRest n s = VG.unfoldrN n (runStateT s) <$> BSL.getContents
{-# INLINE getVecGRest #-}
getVecLn :: Int -> StateT BS.ByteString Maybe a -> IO (V.Vector a)
getVecLn = getVecGLn
{-# INLINE getVecLn #-}
getVecRest :: Int -> StateT BSL.ByteString Maybe a -> IO (V.Vector a)
getVecRest = getVecGRest
{-# INLINE getVecRest #-}
getVecULn :: (VU.Unbox a) =>
  Int -> StateT BS.ByteString Maybe a -> IO (VU.Vector a)
getVecULn = getVecGLn
{-# INLINE getVecULn #-}
getVecURest :: (VU.Unbox a) =>
  Int -> StateT BSL.ByteString Maybe a -> IO (VU.Vector a)
getVecURest = getVecGRest
{-# INLINE getVecURest #-}

rInt :: StateT BSL.ByteString Maybe Int
IL(rInt) = StateT $ BSL.readInt . BSL.dropWhile (<'!')
rIntS :: StateT BS.ByteString Maybe Int
IL(rIntS) = StateT $ BS.readInt . BS.dropWhile (<'!')

unlessM, whenM :: (Monad m) => m Bool -> m () -> m ()
IL(whenM) = (. flip when) . (>>=)
IL(unlessM) = (. flip unless) . (>>=)

IL(wrA) = A.writeArray
IL(rdA) = A.readArray
IL(mdA) = \arr f !i -> do
  ai <- rdA arr i
  let fai = f ai
  wrA arr i fai
  return (ai,fai)
{-# INLINE mdA' #-}
mdA' = \arr f !i -> do
  !ai <- rdA arr i
  let !fai = f ai
  wrA arr i fai
  return (ai,fai)
IL(swapA) = \arr !i !j -> do
  ai <- rdA arr i
  wrA arr i =<< rdA arr j
  wrA arr j ai

#define D(f,r,d)\
  IL(f) :: Integral a=>a->d; f=fromIntegral;\
  IL(r) :: String->d; r=read
#define C(f,r,g,h,d) D(f,r,d);\
  g,h :: RealFrac a=>a->d; IL(g)=floor; IL(h)=ceiling
C(_toInteger_,readInteger,floorInteger,ceilInteger,Integer)
C(toInt,readInt,floorInt,ceilInt,Int)
C(toI8,readI8,floorI8,ceilI8,Int8)
C(toI16,readI16,floorI16,ceilI16,Int16)
C(toI32,readI32,floorI32,ceilI32,Int32)
C(toI64,readI64,floorI64,ceilI64,Int64)
C(toWord,readWord,floorWord,ceilWord,Word)
C(toW8,readW8,floorW8,ceilW8,Word8)
C(toW16,readW16,floorW16,ceilW16,Word16)
C(toW32,readW32,floorW32,ceilW32,Word32)
C(toW64,readW64,floorW64,ceilW64,Word64)
D(toDouble,readDouble,Double)
D(toFloat,readFloat,Float)
#undef D
#undef C

#define N(f,g,h,a,m)\
  IL(f) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> e -> m (a i e);\
  f=A.newArray;\
  IL(g) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> m (a i e);\
  g=A.newArray_;\
  IL(h) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> [e] -> m (a i e);\
  h=A.newListArray
#define C(a,m)
N(newIOA,newIOA_,newIOAL,IOArray,IO)
N(newSTA,newSTA_,newSTAL,STArray s,ST s)
#undef C
#define C(a,m) MArray (a) e (m),
N(newIOUA,newIOUA_,newIOUAL,IOUArray,IO)
N(newSTUA,newSTUA_,newSTUAL,STUArray s,ST s)
#undef C
#undef N

#undef IL

Score : 600 points

Problem Statement

Given is a directed graph G with N vertices and M edges.

The vertices are numbered 1 to N, and the i-th edge is directed from Vertex A_i to Vertex B_i.

It is guaranteed that the graph contains no self-loops or multiple edges.

Determine whether there exists an induced subgraph (see Notes) of G such that the in-degree and out-degree of every vertex are both 1. If the answer is yes, show one such subgraph.

Here the null graph is not considered as a subgraph.

Notes

For a directed graph G = (V, E), we call a directed graph G' = (V', E') satisfying the following conditions an induced subgraph of G:

V' is a (non-empty) subset of V.

E' is the set of all the edges in E that have both endpoints in V'.

Constraints

1 \leq N \leq 1000

0 \leq M \leq 2000

1 \leq A_i,B_i \leq N

A_i \neq B_i

All pairs (A_i, B_i) are distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
:
A_M B_M

Output

If there is no induced subgraph of G that satisfies the condition, print -1.
Otherwise, print an induced subgraph of G that satisfies the condition, in the following format:

K
v_1
v_2
:
v_K

This represents the induced subgraph of G with K vertices whose vertex set is \{v_1, v_2, \ldots, v_K\}. (The order of v_1, v_2, \ldots, v_K does not matter.)
If there are multiple subgraphs of G that satisfy the condition, printing any of them is accepted.

Sample Input 1

4 5
1 2
2 3
2 4
4 1
4 3

Sample Output 1

3
1
2
4

The induced subgraph of G whose vertex set is \{1, 2, 4\} has the edge set \{(1, 2), (2, 4), (4, 1)\}. The in-degree and out-degree of every vertex in this graph are both 1.

Sample Input 2

4 5
1 2
2 3
2 4
1 4
4 3

Sample Output 2

-1

There is no induced subgraph of G that satisfies the condition.

Sample Input 3

6 9
1 2
2 3
3 4
4 5
5 6
5 1
5 2
6 1
6 2

Sample Output 3

4
2
3
4
5


module Main where

import           Control.Applicative
import           Data.Monoid

main :: IO ()
main = show . oddJudge <$> getLine >>= putStrLn

evenJudge :: String -> Bool
evenJudge [] = True
evenJudge (x:xs) = (x == 'L' || x == 'U' || x == 'D') && oddJudge xs

oddJudge :: String -> Bool
oddJudge [] = True
oddJudge (x:xs) = (x == 'R' || x == 'U' || x == 'D') && evenJudge xs

Global OOD problem IDs are disjoint from train, validation, and in-distribution test problem IDs.

Score: 200 points

Problem Statement

Takahashi will do a tap dance. The dance is described by a string S where each character is L, R, U, or D. These characters indicate the positions on which Takahashi should step. He will follow these instructions one by one in order, starting with the first character.

S is said to be easily playable if and only if it satisfies both of the following conditions:

Every character in an odd position (1-st, 3-rd, 5-th, \ldots) is R, U, or D.

Every character in an even position (2-nd, 4-th, 6-th, \ldots) is L, U, or D.

Your task is to print Yes if S is easily playable, and No otherwise.

Constraints

S is a string of length between 1 and 100 (inclusive).

Each character of S is L, R, U, or D.

Input

Input is given from Standard Input in the following format:

S

Output

Print Yes if S is easily playable, and No otherwise.

Sample Input 1

RUDLUDR

Sample Output 1

Yes

Every character in an odd position (1-st, 3-rd, 5-th, 7-th) is R, U, or D.

Every character in an even position (2-nd, 4-th, 6-th) is L, U, or D.

Thus, S is easily playable.

Sample Input 2

DULL

Sample Output 2

No

The 3-rd character is not R, U, nor D, so S is not easily playable.

Sample Input 3

UUUUUUUUUUUUUUU

Sample Output 3

Yes

Sample Input 4

ULURU

Sample Output 4

No

Sample Input 5

RDULULDURURLRDULRLR

Sample Output 5

Yes

Submitted source-code file size, in bytes.

Execution CPU time reported by official Project CodeNet metadata, in milliseconds.

Peak memory usage reported by official Project CodeNet metadata, in kilobytes.


check :: String -> Bool
check s = checkEven s
checkEven [] = True
checkEven (c:s) = (c == 'R' || c == 'U' || c == 'D') && checkOdd s
checkOdd [] = True
checkOdd (c:s) = (c == 'L' || c == 'U' || c == 'D') && checkEven s

main = do
  s <- getLine :: IO String
  putStrLn $ if check s then "Yes" else "No"


import qualified Data.Map as M
import qualified Data.Map as M
import Control.Monad

solve :: [Int] -> M.Map Int Int -> M.Map Int Int
solve inData ans
 | null inData = ans
 | otherwise   = solve (tail inData) newAns
   where newAns = M.insertWith (+) (head inData) 1 ans

main = do
 [n,k,q] <- map read . words <$> getLine :: IO [Int]
 aList   <- replicateM q (read <$> getLine) :: IO [Int]
 mapM_ putStrLn $ map ((\x -> if x > 0 then "Yes" else "No") . (\x -> x + k - q) . snd) $ M.toList $ solve aList $ M.fromList $ zip [1..n] $ repeat 0

Score: 300 points

Problem Statement

Takahashi has decided to hold fastest-finger-fast quiz games. Kizahashi, who is in charge of making the scoreboard, is struggling to write the program that manages the players' scores in a game, which proceeds as follows.

A game is played by N players, numbered 1 to N. At the beginning of a game, each player has K points.

When a player correctly answers a question, each of the other N-1 players receives minus one (-1) point. There is no other factor that affects the players' scores.

At the end of a game, the players with 0 points or lower are eliminated, and the remaining players survive.

In the last game, the players gave a total of Q correct answers, the i-th of which was given by Player A_i.
For Kizahashi, write a program that determines whether each of the N players survived this game.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq K \leq 10^9

1 \leq Q \leq 10^5

1 \leq A_i \leq N\ (1 \leq i \leq Q)

Input

Input is given from Standard Input in the following format:

N K Q
A_1
A_2
.
.
.
A_Q

Output

Print N lines. The i-th line should contain Yes if Player i survived the game, and No otherwise.

Sample Input 1

6 3 4
3
1
3
2

Sample Output 1

No
No
Yes
No
No
No

In the beginning, the players' scores are (3, 3, 3, 3, 3, 3).

Player 3 correctly answers a question. The players' scores are now (2, 2, 3, 2, 2, 2).

Player 1 correctly answers a question. The players' scores are now (2, 1, 2, 1, 1, 1).

Player 3 correctly answers a question. The players' scores are now (1, 0, 2, 0, 0, 0).

Player 2 correctly answers a question. The players' scores are now (0, 0, 1, -1, -1, -1).

Players 1, 2, 4, 5 and 6, who have 0 points or lower, are eliminated, and Player 3 survives this game.

Sample Input 2

6 5 4
3
1
3
2

Sample Output 2

Yes
Yes
Yes
Yes
Yes
Yes

Sample Input 3

10 13 15
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9

Sample Output 3

No
No
No
No
Yes
No
No
No
Yes
No


main = do
  n:k:q:_ <- map read . words <$> getLine :: IO [Integer]
  a <- map read . words <$> getContents :: IO [Integer]
  mapM (putStrLn . (\x -> if 0 < x then "Yes" else "No")) $ map (foldl step (\i -> k) a) [1..n]
      where step f i j = if j == i then f j else (f j - 1)


{-# OPTIONS_GHC -O2 #-}
{-# LANGUAGE BangPatterns #-}
module Main where

import Control.Arrow
import Control.Monad
import Control.Monad.ST
import Control.Monad.State.Strict
import Data.Char
import Data.Function
import Data.List
import Data.Maybe
import qualified Data.Vector as V
import qualified Data.Vector.Unboxed as U
import qualified Data.ByteString.Char8 as C
import qualified Data.ByteString.Unsafe as B

type Parser a = C.ByteString -> Maybe (a, C.ByteString)

parseInt :: Parser Int
parseInt = C.readInt . C.dropWhile isSpace

parseInt2 :: Parser (Int, Int)
parseInt2 = runStateT $
    (,) <$> StateT (C.readInt . C.dropWhile isSpace)
        <*> StateT (C.readInt . B.unsafeTail)

parseInt3 :: Parser (Int, Int, Int)
parseInt3 = runStateT $
    (,,) <$> StateT (C.readInt . C.dropWhile isSpace)
        <*> StateT (C.readInt . B.unsafeTail)
        <*> StateT (C.readInt . B.unsafeTail)

parseInt4 :: Parser (Int, Int, Int, Int)
parseInt4 = runStateT $
    (,,,) <$> StateT (C.readInt . C.dropWhile isSpace)
        <*> StateT (C.readInt . B.unsafeTail)
        <*> StateT (C.readInt . B.unsafeTail)
        <*> StateT (C.readInt . B.unsafeTail)

main :: IO ()
main = do
  !n <- readLn :: IO Int
  print $ n^3

Score : 100 points

Problem Statement

Takahashi is going to set a 3-character password.

