object Main extends App{
  val input = io.StdIn.readLine.split(' ').map(_.toDouble)
  val A = input(0)
  val B = input(1)
  val H = input(2)
  val M = input(3)

  val radH = (H + M / 60d)  / 6.0d
  val radM = M / 30.0d

  val radD = radH - radM

  val posH = (0d, A)
  val posM = (- B*Math.sin(radD * Math.PI), B*Math.cos(radD * Math.PI))

  val distance = Math.sqrt( Math.pow(posH._1 - posM._1 , 2) + Math.pow(posH._2 - posM._2, 2) )

  printf("%.10f",distance)
}


import java.util.Scanner

import scala.annotation.tailrec

object Main {
  def main(args: Array[String]): Unit = {
    //nextLineとnextを同時に使ったら死ぬ
    val sc = new Scanner(System.in)

    val a, b, h, m = sc.nextDouble()
    val hh = (h % 12) * 30 + m / 2
    debug(hh)
    val mm = m * 6
    debug(mm)
    val diff = {
      val ss = math.abs(mm - hh)
      math.min(ss, 360 - ss)
    }
    debug(diff)
    val ans = math.sqrt(
      a * a + b * b - 2 * a * b * math
        .cos(math.toRadians(math.round(diff * 1000) / 1000))
    )
    debug(a * a, b * b, -2 * a * b, math.cos(diff))
    println(ans)
  }

  //region

  def debug(args: Any*): Unit = {
    Console.err.println(args.mkString(" "))
    Console.err.flush()
  }

  def printExecTime(process: => Unit): Unit = {
    val start = System.currentTimeMillis
    process
    debug("処理時間： " + (System.currentTimeMillis - start) + " ミリ秒")
  }

  def bitSearch[T](n: Int, choice: Vector[T]): Vector[Vector[T]] = {
    @tailrec
    def bitSearchLogic(n: Int,
                       choice: Vector[T],
                       acc: Vector[Vector[T]]): Vector[Vector[T]] = {
      if (n == 0) acc
      else {
        val newAcc =
          acc.flatMap(vec => {
            choice.map(c => vec :+ c)
          })
        bitSearchLogic(n - 1, choice, newAcc)
      }
    }

    val firstAcc = choice.map(i => Vector(i))
    bitSearchLogic(n - 1, choice, firstAcc)
  }
  //endregion
}

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

Score: 300 points

Problem Statement

Consider an analog clock whose hour and minute hands are A and B centimeters long, respectively.

An endpoint of the hour hand and an endpoint of the minute hand are fixed at the same point, around which each hand rotates clockwise at constant angular velocity. It takes the hour and minute hands 12 hours and 1 hour to make one full rotation, respectively.

At 0 o'clock, the two hands overlap each other. H hours and M minutes later, what is the distance in centimeters between the unfixed endpoints of the hands?

Constraints

All values in input are integers.

1 \leq A, B \leq 1000

0 \leq H \leq 11

0 \leq M \leq 59

Input

Input is given from Standard Input in the following format:

A B H M

Output

Print the answer without units. Your output will be accepted when its absolute or relative error from the correct value is at most 10^{-9}.

Sample Input 1

3 4 9 0

Sample Output 1

5.00000000000000000000

The two hands will be in the positions shown in the figure below, so the answer is 5 centimeters.

Sample Input 2

3 4 10 40

Sample Output 2

4.56425719433005567605

The two hands will be in the positions shown in the figure below. Note that each hand always rotates at constant angular velocity.

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.


import scala.collection.mutable.{Map => MMap, Queue => MQueue}
import scala.io.StdIn.readLine

object Main extends App
{
	type PathMap = Map[Int, List[Int]]
	val (n, m) = readInts()

	val pathMap: PathMap = (1 to m).foldLeft[PathMap](Map.empty.withDefaultValue(List.empty)) { case (map: PathMap, _) =>
		val (from, to) = readInts()
		val fromList = map(from)
		val toList = map(to)
		map.updated(from, to :: fromList).updated(to, from :: toList)
	}

	val resultPreMap = calc()
	println("Yes")
	(2 to n).foreach(node => println(resultPreMap(node)))

	private def calc(queue: MQueue[Int] = MQueue(1), preMap: MMap[Int, Int] = MMap(1 -> 0)): MMap[Int, Int] =
	{
		while (queue.nonEmpty) {
			val node = queue.dequeue()
			pathMap(node).foreach { nextNode =>
				val pre = preMap.get(nextNode)
				if (pre.isEmpty) {
					queue += nextNode
					preMap += Tuple2(nextNode, node)
				}
			}
		}
		preMap
	}

	private def readInts(): (Int, Int) =
	{
		val line = readLine().split(" ",2).map(_.toInt)
		(line(0), line(1))
	}
}

Score: 400 points

Problem Statement

There is a cave.

The cave has N rooms and M passages. The rooms are numbered 1 to N, and the passages are numbered 1 to M. Passage i connects Room A_i and Room B_i bidirectionally. One can travel between any two rooms by traversing passages. Room 1 is a special room with an entrance from the outside.

It is dark in the cave, so we have decided to place a signpost in each room except Room 1. The signpost in each room will point to one of the rooms directly connected to that room with a passage.

Since it is dangerous in the cave, our objective is to satisfy the condition below for each room except Room 1.

If you start in that room and repeatedly move to the room indicated by the signpost in the room you are in, you will reach Room 1 after traversing the minimum number of passages possible.

Determine whether there is a way to place signposts satisfying our objective, and print one such way if it exists.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

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

1 \leq A_i, B_i \leq N\ (1 \leq i \leq M)

A_i \neq B_i\ (1 \leq i \leq M)

One can travel between any two rooms by traversing passages.

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
:
A_M B_M

Output

If there is no way to place signposts satisfying the objective, print No.

Otherwise, print N lines. The first line should contain Yes, and the i-th line (2 \leq i \leq N) should contain the integer representing the room indicated by the signpost in Room i.

Sample Input 1

4 4
1 2
2 3
3 4
4 2

Sample Output 1

Yes
1
2
2

If we place the signposts as described in the sample output, the following happens:

Starting in Room 2, you will reach Room 1 after traversing one passage: (2) \to 1. This is the minimum number of passages possible.

Starting in Room 3, you will reach Room 1 after traversing two passages: (3) \to 2 \to 1. This is the minimum number of passages possible.

Starting in Room 4, you will reach Room 1 after traversing two passages: (4) \to 2 \to 1. This is the minimum number of passages possible.

Thus, the objective is satisfied.

Sample Input 2

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

Sample Output 2

Yes
6
5
5
1
1

If there are multiple solutions, any of them will be accepted.


import java.util.Scanner

import scala.collection.mutable

object Main extends App {
  val sc = new Scanner(System.in)
  val N, M = sc.nextInt
  val AB = Array.fill(M)((sc.nextInt, sc.nextInt))

  val G = (AB ++ AB.map(_.swap)).groupBy(_._1).map(x => { x._1 -> x._2.map(_._2) })

  val X = Array.fill(N)(false)
  val Y = Array.fill(N)(-1)

  def BFS(x: Int): Unit = {
    val Q = new mutable.Queue[Int]()
    X(x - 1) = true
    Y(x - 1) = 0
    Q += x
    while (Q.nonEmpty) {
      val v = Q.dequeue
      X(v - 1) = true
      G(v).foreach(i => {
        if (!X(i - 1)) {
          X(i - 1) = true
          Y(i - 1) = v
          Q += i
        }
      })
    }
  }
  BFS(1)

  println("Yes")
  Y.tail.foreach(println)
}


import java.util.Scanner

import scala.collection.mutable

object Main {
  def main(args: Array[String]): Unit = {

    val sc = new Scanner(System.in)
    val n = sc.nextInt()
    val m = sc.nextInt()

    val ab = Set.fill(m)(Set(sc.nextInt(), sc.nextInt()))

    val history = mutable.Set.empty[Int]
    val store = mutable.ArrayBuffer.fill(n)(0)
    val stack = mutable.Stack(1)

    def l(p: Int): Unit = {
      history += p
      ab.filter(_.contains(p)).flatten.filter(_ != p).filter(!history.contains(_)).foreach { i =>
        stack.push(i)
        if (store(i - 1) == 0) store(i - 1) = p
      }
      if (stack.nonEmpty) l(stack.pop())
    }

    l(stack.pop())

    println("Yes")
    store.tail.foreach(println)

  }

}


import java.util.Scanner

import scala.io.StdIn._
object Main extends App {

  val s = scala.io.StdIn.readLine()
  val t = scala.io.StdIn.readLine()

  def solve() = {
    if(s + t.last == t) "Yes" else "No"
  }

  println(solve())

}

Score : 100 points

Problem Statement

Takahashi wants to be a member of some web service.

He tried to register himself with the ID S, which turned out to be already used by another user.

Thus, he decides to register using a string obtained by appending one character at the end of S as his ID.

He is now trying to register with the ID T. Determine whether this string satisfies the property above.

Constraints

S and T are strings consisting of lowercase English letters.

1 \leq |S| \leq 10

|T| = |S| + 1

Input

Input is given from Standard Input in the following format:

S
T

Output

If T satisfies the property in Problem Statement, print Yes; otherwise, print No.

Sample Input 1

chokudai
chokudaiz

Sample Output 1

Yes

chokudaiz can be obtained by appending z at the end of chokudai.

Sample Input 2

snuke
snekee

Sample Output 2

No

snekee cannot be obtained by appending one character at the end of snuke.