How many possible passwords are there if each of its characters must be a digit between 1 and N (inclusive)?

Constraints

1 \leq N \leq 9

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of possible passwords.

Sample Input 1

2

Sample Output 1

8

There are eight possible passwords: 111, 112, 121, 122, 211, 212, 221, and 222.

Sample Input 2

1

Sample Output 2

1

There is only one possible password if you can only use one kind of character.


main :: IO ()
main = do
    n <- readLn :: IO Int
    a <- map read . words <$> getLine :: IO [Int]
    b <- map read . words <$> getLine :: IO [Int]
    c <- map read . words <$> getLine :: IO [Int]
    print $ sum [ let ai = a!!(i-1)
                      bi = b!!(ai-1)
                      ci = if i /= 1 && ai == (a!!(i-2) + 1)
                              then c!!(ai-2) else 0
             in bi + ci | i <- [1..n] ]

Score : 200 points

Problem Statement

Takahashi went to an all-you-can-eat buffet with N kinds of dishes and ate all of them (Dish 1, Dish 2, \ldots, Dish N) once.

The i-th dish (1 \leq i \leq N) he ate was Dish A_i.

When he eats Dish i (1 \leq i \leq N), he gains B_i satisfaction points.

Additionally, when he eats Dish i+1 just after eating Dish i (1 \leq i \leq N - 1), he gains C_i more satisfaction points.

Find the sum of the satisfaction points he gained.

Constraints

All values in input are integers.

2 \leq N \leq 20

1 \leq A_i \leq N

A_1, A_2, ..., A_N are all different.

1 \leq B_i \leq 50

1 \leq C_i \leq 50

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N
B_1 B_2 ... B_N
C_1 C_2 ... C_{N-1}

Output

Print the sum of the satisfaction points Takahashi gained, as an integer.

Sample Input 1

3
3 1 2
2 5 4
3 6

Sample Output 1

14

Takahashi gained 14 satisfaction points in total, as follows:

First, he ate Dish 3 and gained 4 satisfaction points.

Next, he ate Dish 1 and gained 2 satisfaction points.

Lastly, he ate Dish 2 and gained 5 + 3 = 8 satisfaction points.

Sample Input 2

4
2 3 4 1
13 5 8 24
45 9 15

Sample Output 2

74

Sample Input 3

2
1 2
50 50
50

Sample Output 3

150


import Data.List
main=do
 getLine
 a<-map read.words<$>getLine
 print$(head a)+(last a)+sum[min x y|(x,y)<-zip a$tail a]

Score : 300 points

Problem Statement

There is an integer sequence A of length N whose values are unknown.

Given is an integer sequence B of length N-1 which is known to satisfy the following:

B_i \geq \max(A_i, A_{i+1})

Find the maximum possible sum of the elements of A.

Constraints

All values in input are integers.

2 \leq N \leq 100

0 \leq B_i \leq 10^5

Input

Input is given from Standard Input in the following format:

N
B_1 B_2 ... B_{N-1}

Output

Print the maximum possible sum of the elements of A.

Sample Input 1

3
2 5

Sample Output 1

9

A can be, for example, ( 2 , 1 , 5 ), ( -1 , -2 , -3 ), or ( 2 , 2 , 5 ). Among those candidates, A = ( 2 , 2 , 5 ) has the maximum possible sum.

Sample Input 2

2
3

Sample Output 2

6

Sample Input 3

6
0 153 10 10 23

Sample Output 3

53


solve :: [Int] -> Int -> [Int]
solve [] prev = [prev]
solve (b:bs) prev = prev' : solve bs b
  where prev' = if prev <= b then prev else b

main :: IO ()
main = do
  getLine
  bs <- reverse . map read . words <$> getLine
  print . sum . solve bs $ head bs


import System.IO (hSetBuffering, stdin, BufferMode (NoBuffering))

main :: IO ()
main = do
  [n, k] <- map read . words <$> getLine :: IO [Int]
  hSetBuffering stdin NoBuffering
  ds <- getContents :: IO String
  let
    s = length . filter id . (zipWith (==) <$> id <*> tail) $ ds :: Int
  print $ (s + 2 * k) `min` (n - 1)

Score : 400 points

Problem Statement

There are N people standing in a queue from west to east.

Given is a string S of length N representing the directions of the people.
The i-th person from the west is facing west if the i-th character of S is L, and east if that character of S is R.

A person is happy if the person in front of him/her is facing the same direction.
If no person is standing in front of a person, however, he/she is not happy.

You can perform the following operation any number of times between 0 and K (inclusive):

Operation: Choose integers l and r such that 1 \leq l \leq r \leq N, and rotate by 180 degrees the part of the queue: the l-th, (l+1)-th, ..., r-th persons. That is, for each i = 0, 1, ..., r-l, the (l + i)-th person from the west will stand the (r - i)-th from the west after the operation, facing east if he/she is facing west now, and vice versa.

What is the maximum possible number of happy people you can have?

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

|S| = N

Each character of S is L or R.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of happy people after at most K operations.

Sample Input 1

6 1
LRLRRL

Sample Output 1

3

If we choose (l, r) = (2, 5), we have LLLRLL, where the 2-nd, 3-rd, and 6-th persons from the west are happy.

Sample Input 2

13 3
LRRLRLRRLRLLR

Sample Output 2

9

Sample Input 3

10 1
LLLLLRRRRR

Sample Output 3

9

Sample Input 4

9 2
RRRLRLRLL

Sample Output 4

7


main :: IO ()
main = do
  [n, k] <- map read . words <$> getLine :: IO [Int]
  ds <- getLine :: IO String
  let
    s = length . filter id . (zipWith (==) <$> id <*> tail) $ ds :: Int
  print $ (s + 2 * k) `min` (n - 1)


import qualified Data.Text.Lazy as T
import qualified Data.Text.Lazy.IO as T
import qualified Data.Text.Lazy.Read as T
import Data.List (group)

main :: IO ()
main = do
  let
    f :: Int -> Int -> Int
    f k m = (2 * k) `min` (m - 1)
  [n, k] <- map unsafeTextToInt . T.words <$> T.getLine :: IO [Int]
  ds <- map ('R' ==) . T.unpack <$> T.getLine :: IO [Bool]
  let
    (m, s) = ((,) <$> length <*> sum . map (pred . length)) . group $ ds :: (Int, Int)
  print $ s + f k m

unsafeTextToInt :: T.Text -> Int
unsafeTextToInt s = case T.signed T.decimal s of
  Right (n, _) -> n
{-# INLINE unsafeTextToInt #-}


main = do
  s <- getLine
  t <- getLine
  let ans = length $ filter id $ zipWith (==) s t
  print ans

Score : 100 points

Problem Statement

You will be given a string S of length 3 representing the weather forecast for three days in the past.

The i-th character (1 \leq i \leq 3) of S represents the forecast for the i-th day. S, C, and R stand for sunny, cloudy, and rainy, respectively.

You will also be given a string T of length 3 representing the actual weather on those three days.

The i-th character (1 \leq i \leq 3) of S represents the actual weather on the i-th day. S, C, and R stand for sunny, cloudy, and rainy, respectively.

Print the number of days for which the forecast was correct.

Constraints

S and T are strings of length 3 each.

S and T consist of S, C, and R.

Input

Input is given from Standard Input in the following format:

S
T

Output

Print the number of days for which the forecast was correct.

Sample Input 1

CSS
CSR

Sample Output 1

2

For the first day, it was forecast to be cloudy, and it was indeed cloudy.

For the second day, it was forecast to be sunny, and it was indeed sunny.

For the third day, it was forecast to be sunny, but it was rainy.

Thus, the forecast was correct for two days in this case.

Sample Input 2

SSR
SSR

Sample Output 2

3

Sample Input 3

RRR
SSS

Sample Output 3

0


-- {-# LANGUAGE BangPatterns #-}
-- {-# LANGUAGE ScopedTypeVariables #-}
-- {-# LANGUAGE FlexibleContexts #-}
-- {-# LANGUAGE ViewPatterns #-}
-- {-# LANGUAGE OverloadedStrings #-}

import qualified Data.Vector as V
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
-- import qualified Data.Vector.Algorithms.Merge as VAM
-- import qualified Data.Heap as HP
import qualified Data.ByteString.Char8 as BC
import qualified Data.Map as MP
import qualified Data.Set as ST
import qualified Data.Sequence as SQ
import Data.Array
import Data.Array.Unboxed
import Data.Array.ST
import Data.Array.IO
import Data.Maybe
import Data.Ratio
import Data.Char
import Data.Bits
import Data.List
import Data.Tuple
import Data.Functor
import Data.IORef
import Data.STRef
import Control.Applicative
import Control.Monad
import Control.Monad.ST
import Debug.Trace

--------------------------------------------------------------------------
solve a b = 1 + (ceiling $ (rest / (a - 1)))
    where
        rest = b - a

main = do
    [a,b] <- singleLineToDoubleL
    print $ solve a b
--------------------------------------------------------------------------

-- Input Util
int :: IO Int
int = readLn

double :: IO Double
double = readLn

str :: IO String
str = getLine

strBC :: IO BC.ByteString
strBC = BC.getLine

singleLineToIntL :: IO [Int]
singleLineToIntL = getLine >>= return . map read . words

singleLineToDoubleL :: IO [Double]
singleLineToDoubleL = getLine >>= return . map read . words

singleLineToIntV :: Int -> IO (VU.Vector Int)
singleLineToIntV n = BC.getLine >>= return . VU.unfoldrN n (BC.readInt . BC.dropWhile isSpace)

singleLineToDoubleV :: Int -> IO (VU.Vector Double)
singleLineToDoubleV n = BC.getLine >>= return . VU.fromList . map (read . BC.unpack) . BC.words

multipleLinesToIntL :: Int -> IO [Int]
multipleLinesToIntL n = replicateM n readLn

multipleLinesToDoubleL :: Int -> IO [Double]
multipleLinesToDoubleL n = replicateM n readLn

multipleLinesToIntV :: Int -> IO (VU.Vector Int)
multipleLinesToIntV n = VU.replicateM n readLn

multipleLinesToDoubleV :: Int -> IO (VU.Vector Double)
multipleLinesToDoubleV n = VU.replicateM n (BC.getLine >>= return . read . BC.unpack)

multipleLinesToTupleL :: Int -> IO [(Int, Int)]
multipleLinesToTupleL n = replicateM n (getLine >>= return . (\[a,b] -> (a,b)) . map read . words)

multipleLinesToTupleV :: Int -> IO (VU.Vector (Int, Int))
multipleLinesToTupleV n = VU.replicateM n $ do
    bs <- strBC
    let
        Just (a, bs') = BC.readInt $ BC.dropWhile isSpace bs
        Just (b, _) = BC.readInt $ BC.dropWhile isSpace bs'
    return (a,b)

-- others
strToChr :: String -> Char
strToChr [x] = x

chrToStr :: Char -> String
chrToStr x = [x]

m1 :: Int -> Int
m1 x = subtract 1 x

toDouble :: Int -> Double
toDouble x = fromIntegral x

toInt :: Double -> Int
toInt x = floor x

Score : 200 points

Problem Statement

Takahashi's house has only one socket.

Takahashi wants to extend it with some number of power strips, each with A sockets, into B or more empty sockets.

One power strip with A sockets can extend one empty socket into A empty sockets.

Find the minimum number of power strips required.

Constraints

All values in input are integers.

2 \leq A \leq 20

1 \leq B \leq 20

Input

Input is given from Standard Input in the following format:

A B

Output

Print the minimum number of power strips required.

Sample Input 1

4 10

Sample Output 1

3

3 power strips, each with 4 sockets, extend the socket into 10 empty sockets.

Sample Input 2

8 9

Sample Output 2

2

2 power strips, each with 8 sockets, extend the socket into 15 empty sockets.

Sample Input 3

8 8

Sample Output 3

1


{-# LANGUAGE BangPatterns #-}

main :: IO ()
main = do
  [a, b] <- map read . words <$> getLine
  print $ solve a b

solve :: Int -> Int -> Int
solve a b = go 0 1
  where
    go !acc o
      | o >= b    = acc
      | otherwise = go (acc + 1) (o + a - 1)


module Main (main) where

import qualified Data.ByteString.Char8 as C8
import Data.Maybe

main :: IO ()
main = readInts >>= print . solve

solve [a,b]
  | b == 1 = 0
  | otherwise = q + signum r
  where
    (q,r) = b `divMod` a

-- utils
readInts :: IO [Int]
readInts = map (fst . fromJust . C8.readInt) . C8.words <$> C8.getLine


{-# LANGUAGE TypeApplications #-}

import           Control.Monad
import           Control.Applicative
import           Data.List
import           Data.Char
import           Control.Arrow
import           Data.Bits

main = do
  n <- readLn @Int
  h <- map (read @Int) . words <$> getLine
  print $ solve h 0 0
 where
  solve [_] k m = m
  solve (a : b : rs) k m | a >= b    = solve (b : rs) (k + 1) (max (k + 1) m)
                         | otherwise = solve (b : rs) 0 m

Score : 300 points

Problem Statement

There are N squares arranged in a row from left to right.

The height of the i-th square from the left is H_i.

You will land on a square of your choice, then repeat moving to the adjacent square on the right as long as the height of the next square is not greater than that of the current square.

Find the maximum number of times you can move.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq H_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
H_1 H_2 ... H_N

Output

Print the maximum number of times you can move.

Sample Input 1

5
10 4 8 7 3

Sample Output 1

2

By landing on the third square from the left, you can move to the right twice.

Sample Input 2

7
4 4 5 6 6 5 5

Sample Output 2

3

By landing on the fourth square from the left, you can move to the right three times.

Sample Input 3

4
1 2 3 4

Sample Output 3

0


main :: IO ()
main = do
    n <- readLn :: IO Int
    let x:xs = [1..n]
    print . sum $ map (\(i,j) -> i `mod` j) $ zip [1..n] xs

Score : 400 points

Problem Statement

For an integer N, we will choose a permutation \{P_1, P_2, ..., P_N\} of \{1, 2, ..., N\}.

Then, for each i=1,2,...,N, let M_i be the remainder when i is divided by P_i.

Find the maximum possible value of M_1 + M_2 + \cdots + M_N.

Constraints

N is an integer satisfying 1 \leq N \leq 10^9.

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum possible value of M_1 + M_2 + \cdots + M_N.

Sample Input 1

2

Sample Output 1

1

When the permutation \{P_1, P_2\} = \{2, 1\} is chosen, M_1 + M_2 = 1 + 0 = 1.

Sample Input 2

13

Sample Output 2

78

Sample Input 3

1

Sample Output 3

0


module Main where

import Control.Monad

main :: IO ()
main = do
  n <- readLn :: IO Integer
  print $ n * (n + 1) `div` 2

  -- forM_ [1..10] $ \n -> do
  --   let xs = [1..n]
  --   print $ maximum $ map (sum . modOrig xs) $ perm2 xs n

-- modOrig x y = zipWith mod y x

-- perm2 :: Eq a => [a] -> Int -> [[a]]
-- perm2 _ 0 = [[]]
-- perm2 xs n = [x:ys | x <- xs, ys <- perm2 xs (n-1)]


{-# LANGUAGE OverloadedStrings #-}
import Data.Maybe
import Data.Char
import Data.List
import qualified Data.ByteString.Char8 as BS
import qualified Data.Vector.Unboxed as VU

modInt = 10^9+7

main = do
  [n, k] <- getIntList
  ns <- getIntList
  let ms = mkpair [] ns
  print . foldr (((`mod` modInt).).(+)) 0 . map (f k) $ ms
    where
      both f (a, b) = (f a, f b)
      mkpair _ [] = []
      mkpair ps (q:qs) = both (length . filter (<q)) (ps, qs) : mkpair (q:ps) qs
      f k (p, q) = k * ((p + q) * k + (q - p)) `div` 2

-- IO
readInt :: BS.ByteString -> Int
readInt = fst . fromJust . BS.readInt

getInt :: IO Int
getInt = readInt <$> BS.getLine

getIntList :: IO [Int]
getIntList = unfoldr (BS.readInt . BS.dropWhile isSpace) <$> BS.getLine

getIntVec :: Int -> IO (VU.Vector Int)
getIntVec n = VU.unfoldrN n (BS.readInt . BS.dropWhile isSpace) <$> BS.getLine

Score : 300 points

Problem Statement

We have a sequence of N integers A~=~A_0,~A_1,~...,~A_{N - 1}.

Let B be a sequence of K \times N integers obtained by concatenating K copies of A. For example, if A~=~1,~3,~2 and K~=~2, B~=~1,~3,~2,~1,~3,~2.

Find the inversion number of B, modulo 10^9 + 7.

Here the inversion number of B is defined as the number of ordered pairs of integers (i,~j)~(0 \leq i < j \leq K \times N - 1) such that B_i > B_j.

Constraints

All values in input are integers.

1 \leq N \leq 2000

1 \leq K \leq 10^9

1 \leq A_i \leq 2000

Input

Input is given from Standard Input in the following format:

N K
A_0 A_1 ... A_{N - 1}

Output

Print the inversion number of B, modulo 10^9 + 7.

Sample Input 1

2 2
2 1

Sample Output 1

3

In this case, B~=~2,~1,~2,~1. We have:

B_0 > B_1

B_0 > B_3

B_2 > B_3

Thus, the inversion number of B is 3.

Sample Input 2

3 5
1 1 1

Sample Output 2

0

A may contain multiple occurrences of the same number.

Sample Input 3

10 998244353
10 9 8 7 5 6 3 4 2 1

Sample Output 3

185297239

Be sure to print the output modulo 10^9 + 7.


module Main where

m = 1000000007

main :: IO ()
main = do
  [n, k] <- map (read :: String -> Integer) . words <$> getLine
  as <- map (read :: String -> Int) . words <$> getLine
  print $ solve n k as
  where
    solve n k as =
      let
        x = f as
      in x * (k * (k + 1) `div` 2) `mod` m

f :: [Int] -> Integer
f [] = 0
f (a:as) = fromIntegral (length (filter (a >) as)) + f as


-- https://github.com/minoki/my-atcoder-solutions
{-# LANGUAGE BangPatterns #-}
import Data.Int (Int64)
import Data.Coerce
import qualified Data.ByteString.Char8 as BS
import Control.Exception (assert)

main = do
  n <- readLn
  s <- BS.getLine
  assert (BS.length s == 2 * n) $ return ()
  let loop :: N -> Int -> Int -> N
      loop !acc !k !i | BS.length s == i = if k == 0 then acc else 0
                      | BS.index s i == 'W' = if even k
                                              then loop (acc * fromIntegral k) (k-1) (i+1)
                                              else loop acc (k+1) (i+1)
                      | otherwise = if even k
                                    then loop acc (k+1) (i+1)
                                    else loop (acc * fromIntegral k) (k-1) (i+1)
  print $ loop (product (coerce ([1..fromIntegral n] :: [Int64]) :: [N])) 0 0

--
-- Modular Arithmetic
--

modulo :: Int64
modulo = 10^9+7
addMod, subMod, mulMod :: Int64 -> Int64 -> Int64
addMod !x !y | x + y >= modulo = x + y - modulo
             | otherwise = x + y
subMod !x !y | x >= y = x - y
             | otherwise = x - y + modulo
mulMod !x !y = (x * y) `rem` modulo

newtype N = N { unwrapN :: Int64 } deriving (Eq)
instance Show N where
  show (N x) = show x
instance Num N where
  (+) = coerce addMod
  (-) = coerce subMod
  (*) = coerce mulMod
  fromInteger n = N (fromInteger (n `mod` fromIntegral modulo))
  abs = undefined; signum = undefined

{-# RULES
"^9/Int" forall x. x ^ (9 :: Int) = let u = x; v = u * u * u in v * v * v
"^9/Integer" forall x. x ^ (9 :: Integer) = let u = x; v = u * u * u in v * v * v
 #-}

Score : 500 points

Problem Statement

There are 2N squares arranged from left to right. You are given a string of length 2N representing the color of each of the squares.

The color of the i-th square from the left is black if the i-th character of S is B, and white if that character is W.

You will perform the following operation exactly N times: choose two distinct squares, then invert the colors of these squares and the squares between them. Here, to invert the color of a square is to make it white if it is black, and vice versa.

Throughout this process, you cannot choose the same square twice or more. That is, each square has to be chosen exactly once.

Find the number of ways to make all the squares white at the end of the process, modulo 10^9+7.

Two ways to make the squares white are considered different if and only if there exists i (1 \leq i \leq N) such that the pair of the squares chosen in the i-th operation is different.

Constraints

1 \leq N \leq 10^5

|S| = 2N

Each character of S is B or W.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of ways to make all the squares white at the end of the process, modulo 10^9+7. If there are no such ways, print 0.

Sample Input 1

2
BWWB

Sample Output 1

4

There are four ways to make all the squares white, as follows:

Choose Squares 1, 3 in the first operation, and choose Squares 2, 4 in the second operation.

Choose Squares 2, 4 in the first operation, and choose Squares 1, 3 in the second operation.

Choose Squares 1, 4 in the first operation, and choose Squares 2, 3 in the second operation.

Choose Squares 2, 3 in the first operation, and choose Squares 1, 4 in the second operation.

Sample Input 2

4
BWBBWWWB

Sample Output 2

288

Sample Input 3

5
WWWWWWWWWW

Sample Output 3

0


main = do
 getLine
 a <- map read . words <$> getLine
 print (f2 a)

f1 :: Double -> Double -- 逆数
f1 x = 1 / x

f2 :: [Double] -> Double
f2 xs = 1 / sum (map f1 xs)

Score : 200 points

Problem Statement

Given is a sequence of N integers A_1, \ldots, A_N.

Find the (multiplicative) inverse of the sum of the inverses of these numbers, \frac{1}{\frac{1}{A_1} + \ldots + \frac{1}{A_N}}.

Constraints

1 \leq N \leq 100

1 \leq A_i \leq 1000

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \ldots A_N

Output

Print a decimal number (or an integer) representing the value of \frac{1}{\frac{1}{A_1} + \ldots + \frac{1}{A_N}}.

Your output will be judged correct when its absolute or relative error from the judge's output is at most 10^{-5}.

Sample Input 1

2
10 30

Sample Output 1

7.5

\frac{1}{\frac{1}{10} + \frac{1}{30}} = \frac{1}{\frac{4}{30}} = \frac{30}{4} = 7.5.

Printing 7.50001, 7.49999, and so on will also be accepted.

Sample Input 2

3
200 200 200

Sample Output 2

66.66666666666667

\frac{1}{\frac{1}{200} + \frac{1}{200} + \frac{1}{200}} = \frac{1}{\frac{3}{200}} = \frac{200}{3} = 66.6666....

Printing 6.66666e+1 and so on will also be accepted.

Sample Input 3

1
1000

Sample Output 3

1000

\frac{1}{\frac{1}{1000}} = 1000.

Printing +1000.0 and so on will also be accepted.


import           Control.Monad
import           Data.Maybe
import           Data.List
import qualified Data.ByteString.Char8         as BS

tuplify2 (x : y : _) = (x, y)
tuplify2 _           = undefined

--Input functions with ByteString
readInt = fst . fromJust . BS.readInteger
readIntTuple = tuplify2 . map readInt . BS.words
readIntList = map readInt . BS.words

getInt = readInt <$> BS.getLine
getIntList = readIntList <$> BS.getLine
getIntNList n = map readIntList <$> replicateM (fromIntegral n) BS.getLine
getIntMatrix = map readIntList . BS.lines <$> BS.getContents
getIntTuple = readIntTuple <$> BS.getLine
getIntNTuples n = map readIntTuple <$> replicateM (fromIntegral n) BS.getLine
getIntTuples = map readIntTuple . BS.lines <$> BS.getContents

main = do
    n    <- getInt
    nums <- map read . words <$> getLine
    print (solve nums)

solve xs = foldl1 (\x y -> (x + y) / 2) $ sort xs

Score : 300 points

Problem Statement

You have a pot and N ingredients. Each ingredient has a real number parameter called value, and the value of the i-th ingredient (1 \leq i \leq N) is v_i.