Sample Input 3

a
aa

Sample Output 3

Yes


import java.util.Scanner

object Main extends App {
  val sc = new Scanner(System.in)
  val S, T = sc.nextLine

  val x = if (T.take(S.length) == S) "Yes" else "No"
  println(x)
}


import java.util.Scanner

import scala.io.StdIn._
object Main extends App {

  val Array(a,b,c,k) = scala.io.StdIn.readLine().split(" ").map(_.toInt)

  def solve() = {
    k match {
      case x if x <= a => 1 * x
      case x if x >= a && x <= a+b => a
      case x if x >= a+b && x <= a+b+c => a - (x - a - b)
      case _ => a - c
    }
  }

  println(solve())

}

Score : 200 points

Problem Statement

We have A cards, each of which has an integer 1 written on it. Similarly, we also have B cards with 0s and C cards with -1s.

We will pick up K among these cards. What is the maximum possible sum of the numbers written on the cards chosen?

Constraints

All values in input are integers.

0 \leq A, B, C

1 \leq K \leq A + B + C \leq 2 \times 10^9

Input

Input is given from Standard Input in the following format:

A B C K

Output

Print the maximum possible sum of the numbers written on the cards chosen.

Sample Input 1

2 1 1 3

Sample Output 1

2

Consider picking up two cards with 1s and one card with a 0.
In this case, the sum of the numbers written on the cards is 2, which is the maximum possible value.

Sample Input 2

1 2 3 4

Sample Output 2

0

Sample Input 3

2000000000 0 0 2000000000

Sample Output 3

2000000000


import java.util.Scanner

import scala.collection.mutable.ArrayBuffer
import scala.util.control.Breaks
import scala.math.abs

object Main extends App {

  val sc = new Scanner(System.in)
  val a, b, c, k = sc.nextLong()

  val res = (if (a >= k) {
    a
  } else if (a + b >= k) {
    a
  } else {
    a * 1 + ((k - a - b) * -1)
  })

  println(res)
}


//import java.util.Scanner

import java.io._
import java.nio.file.Files._
import java.nio.file.Path
import java.util.StringTokenizer

import scala.collection.immutable._
import scala.io.Codec
import scala.language.postfixOps

/**
  * Scala implementation of a faster java.util.Scanner
  * See: http://codeforces.com/blog/entry/7018
  */

class Scanner(reader: LineNumberReader) extends Iterable[String] with AutoCloseable {
  def this(reader: BufferedReader) = this(new LineNumberReader(reader))

  def this(reader: Reader) = this(new BufferedReader(reader))

  def this(inputStream: InputStream)(implicit codec: Codec) = this(new InputStreamReader(inputStream, codec.charSet))

  def this(path: Path)(implicit codec: Codec) = this(newBufferedReader(path, codec.charSet))

  def this(file: File)(implicit codec: Codec) = this(file.toPath)(codec)

  def this(str: String) = this(new StringReader(str))

  override def iterator: Iterator[String] = for {
    line <- Iterator.continually(reader.readLine()).takeWhile(_ != null)
    tokenizer = new StringTokenizer(line)
    tokens <- Iterator.continually(tokenizer).takeWhile(_.hasMoreTokens)
  } yield tokens.nextToken()

  private[this] var current = iterator

  def hasNext: Boolean = current.hasNext

  @inline def next(): String = current.next()

  /**
    * This is different from Java's scanner.nextLine
    * The Java one is a misnomer since it actually travel to end of current line
    * This one actually does fetch the next line
    */
  def nextLine(): String = {
    val line = reader.readLine()
    current = iterator
    line
  }

  def lineNumber: Int = reader.getLineNumber

  def nextString(): String = next()

  def nextBoolean(): Boolean = next().toBoolean

  def nextByte(radix: Int = 10): Byte = java.lang.Byte.parseByte(next(), radix)

  def nextShort(radix: Int = 10): Short = java.lang.Short.parseShort(next(), radix)

  def nextInt(radix: Int = 10): Int = java.lang.Integer.parseInt(next(), radix)

  def nextLong(radix: Int = 10): Long = java.lang.Long.parseLong(next(), radix)

  def nextBigInt(radix: Int = 10): BigInt = BigInt(next(), radix)

  def nextFloat(): Float = next().toFloat

  def nextDouble(): Double = next().toDouble

  def nextBigDecimal(): BigDecimal = BigDecimal(next())

  override def close(): Unit = reader.close()
}

class IUnionFind(val size: Int) {

  private case class Node(var parent: Option[Int], var treeSize: Int)

  private val nodes = Array.fill[Node](size)(Node(None, treeSize = 1))

  def union(t1: Int, t2: Int): IUnionFind = {
    if (t1 == t2) return this

    val root1 = root(t1)
    val root2 = root(t2)
    if (root1 == root2) return this

    val node1 = nodes(root1)
    val node2 = nodes(root2)

    if (node1.treeSize < node2.treeSize) {
      node1.parent = Some(t2)
      node2.treeSize += node1.treeSize
    } else {
      node2.parent = Some(t1)
      node1.treeSize += node2.treeSize
    }
    this
  }

  def connected(t1: Int, t2: Int): Boolean = t1 == t2 || root(t1) == root(t2)

  def root(t: Int): Int = nodes(t).parent match {
    case None => t
    case Some(p) =>
      nodes(t).parent = Some(root(p))
      nodes(t).parent.get
  }
}

object Main {
  def solve(sc: => Scanner): Unit = {
    val N, M = sc.nextInt()
    val X = sc.nextLong()
    val A = Array.fill(N)((sc.nextLong(), Array.fill(M)(sc.nextLong())))
    println(
      (0 until 1 << N).map(now =>
        A.zipWithIndex.filter { case (_, idx) => (now & 1 << idx) != 0 }
          .map(now => (now._1._1, now._1._2))
          .foldLeft((0L, Array.fill(M)(0L)))((a, b) =>
            (a._1 + b._1, a._2.zip(b._2).map { case (x, y) => x + y })
          )
      ).filter { case (_, b) => b.forall(_ >= X) }
        .map(_._1)
        .minOption
        .getOrElse(-1)
    )
  }

  @scala.annotation.tailrec
  def recursive(X: Long, now: Long = 100, ans: Long = 0): Long = {
    if (now >= X) ans
    else recursive(X, (now * 1.01).toLong, ans + 1)
  }

  def flip(A: Array[Boolean], B: Array[(Int, Long)]): Array[(Int, Long)] = {
    B.map { case (index, _) =>
      (index, (if (0 < index && A(index - 1)) B(index - 1)._2 else 0L) + (if (index + 1 < B.length && !A(index + 1)) B(index + 1)._2 else 0L))
    }
  }

  val mod: Long = (1e9 + 7).toLong

  @scala.annotation.tailrec
  def factorial(now: Long, ans: Long = 1): Long = {
    if (now == 0) 0 else if (now == 1) ans else factorial(now - 1, now * ans)
  }

  implicit class implicitInt(val N: Int) {
    def times[B](function: Int => B): IndexedSeq[B] = (0 until N).map(function)
  }

  def calc(input: (Long, Long, Long, Long, Long, Long)): Long = {
    input match {
      case (a, b, c, x, y, z) =>
        var ans: Long = Long.MaxValue
        if (b >= 3 && (a >= 2 || c >= 2)) {
          ans = Math.min(ans, Math.max(a - (b - 1), 0) * x + Math.max(c - Math.min(a - 1, b - 2), 0) * z)
          ans = Math.min(ans, Math.max(a - Math.min(c - 1, b - 2), 0) * x + Math.max(c - (b - 1), 0) * z)
        }
        //        println(ans)
        if (b >= 2 && (a >= b && c >= b + 1)) {
          ans = Math.min(ans, y + Math.max(0, a - (c - 1)) * x)
        }
        //        println(ans)
        if (b >= 2 && (a >= b + 1 && c >= b)) {
          ans = Math.min(ans, y + Math.max(0, c - (a - 1)) * z)
        }
        //        println(ans)
        if (a >= 2 && c >= 2)
          if (a != c)
            ans = Math.min(ans, Math.max(0, b - (Math.min(a, c) - 1)) * y)
          else if (a >= 3 && c >= 3)
            ans = Math.min(ans, Math.max(0, b - (Math.min(a, c) - 2)) * y + Math.min(x, z))
        //        println(ans)
        if (ans == Long.MaxValue) ans = -1
        ans
    }
  }

  def time(s: String): Int =
    s.substring(0, 2).toInt * 60 + s.substring(3, 5).toInt

  /*
  def getPermutation(begin: Long = 0): LazyList[Long] =
    LazyList.cons(begin, getPermutation(begin + 1))

  def getFibonacci(prevprev: Long = 0, prev: Long = 1): LazyList[Long] =
    LazyList.cons(prevprev, getFibonacci(prev, prevprev + prev))

  @scala.annotation.tailrec
  def recursive2(X: Set[Long], Y: LazyList[Long]): Long = if (X.contains(Y.head)) X.size else recursive2(X + Y.head, Y.tail)
*/
  def check(i: Int, X: String): Long = {
    if (X == "") 0 else check(i, X.tail) * i + X.head.toString.toLong
  }

  def shift(n: Long): Long = {
    if (n == 0) 0
    else if (n == 1) 1
    else shift(n - 1) << 1
  }

  def unShift(n: Long): Long = {
    if (n == 0) 0
    else unShift(n >> 1) + 1
  }