When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.

After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Constraints

2 \leq N \leq 50

1 \leq v_i \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
v_1 v_2 \ldots v_N

Output

Print a decimal number (or an integer) representing the maximum possible value of the last ingredient remaining.

Your output will be judged correct when its absolute or relative error from the judge's output is at most 10^{-5}.

Sample Input 1

2
3 4

Sample Output 1

3.5

If you start with two ingredients, the only choice is to put both of them in the pot. The value of the ingredient resulting from the ingredients of values 3 and 4 is (3 + 4) / 2 = 3.5.

Printing 3.50001, 3.49999, and so on will also be accepted.

Sample Input 2

3
500 300 200

Sample Output 2

375

You start with three ingredients this time, and you can choose what to use in the first composition. There are three possible choices:

Use the ingredients of values 500 and 300 to produce an ingredient of value (500 + 300) / 2 = 400. The next composition will use this ingredient and the ingredient of value 200, resulting in an ingredient of value (400 + 200) / 2 = 300.

Use the ingredients of values 500 and 200 to produce an ingredient of value (500 + 200) / 2 = 350. The next composition will use this ingredient and the ingredient of value 300, resulting in an ingredient of value (350 + 300) / 2 = 325.

Use the ingredients of values 300 and 200 to produce an ingredient of value (300 + 200) / 2 = 250. The next composition will use this ingredient and the ingredient of value 500, resulting in an ingredient of value (250 + 500) / 2 = 375.

Thus, the maximum possible value of the last ingredient remaining is 375.

Printing 375.0 and so on will also be accepted.

Sample Input 3

5
138 138 138 138 138

Sample Output 3

138


import Control.Monad
import qualified Data.ByteString.Char8 as BS
import Data.List
import Data.Maybe

readInt = fst . fromJust . BS.readInt

readString = map BS.unpack . BS.words

getInt = readInt <$> BS.getLine

getNString n = map readString <$> replicateM n BS.getLine

countS :: String -> [String] -> Integer -> [String] -> (Integer, [String])
countS s' [] n ts = (n, ts)
countS s' (s : ss) n ts
  | s' == s = countS s' ss (n + 1) ts
  | otherwise = countS s' ss n (s : ts)

getS :: [String] -> Integer
getS [] = 0
getS ss = n + getS ts
  where
    (k, ts) = countS (head ss) ss 0 []
    n = (k * (k -1)) `div` 2

main = do
  n <- getInt
  ss <- getNString n
  let ss' = map (\x -> sort (head x)) ss
  print $ getS ss'

Score : 300 points

Problem Statement

We will call a string obtained by arranging the characters contained in a string a in some order, an anagram of a.

For example, greenbin is an anagram of beginner. As seen here, when the same character occurs multiple times, that character must be used that number of times.

Given are N strings s_1, s_2, \ldots, s_N. Each of these strings has a length of 10 and consists of lowercase English characters. Additionally, all of these strings are distinct. Find the number of pairs of integers i, j (1 \leq i < j \leq N) such that s_i is an anagram of s_j.

Constraints

2 \leq N \leq 10^5

s_i is a string of length 10.

Each character in s_i is a lowercase English letter.

s_1, s_2, \ldots, s_N are all distinct.

Input

Input is given from Standard Input in the following format:

N
s_1
s_2
:
s_N

Output

Print the number of pairs of integers i, j (1 \leq i < j \leq N) such that s_i is an anagram of s_j.

Sample Input 1

3
acornistnt
peanutbomb
constraint

Sample Output 1

1

s_1 = acornistnt is an anagram of s_3 = constraint. There are no other pairs i, j such that s_i is an anagram of s_j, so the answer is 1.

Sample Input 2

2
oneplustwo
ninemodsix

Sample Output 2

0

If there is no pair i, j such that s_i is an anagram of s_j, print 0.

Sample Input 3

5
abaaaaaaaa
oneplustwo
aaaaaaaaba
twoplusone
aaaabaaaaa

Sample Output 3

4

Note that the answer may not fit into a 32-bit integer type, though we cannot put such a case here.


{-# LANGUAGE FlexibleContexts #-}
{-# LANGUAGE BangPatterns #-}
{-# LANGUAGE TypeFamilies #-}
import           Data.Maybe
import           Data.Int
import           Data.List
import           Data.Char
  -- import qualified Data.Vector.Unboxing as VU
import qualified Data.Vector as V
import qualified Data.Vector.Mutable as VM
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.ByteString.Char8 as BS
import           Data.List (scanl')
import           Data.IORef
import           Control.Monad
import           Control.Monad.ST
import           Control.Monad.State.Strict
import           Control.Applicative
import           Debug.Trace

readInt = BS.readInt . BS.dropWhile isSpace
getVector f = VU.unfoldr f <$> BS.getLine
getIntVector = getVector readInt
getBSVector = getVector BS.uncons

main = do
  n <- readLn
  ss <- replicateM n $ sort <$> getLine
  print . sum . map (comb' . (\n -> length n)) . group . sort $ ss

comb x y = (loop1 x y 1) `div` (loop2 y 1)
  where
    loop1 _ 0 n' = n'
    loop1 n r n' = loop1 (n-1) (r-1) (n*n')
    loop2 0 n' = n'
    loop2 n n' = loop2 (n-1) (n*n')

comb' x = comb x 2


import Control.Monad
import Data.List
main = do
    n <- getLine >>= return . (read :: String -> Int)
    s <- replicateM n getLine
    putStrLn . show . sum . map ((\x -> x * (x-1) `div` 2) . length) . group . sort $ map sort s


import           Control.Monad
import           Data.List

fact :: Int -> Int
fact 0 = 1
fact n = n * fact (n - 1)

main = do
    n <- readLn :: IO Int
    s <- sort . map sort <$> replicateM n getLine :: IO [String]
    print . sum . map (\i -> (fact i) `div` 2) . map length $ group s


import Control.Monad
import Text.Printf
import Data.List

getInts :: IO [Int]
getInts = map read . words <$> getLine

main :: IO()
main = do
  [n] <- getInts
  sn <- replicateM n $ getLine
  printf "%d" $ slv sn

slv :: [String] -> Integer
slv sn = sum $ map (\x -> (x*(x-1)) `div` 2 ) $ map (fromIntegral.length) $ group $ sort $ map sort sn


main :: IO ()
main = do
  [a, b, c] <- ((map read) . words) <$> getLine
  print $ c - a + b

Score : 100 points

Problem Statement

We have two bottles for holding water.

Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.

Bottle 2 contains C milliliters of water.

We will transfer water from Bottle 2 to Bottle 1 as much as possible.

How much amount of water will remain in Bottle 2?

Constraints

All values in input are integers.

1 \leq B \leq A \leq 20

1 \leq C \leq 20

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.

Sample Input 1

6 4 3

Sample Output 1

1

We will transfer two milliliters of water from Bottle 2 to Bottle 1, and one milliliter of water will remain in Bottle 2.

Sample Input 2

8 3 9

Sample Output 2

4

Sample Input 3

12 3 7

Sample Output 3

0


main :: IO()
main = do
  [a, b, c] <- map read . words <$> getLine
  print $ slv a b c

slv :: Integer -> Integer -> Integer -> Integer
slv a b c | (b + c) - a >= 0 = (b + c) - a
          | otherwise = 0


import qualified Data.Vector.Unboxed as VU
import Control.Applicative

main :: IO ()
main = do
  n <- read <$> getLine :: IO Int
  print $ length $ takeWhile (<= n)
    $ do x <- iterate (*100) 1
         [x..x*10-1]

Score : 200 points

Problem Statement

Given is an integer N. Find the number of positive integers less than or equal to N that have an odd number of digits (in base ten without leading zeros).

Constraints

1 \leq N \leq 10^5

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of positive integers less than or equal to N that have an odd number of digits.

Sample Input 1

11

Sample Output 1

9

Among the positive integers less than or equal to 11, nine integers have an odd number of digits: 1, 2, \ldots, 9.

Sample Input 2

136

Sample Output 2

46

In addition to 1, 2, \ldots, 9, another 37 integers also have an odd number of digits: 100, 101, \ldots, 136.

Sample Input 3

100000

Sample Output 3

90909


{-# LANGUAGE Strict #-}
import qualified Data.ByteString.Char8 as B
import qualified Data.Vector.Unboxed   as V

main :: IO ()
main = do
  n <- readLn
  ps <- readIntsLn
  putStrLn $ if solve n ps then "YES" else "NO"

solve :: Int -> V.Vector Int -> Bool
solve _ ps = l <= 2
  where l = V.length $ V.ifilter (\i x -> (i + 1) /= x) ps

readIntsLn :: IO (V.Vector Int)
readIntsLn = V.unfoldr (B.readInt . B.dropWhile (== ' ')) <$> B.getLine

Score : 200 points

Problem Statement

We have a sequence p = {p_1,\ p_2,\ ...,\ p_N} which is a permutation of {1,\ 2,\ ...,\ N}.

You can perform the following operation at most once: choose integers i and j (1 \leq i < j \leq N), and swap p_i and p_j. Note that you can also choose not to perform it.

Print YES if you can sort p in ascending order in this way, and NO otherwise.

Constraints

All values in input are integers.

2 \leq N \leq 50

p is a permutation of {1,\ 2,\ ...,\ N}.

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 ... p_N

Output

Print YES if you can sort p in ascending order in the way stated in the problem statement, and NO otherwise.

Sample Input 1

5
5 2 3 4 1

Sample Output 1

YES

You can sort p in ascending order by swapping p_1 and p_5.

Sample Input 2

5
2 4 3 5 1

Sample Output 2

NO

In this case, swapping any two elements does not sort p in ascending order.

Sample Input 3

7
1 2 3 4 5 6 7

Sample Output 3

YES

p is already sorted in ascending order, so no operation is needed.


main=do
 n<-readLn
 p<-map read.words<$>getLine
 putStrLn(if(f p n)<3 then"YES"else"NO")
  where f p n =sum[if x/=p!!(x-1) then 1 else 0|x<-[1..n]]


solve :: [Int] -> Int -> Int -> Int
solve [] _ res = res
solve (x:xs) cnt res = solve xs (cnt + 1) res'
  where res' = if x == cnt
               then res
               else res + 1

main :: IO ()
main = do
  getLine
  input <- map read . words <$> getLine
  putStrLn $ if solve input 1 0 <= 2 then "YES" else "NO"


{-# LANGUAGE BangPatterns #-}
import Control.Monad
import Data.Maybe
import Debug.Trace
import Data.List
import qualified Data.Map.Strict as Map
import qualified Data.IntMap.Strict as IntMap
import qualified Data.Set as Set
import qualified Data.IntSet as IntSet
import Data.Functor
import Data.Array
-- import Data.Array.Unboxed
import Control.Monad.ST
import Data.Array.ST

main :: IO ()
main = do
  n <- readLn :: IO Int
  as <- map read . words <$> getLine :: IO [Int]
  bs <- map read . words <$> getLine :: IO [Int]
  print $ solve n as bs

solve :: Int -> [Int] -> [Int] -> Int
solve n as bs = x + min y (last as)
  where
    (x, y) = foldl' f (0, 0) $ zip as bs
    f :: (Int, Int) -> (Int, Int) -> (Int, Int)
    f !(!count, !left) !(!a, !b) =
      (
        count + min a (b + left),
        max 0 $ b - max 0 (a - left)
      )

Score : 300 points

Problem Statement

There are N+1 towns. The i-th town is being attacked by A_i monsters.

We have N heroes. The i-th hero can defeat monsters attacking the i-th or (i+1)-th town, for a total of at most B_i monsters.

What is the maximum total number of monsters the heroes can cooperate to defeat?

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_{N+1}
B_1 B_2 ... B_N

Output

Print the maximum total number of monsters the heroes can defeat.

Sample Input 1

2
3 5 2
4 5

Sample Output 1

9

If the heroes choose the monsters to defeat as follows, they can defeat nine monsters in total, which is the maximum result.

The first hero defeats two monsters attacking the first town and two monsters attacking the second town.

The second hero defeats three monsters attacking the second town and two monsters attacking the third town.

Sample Input 2

3
5 6 3 8
5 100 8

Sample Output 2

22

Sample Input 3

2
100 1 1
1 100

Sample Output 3

3


{-# LANGUAGE BangPatterns #-}

import qualified Data.ByteString.Char8 as BS
import Data.Char
import Debug.Trace
import Data.List (unfoldr)
import Control.Monad
import qualified Data.Vector.Unboxed as UV

main = do
    n <- UV.head . UV.unfoldr BS.readInt <$> BS.getLine
    a <- UV.unfoldr (BS.readInt . BS.dropWhile isSpace) <$> BS.getLine
    b <- UV.unfoldr (BS.readInt . BS.dropWhile isSpace) <$> BS.getLine
    -- let l = UV.length $ UV.filter not $ UV.zipWith (==) p (UV.fromList [1..n])
    -- putStrLn $ if l <= 2 then "YES" else "NO"
    print $ sub3 0 a b

sub3 :: Int -> UV.Vector Int -> UV.Vector Int -> Int
sub3 m0 a b
    | b == UV.empty = 0
    | otherwise     = minX + sub3 m (UV.tail a) (UV.tail b)
    where t1 = UV.head a - m0
          t2 = UV.last $ UV.take 2 a
          r = UV.head b
          minX = min (t1 + t2) r
          m | r <= t1        = 0
             | r <= (t1 + t2) = r - t1
             | otherwise      = t2


solve :: Int -> String
solve r = show $ 3 * r * r

main :: IO()
main = do
    line <- getLine
    let r = read line :: Int
    putStrLn $ solve r

Score : 100 points

Problem Statement

It is known that the area of a regular dodecagon inscribed in a circle of radius a is 3a^2.

Given an integer r, find the area of a regular dodecagon inscribed in a circle of radius r.

Constraints

1 \leq r \leq 100

r is an integer.

Input

Input is given from Standard Input in the following format:

r

Output

Print an integer representing the area of the regular dodecagon.

Sample Input 1

4

Sample Output 1

48

The area of the regular dodecagon is 3 \times 4^2 = 48.

Sample Input 2

15

Sample Output 2

675

Sample Input 3

80

Sample Output 3

19200


main = do
  li <- getLine
  let r = (read li) :: Int
  let ans = 3 * r * r
  print ans


solve :: Int -> Int -> String
solve n d = show $ (n + 2 * d) `div` (2 * d + 1)

main :: IO()
main = do
    line <- getLine
    let n:d:_ = map read $ words line
    putStrLn $ solve n d

Score : 200 points

Problem Statement

There are N apple trees in a row. People say that one of them will bear golden apples.

We want to deploy some number of inspectors so that each of these trees will be inspected.

Each inspector will be deployed under one of the trees. For convenience, we will assign numbers from 1 through N to the trees. An inspector deployed under the i-th tree (1 \leq i \leq N) will inspect the trees with numbers between i-D and i+D (inclusive).

Find the minimum number of inspectors that we need to deploy to achieve the objective.

Constraints

All values in input are integers.

1 \leq N \leq 20

1 \leq D \leq 20

Input

Input is given from Standard Input in the following format:

N D

Output

Print the minimum number of inspectors that we need to deploy to achieve the objective.

Sample Input 1

6 2

Sample Output 1

2

We can achieve the objective by, for example, placing an inspector under Tree 3 and Tree 4.

Sample Input 2

14 3

Sample Output 2

2

Sample Input 3

20 4

Sample Output 3

3


import Control.Monad
import Data.List

main = do
    n <- readLn
    a <- replicateM n (readLn :: IO Int)
    let b = sort a
    let lb = last b
        llb = last (init b)
    mapM_ print $ map (\x -> if x == lb then llb else lb) a

Score : 300 points

Problem Statement

You are given a sequence of length N: A_1, A_2, ..., A_N.
For each integer i between 1 and N (inclusive), answer the following question:

Find the maximum value among the N-1 elements other than A_i in the sequence.

Constraints

2 \leq N \leq 200000

1 \leq A_i \leq 200000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print N lines. The i-th line (1 \leq i \leq N) should contain the maximum value among the N-1 elements other than A_i in the sequence.

Sample Input 1

3
1
4
3

Sample Output 1

4
3
4

The maximum value among the two elements other than A_1, that is, A_2 = 4 and A_3 = 3, is 4.

The maximum value among the two elements other than A_2, that is, A_1 = 1 and A_3 = 3, is 3.

The maximum value among the two elements other than A_3, that is, A_1 = 1 and A_2 = 4, is 4.

Sample Input 2

2
5
5

Sample Output 2

5
5


{-# LANGUAGE
  ScopedTypeVariables, BangPatterns, TupleSections, ExplicitForAll,
  LambdaCase, MultiWayIf, Unsafe, RecordWildCards, FlexibleContexts, CPP,
  NoMonomorphismRestriction, GADTs, PatternGuards, MagicHash, UnboxedTuples,
  RankNTypes, EmptyDataDecls, EmptyCase, ViewPatterns, PolyKinds,
  TypeFamilies #-}
{-# OPTIONS_GHC -O2 #-}

import Prelude
import Data.Bits
import Data.List
import Data.Maybe
import Data.Tuple
import Data.Ord
import Data.Int
import Data.Word
import Data.Char
import Data.Ratio
import Data.Function
import Data.STRef
import Data.IORef
import Data.Monoid
import Data.Functor
import System.IO
import Control.Arrow
import Control.Applicative
import Control.Monad
import Control.Monad.Primitive
import Control.Monad.State.Strict
import Control.Monad.ST
import Control.Monad.ST.Lazy (strictToLazyST, lazyToStrictST)
import qualified Control.Monad.ST.Lazy as STL
-- import Control.Monad.ST.Safe
import Control.DeepSeq
import Data.Coerce
import qualified Data.ByteString.Char8 as BS
import qualified Data.ByteString.Lazy.Char8 as BSL
import qualified Data.ByteString.Builder as BSB
import Data.IntMap (IntMap)
import qualified Data.IntMap.Strict as IMS
import qualified Data.IntMap.Lazy as IML
import Data.IntSet (IntSet)
import qualified Data.IntSet as IS
import qualified Data.Sequence as Seq
import Data.Sequence (Seq)
import qualified Data.Array.IArray as A
import qualified Data.Array.MArray.Safe as A
import qualified Data.Array.MArray as A
import Data.Array (Array)
import Data.Array.Unboxed (UArray)
import Data.Array.IArray (IArray)
import Data.Array.MArray.Safe (MArray)
import Data.Array.IO.Safe (IOArray, IOUArray)
import Data.Array.ST.Safe (STArray, STUArray, runSTArray, runSTUArray)
import qualified Data.Vector as V
import qualified Data.Vector.Mutable as VM
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.Vector.Storable as VS
import qualified Data.Vector.Storable.Mutable as VSM
import qualified Data.Vector.Primitive as VP
import qualified Data.Vector.Primitive.Mutable as VPM
import qualified Data.Vector.Generic as VG
import qualified Data.Vector.Generic.Mutable as VGM
import qualified Data.Attoparsec.ByteString.Char8 as Atto
import Debug.Trace
import GHC.Exts (build, Int(..), Int#,
                 (+#), (*#), (-#), (<#), (>=#), (==#), quotRemInt#,
                 remInt#, uncheckedIShiftRL#, andI#, orI#,
                isTrue#)

main :: IO ()
main = do
  n <- readInt <$> getLine
  as <- VU.unfoldrN n (runStateT rInt) <$> BSL.getContents
  let (m0,m1) = VU.foldl' (\(!m0,!m1) !a ->
                             if a > m0
                             then (a,m0)
                             else (m0,max m1 a)) (minBound,minBound) as
      m0Str = BSB.intDec m0
      m1Str = BSB.intDec m1
  BSB.hPutBuilder stdout
    $ VU.foldr (\x r -> (if x == m0 then m1Str else m0Str) <>
                        (BSB.char7 '\n' <> r)) mempty as
  return ()

#define IL(f) {-# INLINE f #-}; f

IL(evalVals) :: [a] -> [a]
evalVals xs = build $ \c n -> foldr (c $!) n xs
IL(forceVals) :: (NFData a) => [a] -> [a]
forceVals xs = build $ \c n -> foldr (c $!!) n xs

rInt :: StateT BSL.ByteString Maybe Int
IL(rInt) = StateT $ BSL.readInt . BSL.dropWhile (<'!')
rIntS :: StateT BS.ByteString Maybe Int
IL(rIntS) = StateT $ BS.readInt . BS.dropWhile (<'!')

unlessM, whenM :: (Monad m) => m Bool -> m () -> m ()
IL(whenM) = (. flip when) . (>>=)
IL(unlessM) = (. flip unless) . (>>=)

IL(wrA) = A.writeArray
IL(rdA) = A.readArray
IL(mdA) = \arr f !i -> do
  ai <- rdA arr i
  let fai = f ai
  wrA arr i fai
  return (ai,fai)
{-# INLINE mdA' #-}
mdA' = \arr f !i -> do
  !ai <- rdA arr i
  let !fai = f ai
  wrA arr i fai
  return (ai,fai)
IL(swapA) = \arr !i !j -> do
  ai <- rdA arr i
  wrA arr i =<< rdA arr j
  wrA arr j ai

#define D(f,r,d)\
  IL(f) :: Integral a=>a->d; f=fromIntegral;\
  IL(r) :: String->d; r=read
#define C(f,r,g,h,d) D(f,r,d);\
  g,h :: RealFrac a=>a->d; IL(g)=floor; IL(h)=ceiling
C(_toInteger_,readInteger,floorInteger,ceilInteger,Integer)
C(toInt,readInt,floorInt,ceilInt,Int)
C(toI8,readI8,floorI8,ceilI8,Int8)
C(toI16,readI16,floorI16,ceilI16,Int16)
C(toI32,readI32,floorI32,ceilI32,Int32)
C(toI64,readI64,floorI64,ceilI64,Int64)
C(toWord,readWord,floorWord,ceilWord,Word)
C(toW8,readW8,floorW8,ceilW8,Word8)
C(toW16,readW16,floorW16,ceilW16,Word16)
C(toW32,readW32,floorW32,ceilW32,Word32)
C(toW64,readW64,floorW64,ceilW64,Word64)
D(toDouble,readDouble,Double)
D(toFloat,readFloat,Float)
#undef D
#undef C

#define N(f,g,h,a,m)\
  IL(f) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> e -> m (a i e);\
  f=A.newArray;\
  IL(g) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> m (a i e);\
  g=A.newArray_;\
  IL(h) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> [e] -> m (a i e);\
  h=A.newListArray
#define C(a,m)
N(newIOA,newIOA_,newIOAL,IOArray,IO)
N(newSTA,newSTA_,newSTAL,STArray s,ST s)
#undef C
#define C(a,m) MArray (a) e (m),
N(newIOUA,newIOUA_,newIOUAL,IOUArray,IO)
N(newSTUA,newSTUA_,newSTUAL,STUArray s,ST s)
#undef C
#undef N

#undef IL


import qualified Data.IntMap as M
import Data.Array.IArray

solve :: Int -> [Int] -> [Int]
solve n a = go n M.empty
  where
    a' = listArray (1,n) a :: Array Int Int
    go :: Int -> M.IntMap Bool -> [Int]
    go 0 m = [ k | (k,v) <- M.toAscList m, v ]
    go i m = let l = M.size $ M.filterWithKey (\k v -> k `mod` i == 0 && v) m
                 b = if l `mod` 2 == (a'!i) then False else True
                 m' = M.insert i b m
              in go (i-1) m'

main :: IO ()
main = do
    n <- readLn :: IO Int
    a <- map read . words <$> getLine :: IO [Int]
    let ret = solve n a
    print $ length ret
    mapM_ print ret

Score : 400 points

Problem Statement

There are N empty boxes arranged in a row from left to right.
The integer i is written on the i-th box from the left (1 \leq i \leq N).

For each of these boxes, Snuke can choose either to put a ball in it or to put nothing in it.

We say a set of choices to put a ball or not in the boxes is good when the following condition is satisfied:

For every integer i between 1 and N (inclusive), the total number of balls contained in the boxes with multiples of i written on them is congruent to a_i modulo 2.