  @scala.annotation.tailrec
  def gcd(i: Long, j: Long): Long = {
    if (i < j) gcd(j, i)
    else if (j == 0) i
    else gcd(j, i % j)
  }

  def primeFactors(i: Long): List[Long] = primeFactors_(i, 1).sorted

  def primeFactors_(i: Long, j: Long): List[Long] = {
    if (j * j > i) List.empty else if (i % j == 0) primeFactors_(i, j + 1) ++ List[Long](j, i / j) else primeFactors_(i, j + 1)
  }

  def main(args: Array[String]): Unit = {
    val sc: Scanner = new Scanner(System.in)
    solve(sc)
  }
}

/*
object Util {
  def getPermutation(begin: Long = 0): LazyList[Long] =
    LazyList.cons(begin, getPermutation(begin + 1))

  def getPrimeList: LazyList[Long] =
    getPrimeListRecursive(getPermutation(begin = 2))

  private def getPrimeListRecursive(A: LazyList[Long]): LazyList[Long] =
    LazyList.cons(A.head, getPrimeListRecursive(A.tail.filter(_ % A.head != 0)))

  def fib(a: Long = 0, b: Long = 1, mod: Long = Long.MaxValue): LazyList[Long] = a #:: fib(b % mod, (a + b) % mod, mod)
}

 */

object ArabicRoman {

  type =?>[A, B] = PartialFunction[A, B]

  private val codeTable: List[(Int, String)] = List(
    (1000, "M"), (900, "CM"), (500, "D"), (400, "CD"), (100, "C"),
    (90, "XC"), (50, "L"), (40, "XL"), (10, "X"), (9, "IX"), (5, "V"), (4, "IV"), (1, "I"))

  val arabicToRoman: Int =?> String = {
    case src if src >= 1 && src <= 3999 =>

      @scala.annotation.tailrec
      def convert(left: Int, cont: String = "", code: List[(Int, String)] = codeTable): String = {
        val (unitVal, unitChar) = code.head
        left - unitVal match {
          case n if n == 0 => cont + unitChar
          case n if n > 0 => convert(n, cont + unitChar, code)
          case _ => convert(left, cont, code.tail)
        }
      }

      convert(src)
  }

  val romanToArabic: String =?> Int = {
    case src if Option(src).exists { s => {
      s.nonEmpty && """[^MDCLXVI]""".r.findFirstMatchIn(s.toUpperCase).isEmpty
    }
    } =>

      @scala.annotation.tailrec
      def convert(left: String, cont: Int = 0, code: List[(Int, String)] = codeTable): Int = {
        val (unitVal, unitChar) = code.head
        left.splitAt(unitChar.length) match {
          case ("", _) => cont
          case (`unitChar`, tail) => convert(tail, cont + unitVal, code)
          case _ => convert(left, cont, code.tail)
        }
      }

      convert(src.toUpperCase())
  }
}

object PowMod {
  def naivePowMod(a: Long, k: Long, m: Long): Long = {
    var t: Long = 1
    val aMod: Long = a % m

    for (_ <- 1L to k) {
      t = ((t % m) * aMod) % m
    }
    t.toInt
  }

  def powMod(a: Long, k: Long, m: Long): Long = {
    if (k == 0) 1
    else {
      val t = powMod(a, k / 2, m)
      if ((k % 2) == 0) (t * t) % m else (((t * t) % m) * a) % m
    }
  }
}

Score : 300 points

Problem

Takahashi, who is a novice in competitive programming, wants to learn M algorithms.
Initially, his understanding level of each of the M algorithms is 0.

Takahashi is visiting a bookstore, where he finds N books on algorithms.
The i-th book (1\leq i\leq N) is sold for C_i yen (the currency of Japan). If he buys and reads it, his understanding level of the j-th algorithm will increase by A_{i,j} for each j (1\leq j\leq M).
There is no other way to increase the understanding levels of the algorithms.

Takahashi's objective is to make his understanding levels of all the M algorithms X or higher. Determine whether this objective is achievable. If it is achievable, find the minimum amount of money needed to achieve it.

Constraints

All values in input are integers.

1\leq N, M\leq 12

1\leq X\leq 10^5

1\leq C_i \leq 10^5

0\leq A_{i, j} \leq 10^5

Input

Input is given from Standard Input in the following format:

N M X
C_1 A_{1,1} A_{1,2} \cdots A_{1,M}
C_2 A_{2,1} A_{2,2} \cdots A_{2,M}
\vdots
C_N A_{N,1} A_{N,2} \cdots A_{N,M}

Output

If the objective is not achievable, print -1; otherwise, print the minimum amount of money needed to achieve it.

Sample Input 1

3 3 10
60 2 2 4
70 8 7 9
50 2 3 9

Sample Output 1

120

Buying the second and third books makes his understanding levels of all the algorithms 10 or higher, at the minimum cost possible.

Sample Input 2

3 3 10
100 3 1 4
100 1 5 9
100 2 6 5

Sample Output 2

-1

Buying all the books is still not enough to make his understanding levels of all the algorithms 10 or higher.

Sample Input 3

8 5 22
100 3 7 5 3 1
164 4 5 2 7 8
334 7 2 7 2 9
234 4 7 2 8 2
541 5 4 3 3 6
235 4 8 6 9 7
394 3 6 1 6 2
872 8 4 3 7 2

Sample Output 3

1067


import java.io.{BufferedOutputStream, BufferedReader, InputStreamReader, PrintWriter}

case class Book(cost: Int, skills: Array[Int])

object Main {
  def main(args: Array[String]): Unit = {
    val in = new BufferedReader(new InputStreamReader(System.in))
    val out = new PrintWriter(new BufferedOutputStream(System.out))
    val Array(n, m, x) = in.readLine().split("\\s+").map(_.toInt)
    val books = Array.fill(n) {
      val line = in.readLine().split("\\s+").map(_.toInt)
      Book(line(0), line.slice(1, line.length))
    }

    var minCost = Int.MaxValue

    for (i <- 0 until Math.pow(2, n).toInt) {
      val readBooks = books.foldLeft[(List[Book], Int)]((Nil, i)) {
        case ((acc, q), book) =>
          if ((q & 1) == 1) {
            (book :: acc, q >> 1)
          } else {
            (acc, q >> 1)
          }
      }._1
      var cost = 0
      val skill = Array.ofDim[Int](m)
      for (book <- readBooks) {
        cost += book.cost
        for (i <- skill.indices) {
          skill(i) += book.skills(i)
        }
      }
      if (skill.forall(_ >= x)) {
        minCost = Math.min(minCost, cost)
      }
    }

    out.println(
      if (minCost == Int.MaxValue) {
        -1
      } else {
        minCost
      }
    )

    out.flush()
  }
}


object Main extends App {
	val pw = new java.io.PrintWriter(System.out)

	@scala.annotation.tailrec
	def loop(start: Int, end: Int)(f: Int => Unit): Unit =
		if (start < end) {
			f(start)
			loop(start + 1, end)(f)
		}

	val n, m, x = Scanner.nextInt

	case class Book(c: Int, a: Array[Int])
	val bs = Seq.fill(n)(Book(Scanner.nextInt, Array.fill(m)(Scanner.nextInt)))

	val ans = (1 to n).flatMap(bs combinations _).collect{case b if (b.map(_.a).transpose.map(_.sum).forall(_ >= x)) => b.map(_.c).sum}

	pw.println(if (ans.isEmpty) -1 else ans.min)
	pw.flush()
}

object Scanner {
	private val buf = new Array[Byte](1024); private var ptr = 0; private var len = 0

	@inline private def isPrintableChar(c: Int): Boolean = 33 <= c && c <= 126
	@inline private def hasNextByte(): Boolean = if (ptr >= len) { ptr = 0; len = System.in.read(buf); len > 0 } else true
	@inline private def hasNext(): Boolean = {
		while (hasNextByte() && !isPrintableChar(buf(ptr))) ptr += 1
		hasNextByte()
	}
	@inline private def readByte(): Byte = if (hasNextByte()) { val res = buf(ptr); ptr += 1; res } else -1

	def next(): String = {
		if(!hasNext()) ???
		val sb = new StringBuilder; var b = readByte()
		while (isPrintableChar(b)) { sb.append(b.toChar); b = readByte() }
		sb.toString
	}

	def nextInt(): Int = {
		val n = nextLong()
		if (n < Int.MinValue || Int.MaxValue < n) ???
		n.toInt
	}

	def nextLong(): Long = {
		if(!hasNext()) ???
		var minus = false; var b = readByte()
		if (b == '-') { minus = true; b = readByte() }

		@scala.annotation.tailrec
		def go (b: Byte, n: Long = 0): Long = if ('0' <= b && b <= '9') go(readByte(), n * 10 + b - '0') else if (minus) -n else n

		go(b)
	}

	def nextDouble(): Double = next().toDouble
}


import java.io.{BufferedOutputStream, BufferedReader, InputStreamReader, PrintWriter}

object Main {
  def main(args: Array[String]): Unit = {
    val in = new BufferedReader(new InputStreamReader(System.in))
    val out = new PrintWriter(new BufferedOutputStream(System.out))
    val s = in.readLine()

    out.println(
      if (s == "ABC") {
        "ARC"
      } else {
        "ABC"
      }
    )

    out.flush()
  }
}

Score : 100 points

Problem Statement

AtCoder Inc. holds a contest every Saturday.

There are two types of contests called ABC and ARC, and just one of them is held at a time.

The company holds these two types of contests alternately: an ARC follows an ABC and vice versa.

Given a string S representing the type of the contest held last week, print the string representing the type of the contest held this week.