Does there exist a good set of choices? If the answer is yes, find one good set of choices.

Constraints

All values in input are integers.

1 \leq N \leq 2 \times 10^5

a_i is 0 or 1.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

If a good set of choices does not exist, print -1.

If a good set of choices exists, print one such set of choices in the following format:

M
b_1 b_2 ... b_M

where M denotes the number of boxes that will contain a ball, and b_1,\ b_2,\ ...,\ b_M are the integers written on these boxes, in any order.

Sample Input 1

3
1 0 0

Sample Output 1

1
1

Consider putting a ball only in the box with 1 written on it.

There are three boxes with multiples of 1 written on them: the boxes with 1, 2, and 3. The total number of balls contained in these boxes is 1.

There is only one box with a multiple of 2 written on it: the box with 2. The total number of balls contained in these boxes is 0.

There is only one box with a multiple of 3 written on it: the box with 3. The total number of balls contained in these boxes is 0.

Thus, the condition is satisfied, so this set of choices is good.

Sample Input 2

5
0 0 0 0 0

Sample Output 2

0

Putting nothing in the boxes can be a good set of choices.


import Control.Monad

foo :: [Int] -> Int -> Int
foo [] num = num
foo (n:ns) color = foo (bar n ns) (color+1) where
  bar _ [] = []
  bar x (c:cs) = if c > x then bar x cs else c:bar x cs

main = do
  n <- read <$> getLine :: IO Int
  ints <- replicateM n (read <$> getLine) :: IO [Int]
  print $ foo ints 0

Score : 500 points

Problem Statement

You are given a sequence with N integers: A = \{ A_1, A_2, \cdots, A_N \}.
For each of these N integers, we will choose a color and paint the integer with that color. Here the following condition must be satisfied:

If A_i and A_j (i < j) are painted with the same color, A_i < A_j.

Find the minimum number of colors required to satisfy the condition.

Constraints

1 \leq N \leq 10^5

0 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1
:
A_N

Output

Print the minimum number of colors required to satisfy the condition.

Sample Input 1

5
2
1
4
5
3

Sample Output 1

2

We can satisfy the condition with two colors by, for example, painting 2 and 3 red and painting 1, 4, and 5 blue.

Sample Input 2

4
0
0
0
0

Sample Output 2

4

We have to paint all the integers with distinct colors.


{-# LANGUAGE BangPatterns #-}
import Control.Monad
import Data.Maybe
import Debug.Trace
import Data.List
import qualified Data.Map.Strict as Map
import qualified Data.IntMap.Strict as IntMap
import qualified Data.Set as Set
import qualified Data.IntSet as IntSet
import Data.Functor
import Data.Array
-- import Data.Array.Unboxed
import Control.Monad.ST
import Data.Array.ST
import qualified Data.ByteString.Char8 as B
import Data.Maybe (fromJust)

readInt = fst . fromJust . B.readInt
readInts = map (fst . fromJust . B.readInt) . B.words <$> B.getLine :: IO [Int]
readIntegers = map (fst . fromJust . B.readInteger) . B.words <$> B.getLine :: IO [Integer]
readVInts = map (fst . fromJust . B.readInt) . B.lines <$> B.getContents

main :: IO ()
main = do
  n <- readLn :: IO Int
  -- as <- replicateM n $ readLn :: IO [Int]
  as <- readVInts
  print $ solve'' n as

solve'' n (a:as) = length $ foldl' go [a] as
  where
    go bs a
      | a <= mn = insert a bs
      | otherwise = replaceUpperBound (<a) a bs
      -- | otherwise = insert a $ deleteUpperBound (< a) bs
      where
        mn = head bs

replaceUpperBound :: (a -> Bool) -> a -> [a] -> [a]
replaceUpperBound pred new (a:b:as)
  | pred a && not (pred b) = new:b:as
  | not (pred a) = a:b:as
  | otherwise = a:(replaceUpperBound pred new (b:as))
replaceUpperBound pred new (a:[]) = if pred a then [new] else [a]
replaceUpperBound _ _ [] = []


--{-# LANGUAGE BangPatterns #-}
--{-# LANGUAGE ScopedTypeVariables #-}
--{-# LANGUAGE FlexibleContexts #-}
--{-# LANGUAGE ViewPatterns #-}
--{-# LANGUAGE OverloadedStrings #-}
{-# LANGUAGE MultiWayIf #-}

import qualified Data.Vector as V
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.ByteString.Char8 as BC
import qualified Data.Map as MP
import qualified Data.Set as ST
import qualified Data.Sequence as SQ
import Data.Array
import Data.Array.Unboxed
import Data.Array.ST
import Data.Array.IO
import Data.Maybe
import Data.Ord
import Data.Ratio
import Data.Char
import Data.Bits
import Data.List
import Data.Tuple
import Data.Functor
import Data.IORef
import Data.STRef
import Control.Applicative
import Control.Monad
import Control.Monad.ST
import Debug.Trace
-- import qualified Data.Vector.Algorithms.Merge as VAM
-- import qualified Data.Heap as HP
-- import Numeric.Extra
-- import Data.Tuple.Extra
-- import Data.List.Extra

--------------------------------------------------------------------------
solve1 xs st mp = (s == 0) && (k1 == 0) && (v2*2 == v1)
    where
        s = ST.foldl' xor 0 st
        ((k1,v1),mp') = MP.deleteFindMin mp
        ((k2,v2),mp'') = MP.deleteFindMin mp'

solve2 xs st mp = (s == 0) && (v1 == v2 && v2 == v3 && v3 == v1)
    where
        s = ST.foldl' xor 0 st
        ((k1,v1),mp') = MP.deleteFindMin mp
        ((k2,v2),mp'') = MP.deleteFindMin mp'
        ((k3,v3),mp''') = MP.deleteFindMin mp''

main = do
    n <- int
    xs <- sLineToIntL
    let
        st = ST.fromList xs
        mp = foldl' (\mp x -> MP.insertWith (+) x 1 mp) MP.empty xs
        containZero = ST.member 0 st
        allZero = all (==0) xs

    if | allZero -> putStrLn "Yes"
       | ST.size st == 2 -> yesnoL $ solve1 xs st mp
       | ST.size st `mod` 3 /= 0 -> putStrLn "No"
       | ST.size st == 3 -> yesnoL $ solve2 xs st mp
       | otherwise -> putStrLn "No"
--------------------------------------------------------------------------

-- Input Util
int :: IO Int
int = BC.getLine >>= return . bsToInt

double :: IO Double
double = BC.getLine >>= return . bsToDouble

str :: IO String
str = BC.getLine >>= return . BC.unpack

strBS :: IO BC.ByteString
strBS = BC.getLine

-- convert
bsToInt :: BC.ByteString -> Int
bsToInt = fst . fromJust . BC.readInt

bsToDouble :: BC.ByteString -> Double
bsToDouble = read . BC.unpack

bsToIntTuple :: BC.ByteString -> (Int, Int)
bsToIntTuple bs = (a,b)
    where
        Just (a,bs') = parseInt bs
        Just (b,_) = parseInt bs'

bsToIntTriple :: BC.ByteString -> (Int, Int, Int)
bsToIntTriple bs = (a,b,c)
    where
        Just (a,bs') = parseInt bs
        Just (b,bs'') = parseInt bs'
        Just (c,_) = parseInt bs''

parseInt :: BC.ByteString -> Maybe (Int, BC.ByteString)
parseInt = BC.readInt . BC.dropWhile isSpace

-- single line
fromSLineL :: (BC.ByteString -> a) -> IO [a]
fromSLineL f = strBS >>= return . map f . BC.words

-- from single line to List / VU.Vector
sLineToIntL :: IO [Int]
sLineToIntL = fromSLineL bsToInt

sLineToDoubleL :: IO [Double]
sLineToDoubleL = fromSLineL bsToDouble

sLineToStrL :: IO [String]
sLineToStrL = fromSLineL BC.unpack

sLineToStrBSL :: IO [BC.ByteString]
sLineToStrBSL = fromSLineL id

sLineToIntV :: Int -> IO (VU.Vector Int)
sLineToIntV n = BC.getLine >>= return . VU.unfoldrN n parseInt

sLineToDoubleV :: IO (VU.Vector Double)
sLineToDoubleV = BC.getLine >>= return . VU.fromList . map bsToDouble . BC.words

-- multiple lines
fromMLinesL :: Int -> (BC.ByteString -> a) -> IO [a]
fromMLinesL n f = replicateM n (strBS >>= return . f)

fromMLinesV :: VU.Unbox a => Int -> (BC.ByteString -> a) -> IO (VU.Vector a)
fromMLinesV n f = VU.replicateM n (strBS >>= return . f)

-- from multiple lines to List / VU.Vector
mLinesToIntL :: Int -> IO [Int]
mLinesToIntL n = fromMLinesL n bsToInt

mLinesToDoubleL :: Int -> IO [Double]
mLinesToDoubleL n = fromMLinesL n bsToDouble

mLinesToStrL :: Int -> IO [String]
mLinesToStrL n = fromMLinesL n BC.unpack

mLinesToStrBSL :: Int -> IO [BC.ByteString]
mLinesToStrBSL n = fromMLinesL n id

mLinesToTupleL :: Int -> IO [(Int, Int)]
mLinesToTupleL n = fromMLinesL n bsToIntTuple

mLinesToTripleL :: Int -> IO [(Int, Int, Int)]
mLinesToTripleL n = fromMLinesL n bsToIntTriple

mLinesToIntV :: Int -> IO (VU.Vector Int)
mLinesToIntV n = fromMLinesV n bsToInt

mLinesToDoubleV :: Int -> IO (VU.Vector Double)
mLinesToDoubleV n = fromMLinesV n bsToDouble

mLinesToTupleV :: Int -> IO (VU.Vector (Int, Int))
mLinesToTupleV n = fromMLinesV n bsToIntTuple

mLinesToTripleV :: Int -> IO (VU.Vector (Int, Int, Int))
mLinesToTripleV n = fromMLinesV n bsToIntTriple

-- output
yesnoL :: Bool -> IO ()
yesnoL cond
    | cond = putStrLn "Yes"
    | otherwise = putStrLn "No"

yesnoU :: Bool -> IO ()
yesnoU cond
    | cond = putStrLn "YES"
    | otherwise = putStrLn "NO"

-- others
strToChr :: String -> Char
strToChr [x] = x

chrToStr :: Char -> String
chrToStr x = [x]

m1 :: Int -> Int
m1 x = subtract 1 x

toDouble :: Int -> Double
toDouble x = fromIntegral x

toInt :: Double -> Int
toInt x = floor x

Score : 300 points

Problem Statement

Snuke has N hats. The i-th hat has an integer a_i written on it.

There are N camels standing in a circle.
Snuke will put one of his hats on each of these camels.

If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print Yes; otherwise, print No.

The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself.

What is XOR?

The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non-negative integers x_1, x_2, \ldots, x_n is defined as follows:

- When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even.

For example, 3 \oplus 5 = 6.

Constraints

All values in input are integers.

3 \leq N \leq 10^{5}

0 \leq a_i \leq 10^{9}

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 \ldots a_{N}

Output

Print the answer.

Sample Input 1

3
1 2 3

Sample Output 1

Yes

If we put the hats with 1, 2, and 3 in this order, clockwise, the condition will be satisfied for every camel, so the answer is Yes.

Sample Input 2

4
1 2 4 8

Sample Output 2

No

There is no such way to distribute the hats; the answer is No.


import Data.Bits
import Data.List
import Control.Monad
import qualified Data.Vector.Unboxed as V
import Control.Monad.ST
import Data.Maybe (fromMaybe)
import qualified Data.ByteString.Lazy.Char8 as B
import Debug.Trace

main :: IO ()
main = do
  n <- readLn :: IO Int
  as <- map read.words <$> getLine :: IO [Int]
  let as' = (group.sort) as
      las' = length as'
  if n `mod` 3 /= 0 then putStrLn "No"
    else if las' > 3 then putStrLn "No"
      else if las' == 1 && all (== 0) as then putStrLn "Yes"
        else if las' == 2 && (length (as'!!0) == 2*(length $ as'!!1) || length (as'!!1) == 2*(length $ as'!!0)) then putStrLn "Yes"
          else if las' == 3 && length (as'!!0) == length (as'!!1) && length (as'!!1) == length (as'!!2) then putStrLn "Yes"
            else putStrLn "No"


main :: IO()
main = do
    [n,a,b] <- map read . words <$> getLine
    print $ min (n*a) b

Score : 100 points

Problem Statement

N of us are going on a trip, by train or taxi.