Constraints

S is ABC or ARC.

Input

Input is given from Standard Input in the following format:

S

Output

Print the string representing the type of the contest held this week.

Sample Input 1

ABC

Sample Output 1

ARC

They held an ABC last week, so they will hold an ARC this week.


object Main extends App {
  val sc = new java.util.Scanner(System.in)
  val n, m = sc.nextInt
  val h = Seq.fill(n)(sc.nextInt)

  var good_t = (1 to n).toSet

  for (_ <- 1 to m) {
    val a, b = sc.nextInt

    if (h(a - 1) >= h(b - 1)) {
      good_t = good_t - b
    } else {
      good_t = good_t - a
    }
  }

  print(good_t.size)
}

Score : 300 points

Problem Statement

There are N observatories in AtCoder Hill, called Obs. 1, Obs. 2, ..., Obs. N. The elevation of Obs. i is H_i.
There are also M roads, each connecting two different observatories. Road j connects Obs. A_j and Obs. B_j.

Obs. i is said to be good when its elevation is higher than those of all observatories that can be reached from Obs. i using just one road.
Note that Obs. i is also good when no observatory can be reached from Obs. i using just one road.

How many good observatories are there?

Constraints

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq H_i \leq 10^9

1 \leq A_i,B_i \leq N

A_i \neq B_i

Multiple roads may connect the same pair of observatories.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M
H_1 H_2 ... H_N
A_1 B_1
A_2 B_2
:
A_M B_M

Output

Print the number of good observatories.

Sample Input 1

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

Sample Output 1

2

From Obs. 1, you can reach Obs. 3 using just one road. The elevation of Obs. 1 is not higher than that of Obs. 3, so Obs. 1 is not good.

From Obs. 2, you can reach Obs. 3 and 4 using just one road. The elevation of Obs. 2 is not higher than that of Obs. 3, so Obs. 2 is not good.

From Obs. 3, you can reach Obs. 1 and 2 using just one road. The elevation of Obs. 3 is higher than those of Obs. 1 and 2, so Obs. 3 is good.

From Obs. 4, you can reach Obs. 2 using just one road. The elevation of Obs. 4 is higher than that of Obs. 2, so Obs. 4 is good.

Thus, the good observatories are Obs. 3 and 4, so there are two good observatories.

Sample Input 2

6 5
8 6 9 1 2 1
1 3
4 2
4 3
4 6
4 6

Sample Output 2

3


//import java.util.Scanner

import java.io._
import java.nio.file.Files._
import java.nio.file.Path
import java.util.StringTokenizer

import scala.collection.immutable._
import scala.io.Codec

/**
  * Scala implementation of a faster java.util.Scanner
  * See: http://codeforces.com/blog/entry/7018
  */

class Scanner(reader: LineNumberReader) extends Iterable[String] with AutoCloseable {
  def this(reader: BufferedReader) = this(new LineNumberReader(reader))

  def this(reader: Reader) = this(new BufferedReader(reader))

  def this(inputStream: InputStream)(implicit codec: Codec) = this(new InputStreamReader(inputStream, codec.charSet))

  def this(path: Path)(implicit codec: Codec) = this(newBufferedReader(path, codec.charSet))

  def this(file: File)(implicit codec: Codec) = this(file.toPath)(codec)

  def this(str: String) = this(new StringReader(str))

  override def iterator: Iterator[String] = for {
    line <- Iterator.continually(reader.readLine()).takeWhile(_ != null)
    tokenizer = new StringTokenizer(line)
    tokens <- Iterator.continually(tokenizer).takeWhile(_.hasMoreTokens)
  } yield tokens.nextToken()

  private[this] var current = iterator

  def hasNext: Boolean = current.hasNext

  @inline def next(): String = current.next()

  /**
    * This is different from Java's scanner.nextLine
    * The Java one is a misnomer since it actually travel to end of current line
    * This one actually does fetch the next line
    */
  def nextLine(): String = {
    val line = reader.readLine()
    current = iterator
    line
  }

  def lineNumber: Int = reader.getLineNumber

  def nextString(): String = next()

  def nextBoolean(): Boolean = next().toBoolean

  def nextByte(radix: Int = 10): Byte = java.lang.Byte.parseByte(next(), radix)

  def nextShort(radix: Int = 10): Short = java.lang.Short.parseShort(next(), radix)

  def nextInt(radix: Int = 10): Int = java.lang.Integer.parseInt(next(), radix)

  def nextLong(radix: Int = 10): Long = java.lang.Long.parseLong(next(), radix)

  def nextBigInt(radix: Int = 10): BigInt = BigInt(next(), radix)

  def nextFloat(): Float = next().toFloat

  def nextDouble(): Double = next().toDouble

  def nextBigDecimal(): BigDecimal = BigDecimal(next())

  override def close(): Unit = reader.close()
}

class IUnionFind(val size: Int) {

  private case class Node(var parent: Option[Int], var treeSize: Int)

  private val nodes = Array.fill[Node](size)(Node(None, treeSize = 1))

  def union(t1: Int, t2: Int): IUnionFind = {
    if (t1 == t2) return this

    val root1 = root(t1)
    val root2 = root(t2)
    if (root1 == root2) return this

    val node1 = nodes(root1)
    val node2 = nodes(root2)

    if (node1.treeSize < node2.treeSize) {
      node1.parent = Some(t2)
      node2.treeSize += node1.treeSize
    } else {
      node2.parent = Some(t1)
      node1.treeSize += node2.treeSize
    }
    this
  }

  def connected(t1: Int, t2: Int): Boolean = t1 == t2 || root(t1) == root(t2)

  def root(t: Int): Int = nodes(t).parent match {
    case None => t
    case Some(p) =>
      nodes(t).parent = Some(root(p))
      nodes(t).parent.get
  }
}

object Main {
  def solve(sc: => Scanner): Unit = {
    val X = sc.nextLong()
    val A = (0L to 1100L).map(x => (recursive(x), x)).toMap
    val ans =
      A.keys.toList.combinations(2)
        .filter(x => Math.abs(x(0) - x(1)) == X || Math.abs(x(0) + x(1)) == X)
        .toList.head
    val ans1 = A(ans(0))
    val ans2 = A(ans(1))
    if (recursive(ans1) - recursive(ans2) == X) println(ans1 + " " + ans2)
    else if (recursive(ans1) + recursive(ans2) == X) println(ans1 + " " + -ans2)
    else if (-recursive(ans1) + recursive(ans2) == X) println(-ans1 + " " + ans2)
    else if (-recursive(ans1) - recursive(ans2) == X) println(-ans1 + " " + -ans2)
  }

  @scala.annotation.tailrec
  def recursive(input: Long, ret: Long = 1, num: Long = 5): Long = {
    if (num == 0) ret
    else recursive(input, ret * input, num - 1)
  }

  def flip(A: Array[Boolean], B: Array[(Int, Long)]): Array[(Int, Long)] = {
    B.map { case (index, _) =>
      (index, (if (0 < index && A(index - 1)) B(index - 1)._2 else 0L) + (if (index + 1 < B.length && !A(index + 1)) B(index + 1)._2 else 0L))
    }
  }

  val mod: Long = (1e9 + 7).toLong

  @scala.annotation.tailrec
  def factorial(now: Long, ans: Long = 1): Long = {
    if (now == 0) 0 else if (now == 1) ans else factorial(now - 1, now * ans)
  }

  implicit class implicitInt(val N: Int) {
    def times[B](function: Int => B): IndexedSeq[B] = (0 until N).map(function)
  }

  def calc(input: (Long, Long, Long, Long, Long, Long)): Long = {
    input match {
      case (a, b, c, x, y, z) =>
        var ans: Long = Long.MaxValue
        if (b >= 3 && (a >= 2 || c >= 2)) {
          ans = Math.min(ans, Math.max(a - (b - 1), 0) * x + Math.max(c - Math.min(a - 1, b - 2), 0) * z)
          ans = Math.min(ans, Math.max(a - Math.min(c - 1, b - 2), 0) * x + Math.max(c - (b - 1), 0) * z)
        }
        //        println(ans)
        if (b >= 2 && (a >= b && c >= b + 1)) {
          ans = Math.min(ans, y + Math.max(0, a - (c - 1)) * x)
        }
        //        println(ans)
        if (b >= 2 && (a >= b + 1 && c >= b)) {
          ans = Math.min(ans, y + Math.max(0, c - (a - 1)) * z)
        }
        //        println(ans)
        if (a >= 2 && c >= 2)
          if (a != c)
            ans = Math.min(ans, Math.max(0, b - (Math.min(a, c) - 1)) * y)
          else if (a >= 3 && c >= 3)
            ans = Math.min(ans, Math.max(0, b - (Math.min(a, c) - 2)) * y + Math.min(x, z))
        //        println(ans)
        if (ans == Long.MaxValue) ans = -1
        ans
    }
  }

  def time(s: String): Int =
    s.substring(0, 2).toInt * 60 + s.substring(3, 5).toInt

  /*
  def getPermutation(begin: Long = 0): LazyList[Long] =
    LazyList.cons(begin, getPermutation(begin + 1))

  def getFibonacci(prevprev: Long = 0, prev: Long = 1): LazyList[Long] =
    LazyList.cons(prevprev, getFibonacci(prev, prevprev + prev))