The train will cost each of us A yen (the currency of Japan).

The taxi will cost us a total of B yen.

How much is our minimum total travel expense?

Constraints

All values in input are integers.

1 \leq N \leq 20

1 \leq A \leq 50

1 \leq B \leq 50

Input

Input is given from Standard Input in the following format:

N A B

Output

Print an integer representing the minimum total travel expense.

Sample Input 1

4 2 9

Sample Output 1

8

The train will cost us 4 \times 2 = 8 yen, and the taxi will cost us 9 yen, so the minimum total travel expense is 8 yen.

Sample Input 2

4 2 7

Sample Output 2

7

Sample Input 3

4 2 8

Sample Output 3

8


import qualified Data.ByteString.Char8 as BS
import Data.Char
import Data.List

toInts :: IO [Int]
toInts = unfoldr (BS.readInt . BS.dropWhile isSpace) <$> BS.getLine

solver :: Int -> Int -> Int -> Int
solver n a b = min b (a * n)

main :: IO ()
main = do
  n:a:b:[] <- toInts
  print $ solver n a b


{-# LANGUAGE
  ScopedTypeVariables, BangPatterns, TupleSections, ExplicitForAll,
  LambdaCase, MultiWayIf, Unsafe, RecordWildCards, FlexibleContexts, CPP,
  NoMonomorphismRestriction, GADTs, PatternGuards, MagicHash, UnboxedTuples,
  RankNTypes, EmptyDataDecls, EmptyCase, ViewPatterns, PolyKinds,
  TypeFamilies #-}
{-# OPTIONS_GHC -O2 #-}

import Prelude
import Data.Bits
import Data.List
import Data.Maybe
import Data.Tuple
import Data.Ord
import Data.Int
import Data.Word
import Data.Char
import Data.Ratio
import Data.Function
import Data.STRef
import Data.IORef
import Data.Monoid
import Data.Functor
import System.IO
import Control.Arrow
import Control.Applicative
import Control.Monad
import Control.Monad.Primitive
import Control.Monad.State.Strict
import Control.Monad.ST
import Control.Monad.ST.Lazy (strictToLazyST, lazyToStrictST)
import qualified Control.Monad.ST.Lazy as STL
-- import Control.Monad.ST.Safe
import Control.DeepSeq
import Data.Coerce
import qualified Data.ByteString.Char8 as BS
import qualified Data.ByteString.Lazy.Char8 as BSL
import qualified Data.ByteString.Builder as BSB
import Data.IntMap (IntMap)
import qualified Data.IntMap.Strict as IMS
import qualified Data.IntMap.Lazy as IML
import Data.IntSet (IntSet)
import qualified Data.IntSet as IS
import qualified Data.Sequence as Seq
import Data.Sequence (Seq)
import qualified Data.Array.IArray as A
import qualified Data.Array.MArray.Safe as A
import qualified Data.Array.MArray as A
import Data.Array (Array)
import Data.Array.Unboxed (UArray)
import Data.Array.IArray (IArray)
import Data.Array.MArray.Safe (MArray)
import Data.Array.IO.Safe (IOArray, IOUArray)
import Data.Array.ST.Safe (STArray, STUArray, runSTArray, runSTUArray)
import qualified Data.Vector as V
import qualified Data.Vector.Mutable as VM
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.Vector.Storable as VS
import qualified Data.Vector.Storable.Mutable as VSM
import qualified Data.Vector.Primitive as VP
import qualified Data.Vector.Primitive.Mutable as VPM
import qualified Data.Vector.Generic as VG
import qualified Data.Vector.Generic.Mutable as VGM
import qualified Data.Attoparsec.ByteString.Char8 as Atto
import Debug.Trace
import GHC.Exts (build, Int(..), Int#,
                 (+#), (*#), (-#), (<#), (>=#), (==#), quotRemInt#,
                 remInt#, uncheckedIShiftRL#, andI#, orI#,
                isTrue#)

main :: IO ()
main = do
  [n,d] <- map readInt . words <$> getLine
  xs <- V.replicateM n
    $ VU.unfoldrN d (runStateT $ rIntS) <$> BS.getLine
  print $ V.sum $ (`V.imap` xs)
    $ \ !i !xi -> V.length $ (`V.filter` V.drop (i+1) xs)
    $ \ !xj ->
        let sqLen = VU.sum $ VU.map (^2) $ VU.zipWith (-) xi xj
        in floorInt (sqrt (toDouble sqLen)) ^ 2 == sqLen
  return ()

#define IL(f) {-# INLINE f #-}; f

IL(evalVals) :: [a] -> [a]
evalVals xs = build $ \c n -> foldr (c $!) n xs
IL(forceVals) :: (NFData a) => [a] -> [a]
forceVals xs = build $ \c n -> foldr (c $!!) n xs

rInt :: StateT BSL.ByteString Maybe Int
IL(rInt) = StateT $ BSL.readInt . BSL.dropWhile (<'!')
rIntS :: StateT BS.ByteString Maybe Int
IL(rIntS) = StateT $ BS.readInt . BS.dropWhile (<'!')

unlessM, whenM :: (Monad m) => m Bool -> m () -> m ()
IL(whenM) = (. flip when) . (>>=)
IL(unlessM) = (. flip unless) . (>>=)

IL(wrA) = A.writeArray
IL(rdA) = A.readArray
IL(mdA) = \arr f !i -> do
  ai <- rdA arr i
  let fai = f ai
  wrA arr i fai
  return (ai,fai)
{-# INLINE mdA' #-}
mdA' = \arr f !i -> do
  !ai <- rdA arr i
  let !fai = f ai
  wrA arr i fai
  return (ai,fai)
IL(swapA) = \arr !i !j -> do
  ai <- rdA arr i
  wrA arr i =<< rdA arr j
  wrA arr j ai

#define D(f,r,d)\
  IL(f) :: Integral a=>a->d; f=fromIntegral;\
  IL(r) :: String->d; r=read
#define C(f,r,g,h,d) D(f,r,d);\
  g,h :: RealFrac a=>a->d; IL(g)=floor; IL(h)=ceiling
C(_toInteger_,readInteger,floorInteger,ceilInteger,Integer)
C(toInt,readInt,floorInt,ceilInt,Int)
C(toI8,readI8,floorI8,ceilI8,Int8)
C(toI16,readI16,floorI16,ceilI16,Int16)
C(toI32,readI32,floorI32,ceilI32,Int32)
C(toI64,readI64,floorI64,ceilI64,Int64)
C(toWord,readWord,floorWord,ceilWord,Word)
C(toW8,readW8,floorW8,ceilW8,Word8)
C(toW16,readW16,floorW16,ceilW16,Word16)
C(toW32,readW32,floorW32,ceilW32,Word32)
C(toW64,readW64,floorW64,ceilW64,Word64)
D(toDouble,readDouble,Double)
D(toFloat,readFloat,Float)
#undef D
#undef C

#define N(f,g,h,a,m)\
  IL(f) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> e -> m (a i e);\
  f=A.newArray;\
  IL(g) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> m (a i e);\
  g=A.newArray_;\
  IL(h) :: forall e i s. (C(a,m)A.Ix i) => (i,i) -> [e] -> m (a i e);\
  h=A.newListArray
#define C(a,m)
N(newIOA,newIOA_,newIOAL,IOArray,IO)
N(newSTA,newSTA_,newSTAL,STArray s,ST s)
#undef C
#define C(a,m) MArray (a) e (m),
N(newIOUA,newIOUA_,newIOUAL,IOUArray,IO)
N(newSTUA,newSTUA_,newSTUAL,STUArray s,ST s)
#undef C
#undef N

#undef IL

Score : 200 points

Problem Statement

There are N points in a D-dimensional space.

The coordinates of the i-th point are (X_{i1}, X_{i2}, ..., X_{iD}).

The distance between two points with coordinates (y_1, y_2, ..., y_D) and (z_1, z_2, ..., z_D) is \sqrt{(y_1 - z_1)^2 + (y_2 - z_2)^2 + ... + (y_D - z_D)^2}.

How many pairs (i, j) (i < j) are there such that the distance between the i-th point and the j-th point is an integer?

Constraints

All values in input are integers.

2 \leq N \leq 10

1 \leq D \leq 10

-20 \leq X_{ij} \leq 20

No two given points have the same coordinates. That is, if i \neq j, there exists k such that X_{ik} \neq X_{jk}.

Input

Input is given from Standard Input in the following format:

N D
X_{11} X_{12} ... X_{1D}
X_{21} X_{22} ... X_{2D}
\vdots
X_{N1} X_{N2} ... X_{ND}

Output

Print the number of pairs (i, j) (i < j) such that the distance between the i-th point and the j-th point is an integer.

Sample Input 1

3 2
1 2
5 5
-2 8

Sample Output 1

1

The number of pairs with an integer distance is one, as follows:

The distance between the first point and the second point is \sqrt{|1-5|^2 + |2-5|^2} = 5, which is an integer.

The distance between the second point and the third point is \sqrt{|5-(-2)|^2 + |5-8|^2} = \sqrt{58}, which is not an integer.

The distance between the third point and the first point is \sqrt{|-2-1|^2+|8-2|^2} = 3\sqrt{5}, which is not an integer.

Sample Input 2

3 4
-3 7 8 2
-12 1 10 2
-2 8 9 3

Sample Output 2

2

Sample Input 3

5 1
1
2
3
4
5

Sample Output 3

10


import Data.Maybe
import Control.Monad
import qualified Data.ByteString.Char8 as C

readInteger' = fst.fromJust.C.readInteger
readLnInteger = map readInteger'.C.words<$>C.getLine
readCsInteger n = concat<$>replicateM n readLnInteger

x ? (a,b) = if x then a else b
half = flip div 2
aliq = ((0==).).mod

-- さんすうができなくてつらい

main = do
 [n] <- readLnInteger
 inputs <- readLnInteger
 putStr.unwords.map show $ solve n inputs

solve 0 s = []
solve n s@(a:xs) = [calc s] ++ solve (n-1) (s++[a])

calc [a] = a
calc (a:b:x) = a - b + (null x)?(0,calc x)

Score : 400 points

Problem Statement

There are N mountains in a circle, called Mountain 1, Mountain 2, ..., Mountain N in clockwise order. N is an odd number.

Between these mountains, there are N dams, called Dam 1, Dam 2, ..., Dam N. Dam i (1 \leq i \leq N) is located between Mountain i and i+1 (Mountain N+1 is Mountain 1).

When Mountain i (1 \leq i \leq N) receives 2x liters of rain, Dam i-1 and Dam i each accumulates x liters of water (Dam 0 is Dam N).

One day, each of the mountains received a non-negative even number of liters of rain.

As a result, Dam i (1 \leq i \leq N) accumulated a total of A_i liters of water.

Find the amount of rain each of the mountains received. We can prove that the solution is unique under the constraints of this problem.

Constraints

All values in input are integers.

3 \leq N \leq 10^5-1

N is an odd number.

0 \leq A_i \leq 10^9

The situation represented by input can occur when each of the mountains receives a non-negative even number of liters of rain.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print N integers representing the number of liters of rain Mountain 1, Mountain 2, ..., Mountain N received, in this order.

Sample Input 1

3
2 2 4

Sample Output 1

4 0 4

If we assume Mountain 1, 2, and 3 received 4, 0, and 4 liters of rain, respectively, it is consistent with this input, as follows:

Dam 1 should have accumulated \frac{4}{2} + \frac{0}{2} = 2 liters of water.

Dam 2 should have accumulated \frac{0}{2} + \frac{4}{2} = 2 liters of water.

Dam 3 should have accumulated \frac{4}{2} + \frac{4}{2} = 4 liters of water.