  @scala.annotation.tailrec
  def recursive2(X: Set[Long], Y: LazyList[Long]): Long = if (X.contains(Y.head)) X.size else recursive2(X + Y.head, Y.tail)
*/
  def check(i: Int, X: String): Long = {
    if (X == "") 0 else check(i, X.tail) * i + X.head.toString.toLong
  }

  def shift(n: Long): Long = {
    if (n == 0) 0
    else if (n == 1) 1
    else shift(n - 1) << 1
  }

  def unShift(n: Long): Long = {
    if (n == 0) 0
    else unShift(n >> 1) + 1
  }

  @scala.annotation.tailrec
  def gcd(i: Long, j: Long): Long = {
    if (i < j) gcd(j, i)
    else if (j == 0) i
    else gcd(j, i % j)
  }

  def primeFactors(i: Long): List[Long] = primeFactors_(i, 1).sorted

  def primeFactors_(i: Long, j: Long): List[Long] = {
    if (j * j > i) List.empty else if (i % j == 0) primeFactors_(i, j + 1) ++ List[Long](j, i / j) else primeFactors_(i, j + 1)
  }

  def main(args: Array[String]): Unit = {
    val sc: Scanner = new Scanner(System.in)
    solve(sc)
  }
}

/*
object Util {
  def getPermutation(begin: Long = 0): LazyList[Long] =
    LazyList.cons(begin, getPermutation(begin + 1))

  def getPrimeList: LazyList[Long] =
    getPrimeListRecursive(getPermutation(begin = 2))

  private def getPrimeListRecursive(A: LazyList[Long]): LazyList[Long] =
    LazyList.cons(A.head, getPrimeListRecursive(A.tail.filter(_ % A.head != 0)))

  def fib(a: Long = 0, b: Long = 1, mod: Long = Long.MaxValue): LazyList[Long] = a #:: fib(b % mod, (a + b) % mod, mod)
}

 */

object ArabicRoman {

  type =?>[A, B] = PartialFunction[A, B]

  private val codeTable: List[(Int, String)] = List(
    (1000, "M"), (900, "CM"), (500, "D"), (400, "CD"), (100, "C"),
    (90, "XC"), (50, "L"), (40, "XL"), (10, "X"), (9, "IX"), (5, "V"), (4, "IV"), (1, "I"))

  val arabicToRoman: Int =?> String = {
    case src if src >= 1 && src <= 3999 =>

      @scala.annotation.tailrec
      def convert(left: Int, cont: String = "", code: List[(Int, String)] = codeTable): String = {
        val (unitVal, unitChar) = code.head
        left - unitVal match {
          case n if n == 0 => cont + unitChar
          case n if n > 0 => convert(n, cont + unitChar, code)
          case _ => convert(left, cont, code.tail)
        }
      }

      convert(src)
  }

  val romanToArabic: String =?> Int = {
    case src if Option(src).exists { s => {
      s.nonEmpty && """[^MDCLXVI]""".r.findFirstMatchIn(s.toUpperCase).isEmpty
    }
    } =>

      @scala.annotation.tailrec
      def convert(left: String, cont: Int = 0, code: List[(Int, String)] = codeTable): Int = {
        val (unitVal, unitChar) = code.head
        left.splitAt(unitChar.length) match {
          case ("", _) => cont
          case (`unitChar`, tail) => convert(tail, cont + unitVal, code)
          case _ => convert(left, cont, code.tail)
        }
      }

      convert(src.toUpperCase())
  }
}

object PowMod {
  def naivePowMod(a: Long, k: Long, m: Long): Long = {
    var t: Long = 1
    val aMod: Long = a % m

    for (_ <- 1L to k) {
      t = ((t % m) * aMod) % m
    }
    t.toInt
  }

  def powMod(a: Long, k: Long, m: Long): Long = {
    if (k == 0) 1
    else {
      val t = powMod(a, k / 2, m)
      if ((k % 2) == 0) (t * t) % m else (((t * t) % m) * a) % m
    }
  }
}

Score : 400 points

Problem Statement

Give a pair of integers (A, B) such that A^5-B^5 = X.
It is guaranteed that there exists such a pair for the given integer X.

Constraints

1 \leq X \leq 10^9

X is an integer.

There exists a pair of integers (A, B) satisfying the condition in Problem Statement.

Input

Input is given from Standard Input in the following format:

X

Output

Print A and B, with space in between.
If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.

A B

Sample Input 1

33

Sample Output 1

2 -1

For A=2 and B=-1, A^5-B^5 = 33.

Sample Input 2

1

Sample Output 2

0 -1


import java.util

import scala.io.StdIn.readLine
import scala.io.StdIn.readInt
import java.util.Scanner
import java.util.ArrayList

object Main extends App {
  val sc = new Scanner(System.in)
  val x = sc.nextLong()
  val f = (x:Long)=>x*x*x*x*x
  for(i <- -1000 to 1000 ;j<- -1000 to 1000){
    if(f(i) - f(j) == x){
      println(f"$i $j")
      System.exit(0)
    }
  }

}


import scala.io.StdIn._
object Main {
  def main(args:Array[String]) {
    val X = readLine().toLong

    var A = 0d
    var B = 0
    var flag = true
    var i = 1
    while(flag) {
      if(i % 2 == 0) B += 1
      else B *= -1
      A = Math.pow(X + Math.pow(B, 5), 1d / 5)
      if(A - A.toInt == 0) flag = false
      i += 1
    }
    println(A.toInt + " " + B)
  }
}


//import java.util.Scanner

import java.io._
import java.nio.file.Files._
import java.nio.file.Path
import java.util.StringTokenizer

import scala.collection.immutable._
import scala.io.Codec
import scala.language.postfixOps

/**
  * Scala implementation of a faster java.util.Scanner
  * See: http://codeforces.com/blog/entry/7018
  */

class Scanner(reader: LineNumberReader) extends Iterable[String] with AutoCloseable {
  def this(reader: BufferedReader) = this(new LineNumberReader(reader))

  def this(reader: Reader) = this(new BufferedReader(reader))

  def this(inputStream: InputStream)(implicit codec: Codec) = this(new InputStreamReader(inputStream, codec.charSet))

  def this(path: Path)(implicit codec: Codec) = this(newBufferedReader(path, codec.charSet))

  def this(file: File)(implicit codec: Codec) = this(file.toPath)(codec)

  def this(str: String) = this(new StringReader(str))

  override def iterator: Iterator[String] = for {
    line <- Iterator.continually(reader.readLine()).takeWhile(_ != null)
    tokenizer = new StringTokenizer(line)
    tokens <- Iterator.continually(tokenizer).takeWhile(_.hasMoreTokens)
  } yield tokens.nextToken()

  private[this] var current = iterator

  def hasNext: Boolean = current.hasNext

  @inline def next(): String = current.next()

  /**
    * This is different from Java's scanner.nextLine
    * The Java one is a misnomer since it actually travel to end of current line
    * This one actually does fetch the next line
    */
  def nextLine(): String = {
    val line = reader.readLine()
    current = iterator
    line
  }

  def lineNumber: Int = reader.getLineNumber

  def nextString(): String = next()

  def nextBoolean(): Boolean = next().toBoolean

  def nextByte(radix: Int = 10): Byte = java.lang.Byte.parseByte(next(), radix)

  def nextShort(radix: Int = 10): Short = java.lang.Short.parseShort(next(), radix)

  def nextInt(radix: Int = 10): Int = java.lang.Integer.parseInt(next(), radix)

  def nextLong(radix: Int = 10): Long = java.lang.Long.parseLong(next(), radix)

  def nextBigInt(radix: Int = 10): BigInt = BigInt(next(), radix)

  def nextFloat(): Float = next().toFloat

  def nextDouble(): Double = next().toDouble

  def nextBigDecimal(): BigDecimal = BigDecimal(next())

  override def close(): Unit = reader.close()
}

class IUnionFind(val size: Int) {

  private case class Node(var parent: Option[Int], var treeSize: Int)

  private val nodes = Array.fill[Node](size)(Node(None, treeSize = 1))

  def union(t1: Int, t2: Int): IUnionFind = {
    if (t1 == t2) return this

    val root1 = root(t1)
    val root2 = root(t2)
    if (root1 == root2) return this

    val node1 = nodes(root1)
    val node2 = nodes(root2)

    if (node1.treeSize < node2.treeSize) {
      node1.parent = Some(t2)
      node2.treeSize += node1.treeSize
    } else {
      node2.parent = Some(t1)
      node1.treeSize += node2.treeSize
    }
    this
  }

  def connected(t1: Int, t2: Int): Boolean = t1 == t2 || root(t1) == root(t2)

  def root(t: Int): Int = nodes(t).parent match {
    case None => t
    case Some(p) =>
      nodes(t).parent = Some(root(p))
      nodes(t).parent.get
  }
}

object Main {
  def solve(sc: => Scanner): Unit = {
    val K, A, B = sc.nextLong()
    println(if (B / K - A / K != 0 || B % K == 0) "OK" else "NG")
  }

  @scala.annotation.tailrec
  def recursive(N: Long, ret: List[Long] = List(), now: Long = 2): List[Long] = {
    if (N < now * now) ret :+ N
    else if (N % now == 0) recursive(N / now, ret :+ now, now)
    else recursive(N, ret, now + 1)
  }

  @scala.annotation.tailrec
  def recursive2(N: Long, now: Long = 1): Long = {
    if (N >= now) recursive2(N - now, now + 1)
    else now - 1
  }

  def flip(A: Array[Boolean], B: Array[(Int, Long)]): Array[(Int, Long)] = {
    B.map { case (index, _) =>
      (index, (if (0 < index && A(index - 1)) B(index - 1)._2 else 0L) + (if (index + 1 < B.length && !A(index + 1)) B(index + 1)._2 else 0L))
    }
  }

  val mod: Long = (1e9 + 7).toLong

  @scala.annotation.tailrec
  def factorial(now: Long, ans: Long = 1): Long = {
    if (now == 0) 0 else if (now == 1) ans else factorial(now - 1, now * ans)
  }

  implicit class implicitInt(val N: Int) {
    def times[B](function: Int => B): IndexedSeq[B] = (0 until N).map(function)
  }

  def calc(input: (Long, Long, Long, Long, Long, Long)): Long = {
    input match {
      case (a, b, c, x, y, z) =>
        var ans: Long = Long.MaxValue
        if (b >= 3 && (a >= 2 || c >= 2)) {
          ans = Math.min(ans, Math.max(a - (b - 1), 0) * x + Math.max(c - Math.min(a - 1, b - 2), 0) * z)
          ans = Math.min(ans, Math.max(a - Math.min(c - 1, b - 2), 0) * x + Math.max(c - (b - 1), 0) * z)
        }
        //        println(ans)
        if (b >= 2 && (a >= b && c >= b + 1)) {
          ans = Math.min(ans, y + Math.max(0, a - (c - 1)) * x)
        }
        //        println(ans)
        if (b >= 2 && (a >= b + 1 && c >= b)) {
          ans = Math.min(ans, y + Math.max(0, c - (a - 1)) * z)
        }
        //        println(ans)
        if (a >= 2 && c >= 2)
          if (a != c)
            ans = Math.min(ans, Math.max(0, b - (Math.min(a, c) - 1)) * y)
          else if (a >= 3 && c >= 3)
            ans = Math.min(ans, Math.max(0, b - (Math.min(a, c) - 2)) * y + Math.min(x, z))
        //        println(ans)
        if (ans == Long.MaxValue) ans = -1
        ans
    }
  }

  def time(s: String): Int =
    s.substring(0, 2).toInt * 60 + s.substring(3, 5).toInt

  /*
  def getPermutation(begin: Long = 0): LazyList[Long] =
    LazyList.cons(begin, getPermutation(begin + 1))

  def getFibonacci(prevprev: Long = 0, prev: Long = 1): LazyList[Long] =
    LazyList.cons(prevprev, getFibonacci(prev, prevprev + prev))

  @scala.annotation.tailrec
  def recursive2(X: Set[Long], Y: LazyList[Long]): Long = if (X.contains(Y.head)) X.size else recursive2(X + Y.head, Y.tail)
*/
  def check(i: Int, X: String): Long = {
    if (X == "") 0 else check(i, X.tail) * i + X.head.toString.toLong
  }

  def shift(n: Long): Long = {
    if (n == 0) 0
    else if (n == 1) 1
    else shift(n - 1) << 1
  }

  def unShift(n: Long): Long = {
    if (n == 0) 0
    else unShift(n >> 1) + 1
  }

  @scala.annotation.tailrec
  def gcd(i: Long, j: Long): Long = {
    if (i < j) gcd(j, i)
    else if (j == 0) i
    else gcd(j, i % j)
  }

  def primeFactors(i: Long): List[Long] = primeFactors_(i, 1).sorted

  def primeFactors_(i: Long, j: Long): List[Long] = {
    if (j * j > i) List.empty else if (i % j == 0) primeFactors_(i, j + 1) ++ List[Long](j, i / j) else primeFactors_(i, j + 1)
  }

  def main(args: Array[String]): Unit = {
    val sc: Scanner = new Scanner(System.in)
    solve(sc)
  }
}

/*
object Util {
  def getPermutation(begin: Long = 0): LazyList[Long] =
    LazyList.cons(begin, getPermutation(begin + 1))

  def getPrimeList: LazyList[Long] =
    getPrimeListRecursive(getPermutation(begin = 2))

  private def getPrimeListRecursive(A: LazyList[Long]): LazyList[Long] =
    LazyList.cons(A.head, getPrimeListRecursive(A.tail.filter(_ % A.head != 0)))

  def fib(a: Long = 0, b: Long = 1, mod: Long = Long.MaxValue): LazyList[Long] = a #:: fib(b % mod, (a + b) % mod, mod)
}

 */

object ArabicRoman {

  type =?>[A, B] = PartialFunction[A, B]

  private val codeTable: List[(Int, String)] = List(
    (1000, "M"), (900, "CM"), (500, "D"), (400, "CD"), (100, "C"),
    (90, "XC"), (50, "L"), (40, "XL"), (10, "X"), (9, "IX"), (5, "V"), (4, "IV"), (1, "I"))

  val arabicToRoman: Int =?> String = {
    case src if src >= 1 && src <= 3999 =>

      @scala.annotation.tailrec
      def convert(left: Int, cont: String = "", code: List[(Int, String)] = codeTable): String = {
        val (unitVal, unitChar) = code.head
        left - unitVal match {
          case n if n == 0 => cont + unitChar
          case n if n > 0 => convert(n, cont + unitChar, code)
          case _ => convert(left, cont, code.tail)
        }
      }

      convert(src)
  }

  val romanToArabic: String =?> Int = {
    case src if Option(src).exists { s => {
      s.nonEmpty && """[^MDCLXVI]""".r.findFirstMatchIn(s.toUpperCase).isEmpty
    }
    } =>

      @scala.annotation.tailrec
      def convert(left: String, cont: Int = 0, code: List[(Int, String)] = codeTable): Int = {
        val (unitVal, unitChar) = code.head
        left.splitAt(unitChar.length) match {
          case ("", _) => cont
          case (`unitChar`, tail) => convert(tail, cont + unitVal, code)
          case _ => convert(left, cont, code.tail)
        }
      }

      convert(src.toUpperCase())
  }
}

object PowMod {
  def naivePowMod(a: Long, k: Long, m: Long): Long = {
    var t: Long = 1
    val aMod: Long = a % m

    for (_ <- 1L to k) {
      t = ((t % m) * aMod) % m
    }
    t.toInt
  }

  def powMod(a: Long, k: Long, m: Long): Long = {
    if (k == 0) 1
    else {
      val t = powMod(a, k / 2, m)
      if ((k % 2) == 0) (t * t) % m else (((t * t) % m) * a) % m
    }
  }
}

Score: 100 points

Problem Statement

Takahashi the Jumbo will practice golf.

His objective is to get a carry distance that is a multiple of K, while he can only make a carry distance of between A and B (inclusive).

If he can achieve the objective, print OK; if he cannot, print NG.

Constraints

All values in input are integers.

1 \leq A \leq B \leq 1000

1 \leq K \leq 1000

Input

Input is given from Standard Input in the following format:

K
A B

Output

If he can achieve the objective, print OK; if he cannot, print NG.

Sample Input 1

7
500 600

Sample Output 1

OK

Among the multiples of 7, for example, 567 lies between 500 and 600.

Sample Input 2

4
5 7

Sample Output 2

NG

No multiple of 4 lies between 5 and 7.

Sample Input 3

1
11 11

Sample Output 3

OK


object Main extends App {
  import io.StdIn._

  val k = readInt()
  val Array(a, b) =
    readLine()
      .split(" ")
      .map(_.toInt)

  var m = a
  var pass = false
  while (m < b && !pass) {
    if (m % k == 0)
      pass = true

    m += 1
  }

  println {
    if (pass) "OK"
    else "NO"
  }
}


object Main extends App {
	val pw = new java.io.PrintWriter(System.out)

	@scala.annotation.tailrec
	def loop(start: Int, end: Int)(f: Int => Unit): Unit =
		if (start < end) {
			f(start)
			loop(start + 1, end)(f)
		}

	val k, a, b = Scanner.nextInt

	pw.println( if (a == b || (1 to (b-a) filter (_ % k == 0)).nonEmpty) "OK" else "NG" )
	pw.flush()
}

object Scanner {
	private val buf = new Array[Byte](1024); private var ptr = 0; private var len = 0

	@inline private def isPrintableChar(c: Int): Boolean = 33 <= c && c <= 126
	@inline private def hasNextByte(): Boolean = if (ptr >= len) { ptr = 0; len = System.in.read(buf); len > 0 } else true
	@inline private def hasNext(): Boolean = {
		while (hasNextByte() && !isPrintableChar(buf(ptr))) ptr += 1
		hasNextByte()
	}
	@inline private def readByte(): Byte = if (hasNextByte()) { val res = buf(ptr); ptr += 1; res } else -1

	def next(): String = {
		if(!hasNext()) ???
		val sb = new StringBuilder; var b = readByte()
		while (isPrintableChar(b)) { sb.append(b.toChar); b = readByte() }
		sb.toString
	}

	def nextInt(): Int = {
		val n = nextLong()
		if (n < Int.MinValue || Int.MaxValue < n) ???
		n.toInt
	}

	def nextLong(): Long = {
		if(!hasNext()) ???
		var minus = false; var b = readByte()
		if (b == '-') { minus = true; b = readByte() }

		@scala.annotation.tailrec
		def go (b: Byte, n: Long = 0): Long = if ('0' <= b && b <= '9') go(readByte(), n * 10 + b - '0') else if (minus) -n else n

		go(b)
	}

	def nextDouble(): Double = next().toDouble
}


object Main extends App{
  val k = scala.io.StdIn.readInt
  val Array(a,b) = scala.io.StdIn.readLine.split(" ").map(_.toInt)
  if((a to b).map(_%k).contains(0)) println ("OK")
  else println("NG")
}


import scala.util.control.Breaks

object Main extends App {
  val k =  scala.io.StdIn.readInt()
  val ab =  scala.io.StdIn.readLine()
  val abArray = ab.split(" ").map(_.toInt).toSeq
  val a = abArray(0)
  val b = abArray(1)
  var okFlg = false

  val break = new Breaks
  break.breakable {
    (a to b).foreach { i =>
      if (i % k == 0) {
        okFlg = true
        break.break()
      }
    }
  }
  if (okFlg) println("OK") else println("NG")
}


object Main {
  import scala.io.StdIn.readLine
  def main(args: Array[String]): Unit = {
    val x = readLine.trim().toLong
    var current = 100L

    var i = 0
    while(current < x) {
      current += (current * 0.01).toLong
      i += 1
    }
    println(i)
  }
}

Score : 200 points

Problem Statement

Takahashi has a deposit of 100 yen (the currency of Japan) in AtCoder Bank.