Sample Input 2

5
3 8 7 5 5

Sample Output 2

2 4 12 2 8

Sample Input 3

3
1000000000 1000000000 0

Sample Output 3

0 2000000000 0


main = interact$show.f.tail.(map read).words
f::[Int]->Int
f(_:_:[])=0
f(a:b:c:d)
 |z a b==z b c=1+f(b:c:d)
 |True=f$b:c:d
z=compare

Score : 200 points

Problem Statement

We have a permutation p = {p_1,\ p_2,\ ...,\ p_n} of {1,\ 2,\ ...,\ n}.

Print the number of elements p_i (1 < i < n) that satisfy the following condition:

p_i is the second smallest number among the three numbers p_{i - 1}, p_i, and p_{i + 1}.

Constraints

All values in input are integers.

3 \leq n \leq 20

p is a permutation of {1,\ 2,\ ...,\ n}.

Input

Input is given from Standard Input in the following format:

n
p_1 p_2 ... p_n

Output

Print the number of elements p_i (1 < i < n) that satisfy the condition.

Sample Input 1

5
1 3 5 4 2

Sample Output 1

2

p_2 = 3 is the second smallest number among p_1 = 1, p_2 = 3, and p_3 = 5. Also, p_4 = 4 is the second smallest number among p_3 = 5, p_4 = 4, and p_5 = 2. These two elements satisfy the condition.

Sample Input 2

9
9 6 3 2 5 8 7 4 1

Sample Output 2

5


main = do
  getLine
  p <- map (read :: String -> Int) . words <$> getLine
  let d_p = zipWith (<) p (tail p)
  let d2_p = zipWith (==) d_p (tail d_p)
  print . length $ filter (== True) d2_p


import Data.List
main :: IO()
main = do
  _ <- getLine
  ps <- map read . words <$> getLine
  print $ slv ps 0

slv :: [Integer] -> Integer -> Integer
slv p ret | (length p) < 3 = ret
          | sec == p !! 1 = slv (tail p) (ret + 1)
          | otherwise = slv (tail p) ret
  where sec = (sort (take 3 p)) !! 1


main = do
    _ <- getLine
    s <- getLine
    let ps = map read $ words s :: [Int]
    let ans = count ps
    print ans

count :: [Int] -> Int
count ps
    | length ps < 3 = 0
    | otherwise = do
        let countup = (p1 < p2 && p2 < p3) || (p1 > p2 && p2 > p3)
        let base = if countup then 1 else 0
        base + count (p2:p3:rest)
    where
        p1:p2:p3:rest = ps


import Control.Monad
import Data.Maybe
import Data.List
import qualified Data.ByteString.Char8 as BS
import qualified Data.Vector as V

main = abc132c

abc132c = do
    n  <- getIntList
    dL <- getIntList
    let (small, large) = getMedian dL
    print $ large - small

getMedian :: [Integer] -> (Integer, Integer)
getMedian dL = let sorted = V.fromList (sort dL)
                   num = V.length sorted
                in (sorted V.! (div num 2 - 1), sorted V.! div num 2)

readInt = fst . fromJust . BS.readInteger
readIntList = map readInt . BS.words
getIntList = readIntList <$> BS.getLine

Score : 300 points

Problem Statement

Takahashi made N problems for competitive programming.
The problems are numbered 1 to N, and the difficulty of Problem i is represented as an integer d_i (the higher, the harder).

He is dividing the problems into two categories by choosing an integer K, as follows:

A problem with difficulty K or higher will be for ARCs.

A problem with difficulty lower than K will be for ABCs.

How many choices of the integer K make the number of problems for ARCs and the number of problems for ABCs the same?

Problem Statement

2 \leq N \leq 10^5

N is an even number.

1 \leq d_i \leq 10^5

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
d_1 d_2 ... d_N

Output

Print the number of choices of the integer K that make the number of problems for ARCs and the number of problems for ABCs the same.

Sample Input 1

6
9 1 4 4 6 7

Sample Output 1

2

If we choose K=5 or 6, Problem 1, 5, and 6 will be for ARCs, Problem 2, 3, and 4 will be for ABCs, and the objective is achieved.
Thus, the answer is 2.

Sample Input 2

8
9 1 14 5 5 4 4 14

Sample Output 2

0

There may be no choice of the integer K that make the number of problems for ARCs and the number of problems for ABCs the same.

Sample Input 3

14
99592 10342 29105 78532 83018 11639 92015 77204 30914 21912 34519 80835 100000 1

Sample Output 3

42685


import Data.List

hoge :: [Int] -> Int
hoge [] = 0
hoge ints = bar $ sort ints
  where
    bar ls
      | length ls `mod` 2 == 0 = ls!!(length ls `div` 2) - ls!!(length ls `div` 2 - 1)
      | otherwise = 0

main = do
  _ <- getLine
  ints <- map read . words <$> getLine :: IO [Int]
  print $ hoge ints


import           Data.List
main = do
    n <- readLn :: IO Int
    d <- sort . map read . words <$> getLine :: IO [Int]
    print $ (d !! (n `div` 2)) - (d !! ((n `div` 2) - 1))


import Control.Applicative

m = 10^9 + 7
modm n = mod n m

mul' x y = modm $ modm x * modm y

power' x 0 = 1
power' x 1 = modm x
power' x n
  | (n `mod` 2) == 0 = let i = power' x (n `div` 2) in mul' i i
  | otherwise = mul' x $ power' x (n - 1)

div' x y = mul' x $ power' y (m - 2)

product' xs = foldl mul' 1 xs

factorial' n = product' [1..n]

npr' n r = product' [n - i | i <- [0..(r - 1)]]

ncr' n r = modm $ (npr' n r) `div'` (factorial' r)

main = do
  [n, k] <- (map read). words <$> getLine :: IO [Int]
  mapM_ (putStrLn.show) $ (flip map) [1..k] $ \x -> modm $ product' [(ncr' (n - k + 1) x), (ncr' (k - 1) (x - 1))]

Score : 400 points

Problem Statement

There are K blue balls and N-K red balls. The balls of the same color cannot be distinguished. Snuke and Takahashi are playing with these balls.

First, Snuke will arrange the N balls in a row from left to right.

Then, Takahashi will collect only the K blue balls. In one move, he can collect any number of consecutive blue balls. He will collect all the blue balls in the fewest moves possible.

How many ways are there for Snuke to arrange the N balls in a row so that Takahashi will need exactly i moves to collect all the blue balls? Compute this number modulo 10^9+7 for each i such that 1 \leq i \leq K.

Constraints

1 \leq K \leq N \leq 2000

Input

Input is given from Standard Input in the following format:

N K

Output

Print K lines. The i-th line (1 \leq i \leq K) should contain the number of ways to arrange the N balls so that Takahashi will need exactly i moves to collect all the blue balls, modulo 10^9+7.

Sample Input 1

5 3

Sample Output 1

3
6
1

There are three ways to arrange the balls so that Takahashi will need exactly one move: (B, B, B, R, R), (R, B, B, B, R), and (R, R, B, B, B). (R and B stands for red and blue, respectively).

There are six ways to arrange the balls so that Takahashi will need exactly two moves: (B, B, R, B, R), (B, B, R, R, B), (R, B, B, R, B), (R, B, R, B, B), (B, R, B, B, R), and (B, R, R, B, B).

There is one way to arrange the balls so that Takahashi will need exactly three moves: (B, R, B, R, B).

Sample Input 2

2000 3

Sample Output 2

1998
3990006
327341989

Be sure to print the numbers of arrangements modulo 10^9+7.


import Control.Monad
import Data.Array.ST
import Data.Array.Unboxed

dex :: Int
dex = 1000000007

main :: IO ()
main = do
  [n,k] <- map read . words <$> getLine
  solve n k $ pascal n k

solve :: Int -> Int -> UArray (Int,Int) Int -> IO ()
solve n0 k0 b1 = forM_ [1..k0] $ \i -> print $ ((b1!(n0-k0+1,i)) * (b1!(k0-1,i-1))) `mod` dex

pascal :: Int -> Int -> UArray (Int,Int) Int
pascal n0 k0 = runSTUArray $ do
  dptbl <- newArray ((0,0),(n0,k0)) 0
  forM_ [(n',k') | n' <- [0..n0], k' <- [0..k0], n'>=k'] $ \(n,k) ->
    if k == 0 || k == n
    then writeArray dptbl (n,k) 1
    else liftM2 (\x y -> (x+y)`mod`dex) (readArray dptbl (n-1,k-1)) (readArray dptbl (n-1,k)) >>= writeArray dptbl (n,k)
  return dptbl


import qualified Data.Text as T
import qualified Data.Text.IO as T
import qualified Data.Text.Read as T
import Data.List (unfoldr)

main :: IO ()
main = do
  (n : k : _) <- map unsafeSignedDecimal . T.words <$> T.getLine :: IO [Integer]
  let
    as = unfoldr next (1, n - k + 1)
    next (i, a) = Just (a, (succ i, ((a * (n - k + 1 - i) * (k - i)) `div` ((i + 1) * i))))
  mapM_ (print . (`mod` modulus)) (take (fromInteger k) as)

modulus = 10 ^ 9 + 7 :: Integer

unsafeSignedDecimal :: T.Text -> Integer
unsafeSignedDecimal s = case T.signed T.decimal s of
  Right (n, _) -> n


import qualified Data.Map as Map
import Data.Maybe

kenkenpa :: Map.Map Int Int -> Int -> [Int]
kenkenpa map start = (:[]) $ fromMaybe (-1) (Map.lookup start map)

kenkenpa3 :: Map.Map Int Int -> Int -> [Int]
kenkenpa3 map start = kenkenpa map start >>= kenkenpa map >>= kenkenpa map

steps :: Map.Map Int Int -> [Int] -> Int -> Int -> Int -> Int
steps map memo start dest num =
  let
    ls = kenkenpa3 map start
    next = head ls
  in
    if next == dest then num
    else if next == (-1) || next `elem` memo then (-1)
    else steps map (next:memo) next dest (num+1)

main = do
  _ <- getLine
  s <- map  ((\ls -> (head ls,last ls)) . map read . words) . lines <$> getContents :: IO [(Int,Int)]
  let map = Map.fromList (init s)
  let (start,dest) = last s
  print $ steps map [start] start dest 1

Score : 500 points

Problem Statement

Ken loves ken-ken-pa (Japanese version of hopscotch). Today, he will play it on a directed graph G.
G consists of N vertices numbered 1 to N, and M edges. The i-th edge points from Vertex u_i to Vertex v_i.

First, Ken stands on Vertex S. He wants to reach Vertex T by repeating ken-ken-pa. In one ken-ken-pa, he does the following exactly three times: follow an edge pointing from the vertex on which he is standing.

Determine if he can reach Vertex T by repeating ken-ken-pa. If the answer is yes, find the minimum number of ken-ken-pa needed to reach Vertex T. Note that visiting Vertex T in the middle of a ken-ken-pa does not count as reaching Vertex T by repeating ken-ken-pa.

Constraints

2 \leq N \leq 10^5

0 \leq M \leq \min(10^5, N (N-1))

1 \leq u_i, v_i \leq N(1 \leq i \leq M)

u_i \neq v_i (1 \leq i \leq M)

If i \neq j, (u_i, v_i) \neq (u_j, v_j).

1 \leq S, T \leq N

S \neq T

Input

Input is given from Standard Input in the following format:

N M
u_1 v_1
:
u_M v_M
S T

Output

If Ken cannot reach Vertex T from Vertex S by repeating ken-ken-pa, print -1.
If he can, print the minimum number of ken-ken-pa needed to reach vertex T.

Sample Input 1

4 4
1 2
2 3
3 4
4 1
1 3

Sample Output 1

2

Ken can reach Vertex 3 from Vertex 1 in two ken-ken-pa, as follows: 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 in the first ken-ken-pa, then 4 \rightarrow 1 \rightarrow 2 \rightarrow 3 in the second ken-ken-pa. This is the minimum number of ken-ken-pa needed.

Sample Input 2

3 3
1 2
2 3
3 1
1 2

Sample Output 2

-1

Any number of ken-ken-pa will bring Ken back to Vertex 1, so he cannot reach Vertex 2, though he can pass through it in the middle of a ken-ken-pa.

Sample Input 3

2 0
1 2

Sample Output 3

-1

Vertex S and Vertex T may be disconnected.

Sample Input 4

6 8
1 2
2 3
3 4
4 5
5 1
1 4
1 5
4 6
1 6

Sample Output 4

2