The bank pays an annual interest rate of 1 % compounded annually. (A fraction of less than one yen is discarded.)

Assuming that nothing other than the interest affects Takahashi's balance, in how many years does the balance reach X yen or above for the first time?

Constraints

101 \le X \le 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X

Output

Print the number of years it takes for Takahashi's balance to reach X yen or above for the first time.

Sample Input 1

103

Sample Output 1

3

The balance after one year is 101 yen.

The balance after two years is 102 yen.

The balance after three years is 103 yen.

Thus, it takes three years for the balance to reach 103 yen or above.

Sample Input 2

1000000000000000000

Sample Output 2

3760

Sample Input 3

1333333333

Sample Output 3

1706


import scala.io.StdIn._

object Helpers {
  def readIntLine = readLine.split(" ").map(_.toInt)
  def readInts(n: Int) = Array.fill(n)(readInt)

  def splitDigits(i: Int): Seq[Int] = i.toString.map(_.asDigit)

  def log[A](x: A): A = {println(x); x}

  implicit class IntImplicits(val i: Int) {
    def isEven = i % 2 == 0
    def isOdd = !isEven

    def ** (exp: Int): Int = scala.math.pow(i, exp).intValue
  }

  implicit class LongImplicits(val i: Long) {
    def isEven = i % 2 == 0
    def isOdd = !isEven

    def ** (exp: Long): Long = scala.math.pow(i, exp).longValue
  }
}

object Main extends App {
  import Helpers._

  val Array(n, m, q) = readIntLine
  val as = for (_ <- 0 until q) yield readIntLine

  def mycomb[T](n: Int, l: List[T]): List[List[T]] =
    n match {
      case 0 => List(List())
      case _ => for(el <- l;
                sl <- mycomb(n-1, l dropWhile { _ != el } ))
                yield el :: sl
  }

  def comb[T](n: Int, l: List[T]): List[List[T]] = mycomb(n, l.distinct)

  val candidate = comb(n, (1 to m).toList)

  println((for {
    x <- candidate
  } yield (for {
    a <- as
    if x(a(1) - 1) - x(a(0) - 1) == a(2)
  } yield a(3)).sum).max)
}

Score : 300 points

Problem Statement

Given are positive integers N, M, Q, and Q quadruples of integers ( a_i , b_i , c_i , d_i ).

Consider a sequence A satisfying the following conditions:

A is a sequence of N positive integers.

1 \leq A_1 \leq A_2 \le \cdots \leq A_N \leq M.

Let us define a score of this sequence as follows:

The score is the sum of d_i over all indices i such that A_{b_i} - A_{a_i} = c_i. (If there is no such i, the score is 0.)

Find the maximum possible score of A.

Constraints

All values in input are integers.

2 ≤ N ≤ 10

1 \leq M \leq 10

1 \leq Q \leq 50

1 \leq a_i < b_i \leq N ( i = 1, 2, ..., Q )

0 \leq c_i \leq M - 1 ( i = 1, 2, ..., Q )

(a_i, b_i, c_i) \neq (a_j, b_j, c_j) (where i \neq j)

1 \leq d_i \leq 10^5 ( i = 1, 2, ..., Q )

Input

Input is given from Standard Input in the following format:

N M Q
a_1 b_1 c_1 d_1
:
a_Q b_Q c_Q d_Q

Output

Print the maximum possible score of A.

Sample Input 1

3 4 3
1 3 3 100
1 2 2 10
2 3 2 10

Sample Output 1

110

When A = \{1, 3, 4\}, its score is 110. Under these conditions, no sequence has a score greater than 110, so the answer is 110.

Sample Input 2

4 6 10
2 4 1 86568
1 4 0 90629
2 3 0 90310
3 4 1 29211
3 4 3 78537
3 4 2 8580
1 2 1 96263
1 4 2 2156
1 2 0 94325
1 4 3 94328

Sample Output 2

357500

Sample Input 3

10 10 1
1 10 9 1

Sample Output 3

1


import scala.io.StdIn

object Main extends App {

  val Array(a, b, n) = StdIn.readLine().split(" ").map(_.toLong)
  val x = n / (b - 1) * (b - 1)
  println(f(x))

  def f(x: Long) = (a * x) / b - a * (x / b)

  def readIntPair: (Int, Int) = {
    val Array(x, y) = StdIn.readLine().split(" ").map(_.toInt)
    (x, y)
  }

  def readIntArr = StdIn.readLine().split(" ").map(_.toInt)
}

Score : 400 points

Problem Statement

Given are integers A, B, and N.

Find the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N.

Here floor(t) denotes the greatest integer not greater than the real number t.

Constraints

1 ≤ A ≤ 10^{6}

1 ≤ B ≤ 10^{12}

1 ≤ N ≤ 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B N

Output

Print the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N, as an integer.

Sample Input 1

5 7 4

Sample Output 1

2

When x=3, floor(Ax/B)-A×floor(x/B) = floor(15/7) - 5×floor(3/7) = 2. This is the maximum value possible.

Sample Input 2

11 10 9

Sample Output 2

9


import java.security.KeyStore.TrustedCertificateEntry
import java.util
import java.util.regex.Pattern
import java.util.{ArrayList, Collections, Comparator, List => JList}

import scala.collection.immutable.TreeMap
import scala.math._
import scala.collection.mutable
import scala.collection.mutable.{ArrayBuffer, ListBuffer}
import scala.util.{Failure, Success, Try}

object Main{
  val LS = "\n"
  def main(args: Array[String]): Unit = {
    if (sys.env.getOrElse("TEST", "")=="1"){
      println(test());
    }else{
      val input=io.Source.stdin.getLines().mkString(LS);
      println(solve(input).trim());
    }
  }

  def init (nRows: Int, nCols: Int, default: Int) = Array.tabulate(nRows,nCols)( (x,y) =>  default)

  def sortIgnoreCase1(list: JList[String]): JList[String] = {
    val r = new ArrayList[String](list)
    Collections.sort(r, new Comparator[String] {
      override def compare(o1: String, o2: String): Int = o1.compareToIgnoreCase(o2)
    })
    r
  }

//  def solve(input: String): String = {
//    val Array(n, q) = input.split(LINE_SEPARATOR).head.split(" ").map(_.toInt)
//    val a = input.split(LINE_SEPARATOR).tail.take(n).map(_.split(" ").head.toInt)
//    val b = input.split(LINE_SEPARATOR).tail.take(n).map(_.split(" ").tail.head.toInt)
//    val c = input.split(LINE_SEPARATOR).drop(n + 1).map(_.split(" ").head.toInt)
//    val d = input.split(LINE_SEPARATOR).drop(n + 1).map(_.split(" ").tail.head.toInt)
//    val k = 2 * pow(10, 5).toInt
//    val INF = pow(10, 9).toInt + 1
//    val pqMap = (0 until k).map(_ => new mutable.TreeMap[Int, Int]).toArray
//    def add(k: Int, ai: Int): Unit = {
//      val count = pqMap(k).getOrElse(ai, 0)
//      pqMap(k) += ai -> (count + 1)
//    }
//    def delete(k: Int, ai: Int): Unit = {
//      val count = pqMap(k).getOrElse(ai, 0)
//      if (count == 1) pqMap(k) -= ai else pqMap(k) += ai -> (count - 1)
//    }
//    val i2k = (0 until n).map{i =>
//      b(i)
//    }.toArray
//    (0 until n).foreach{i =>
//      add(b(i) - 1, a(i))
//    }
//    val maxList = pqMap.map(pq => if (pq.nonEmpty) pq.last._1 else INF)
//
//    (0 until q).map{j =>
//      add(d(j) - 1, a(c(j) - 1))
//      delete(i2k(c(j) - 1) - 1, a(c(j) - 1))
//      maxList(i2k(c(j) - 1) - 1) =
//        if (pqMap(i2k(c(j) - 1) - 1).nonEmpty) pqMap(i2k(c(j) - 1) - 1).last._1 else INF
//      i2k(c(j) - 1) = d(j)
//      maxList(d(j) - 1) = if (pqMap(d(j) - 1).nonEmpty) pqMap(d(j) - 1).last._1 else INF
//      maxList.min
//    }.mkString(LINE_SEPARATOR)
//  }

  def solve(input: String): String = {
    val n = input.split(LS).head.toInt
    input.split(LS).tail.distinct.size.toString
  }

  def divideDigits(num: Long, base: Long = 10L): List[Long] = {
    def proc(num: Long, base: Long, flg: Boolean): List[Long] = {
      var flag = flg
      val quotient = num / base
      val remainder = if (num % base == 0) {
        flag = true
        base
      } else if (flag) {
        flag = false
        val temp = num % base - 1
        if (temp == 0) {
          flag = true
          base - 1
        } else temp
      } else {
        flag = flg
        num % base
      }
      quotient match {
        case 0 =>
          if (flag) {
            List.empty
          } else List(remainder)
        case _ => remainder :: proc(quotient, base, flag)
      }
    }

    proc(num, base, false).reverse//.filterNot(_ == 0)
  }

  def medianCalculator(seq: Seq[Long]): Long = {
    //In order if you are not sure that 'seq' is sorted
    val sortedSeq = seq.sortWith(_ < _)

    if (seq.size % 2 == 1) sortedSeq(sortedSeq.size / 2)
    else {
      val (up, down) = sortedSeq.splitAt(seq.size / 2)
      (up.last + down.head) / 2
    }
  }

  def factorize(x: Long): List[Long] = {
    def foo(x: Long, a: Long = 2, list: List[Long] = Nil): List[Long] = a*a > x match {
      case false if x % a == 0 => foo(x / a, a    , a :: list)
      case false               => foo(x    , a + 1, list)
      case true                => x :: list
    }
    foo(x)
  }

  def divisors(n: Int): Vector[Int] = {
    val halfList = (1 to sqrt(n).toInt).toVector.filter(n % _ == 0).map(_.toInt)
    val reversedList = if (sqrt(n).floor == sqrt(n)) halfList.reverse.tail.map(n/_) else halfList.reverse.map(n/_)
    halfList ++ reversedList
  }

  def gcd(m: Long, n: Long): Long = n match {
    case 0 => m
    case _ => gcd(n, m % n)
  }
  def lcm(a: Long, b: Long): Long = a * b / gcd(a, b)

  val tests=List(
    """3
      |apple
      |orange
      |apple""".stripMargin -> """2
                                     |
                                                |
                             |
                                   |
                                      |""".stripMargin
    ,
    """5
      |grape
      |grape
      |grape
      |grape
      |grape""".stripMargin -> """1
                             |""".stripMargin
  )

  def test():String= {
    return tests.map{case (i,o)=>(i.trim(),o.trim())}
      .zipWithIndex.map{
      case ((input,outputExpect),i)=>{
        val output=solve(input).trim();
        s"test${i+1}:"+(if(output==outputExpect){
          "Passed"
        }else{
          s"Failed\nexpect:\n${outputExpect}\noutput:\n${output}"
        })
      }}
      .mkString("\n");
  }

}

Score : 300 points

Problem Statement

You drew lottery N times. In the i-th draw, you got an item of the kind represented by a string S_i.

How many kinds of items did you get?

Constraints

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

S_i consists of lowercase English letters and has a length between 1 and 10 (inclusive).

Input

Input is given from Standard Input in the following format:

N
S_1
:
S_N

Output

Print the number of kinds of items you got.

Sample Input 1

3
apple
orange
apple

Sample Output 1

2

You got two kinds of items: apple and orange.

Sample Input 2

5
grape
grape
grape
grape
grape

Sample Output 2

1

Sample Input 3

4
aaaa
a
aaa
aa

Sample Output 3

4


object Main extends App {
	val N = scala.io.StdIn.readInt
	val set = scala.collection.mutable.Set[String]()

	for(i<-0 until N){
		val s = scala.io.StdIn.readLine
		set += s
	}

	println(set.size)
}


import scala.io.StdIn._
object Main {
  def main(args:Array[String]) {
    val N = readLine().toInt

    var list = for(i <- 0 until N) yield readLine()
    println(list.distinct.size)
  }
}


object Main extends App {
	val N = scala.io.StdIn.readLong

	val a = N / 3L
	val b = N / 5L
	val c = N / 15L

	println(N*(N+1L)/2L - 3L*a*(a+1L)/2L - 5L*b*(b+1L)/2L + 15L*c*(c+1L)/2L )
}

Score : 200 points

Problem Statement

Let us define the FizzBuzz sequence a_1,a_2,... as follows:

If both 3 and 5 divides i, a_i=\mbox{FizzBuzz}.

If the above does not hold but 3 divides i, a_i=\mbox{Fizz}.

If none of the above holds but 5 divides i, a_i=\mbox{Buzz}.

If none of the above holds, a_i=i.

Find the sum of all numbers among the first N terms of the FizzBuzz sequence.

Constraints

1 \leq N \leq 10^6

Input

Input is given from Standard Input in the following format:

N

Output

Print the sum of all numbers among the first N terms of the FizzBuzz sequence.

Sample Input 1

15

Sample Output 1

60

The first 15 terms of the FizzBuzz sequence are:

1,2,\mbox{Fizz},4,\mbox{Buzz},\mbox{Fizz},7,8,\mbox{Fizz},\mbox{Buzz},11,\mbox{Fizz},13,14,\mbox{FizzBuzz}

Among them, numbers are 1,2,4,7,8,11,13,14, and the sum of them is 60.

Sample Input 2

1000000

Sample Output 2

266666333332

Watch out for overflow.


import scala.math.abs

object Main extends App {
  val sc = new java.util.Scanner(System.in)
  var n = sc.nextLong()
  var k = sc.nextLong()
  val ans = (n % k).min(abs(n % k - k))
  println(ans)
}

Score : 300 points

Problem Statement

Given any integer x, Aoki can do the operation below.

Operation: Replace x with the absolute difference of x and K.

You are given the initial value of an integer N. Find the minimum possible value taken by N after Aoki does the operation zero or more times.

Constraints

0 ≤ N ≤ 10^{18}

1 ≤ K ≤ 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the minimum possible value taken by N after Aoki does the operation zero or more times.

Sample Input 1

7 4

Sample Output 1

1

Initially, N=7.

After one operation, N becomes |7-4| = 3.

After two operations, N becomes |3-4| = 1, which is the minimum value taken by N.

Sample Input 2

2 6

Sample Output 2

2

N=2 after zero operations is the minimum.

Sample Input 3

1000000000000000000 1

Sample Output 3

0


object Main extends App {
  val in = new java.util.Scanner(System.in)
  val N, K = in.nextLong
  val mod = N % K
  println(mod min K - mod)
}


object Main {
  import java.io.{BufferedReader, InputStream, InputStreamReader}
  import java.util.StringTokenizer
  import scala.reflect.ClassTag

  def main(args: Array[String]): Unit = {
    val out = new java.io.PrintWriter(System.out)
    new Main(out, new InputReader(System.in)).solve()
    out.flush()
  }

  private[this] val oj = System.getenv("MY_DEBUG") == null
  def DEBUG(f: => Unit): Unit = {
    if (!oj){ f }
  }
  def debug(as: Array[Boolean]): Unit = if (!oj){ debug(as.map(x => if(x) "1" else "0").mkString) }
  def debug(as: Array[Int]): Unit = if (!oj){ debug(as.mkString(" ")) }
  def debug(as: Array[Long]): Unit =if (!oj){ debug(as.mkString(" ")) }
  def debugDim(m: Array[Array[Int]]): Unit = if (!oj){
    REP(m.length) { i =>
      debug(m(i))
    }
  }
  def debugDimFlip(m: Array[Array[Long]]): Unit = if (!oj){
    REP(m(0).length) { j =>
      REP(m.length) { i =>
        System.err.print(m(i)(j))
        System.err.print(" ")
      }
      System.err.println()
    }
  }
  def debug(s: => String): Unit = {
    if (!oj){ System.err.println(s) }
  }
  def isDebug[A](debug: => A, online: => A): A = {
    if (oj) online else debug
  }

  class InputReader(val stream: InputStream) {
    private[this] val reader = new BufferedReader(new InputStreamReader(stream), 32768)
    private[this] var tokenizer: StringTokenizer = _

    private[this] def next(): String = {
      while (tokenizer == null || !tokenizer.hasMoreTokens)
        tokenizer = new StringTokenizer(reader.readLine)
      tokenizer.nextToken
    }

    def nextInt(): Int = Integer.parseInt(next())
    def nextLong(): Long = java.lang.Long.parseLong(next())
    def nextChar(): Char = next().charAt(0)

    def ni(): Int = nextInt()
    def nl(): Long = nextLong()
    def nc(): Char = nextChar()
    def ns(): String = next()
    def ns(n: Int): Array[Char] = ns().toCharArray
    def na(n: Int, offset: Int = 0): Array[Int] = map(n)(_ => ni() + offset)
    def na2(n: Int, offset: Int = 0): (Array[Int], Array[Int]) = {
      val A1, A2 = Array.ofDim[Int](n)
      REP(n) { i =>
        A1(i) = ni() + offset
        A2(i) = ni() + offset
      }
      (A1, A2)
    }
    def nm(n: Int, m: Int): Array[Array[Int]] = {
      val A = Array.ofDim[Int](n, m)
      REP(n) { i =>
        REP(m) { j =>
          A(i)(j) = ni()
        }
      }
      A
    }
    def nal(n: Int): Array[Long] = map(n)(_ => nl())
    def nm_c(n: Int, m: Int): Array[Array[Char]] = map(n) (_ => ns(m))
  }

  def REP(n: Int, offset: Int = 0)(f: Int => Unit): Unit = {
    var i = offset
    val N = n + offset
    while(i < N) { f(i); i += 1 }
  }
  def REP_r(n: Int, offset: Int = 0)(f: Int => Unit): Unit = {
    var i = n - 1 + offset
    while(i >= offset) { f(i); i -= 1 }
  }
  def TO(from: Int, to: Int)(f: Int => Unit): Unit = {
    REP(to - from + 1, from)(f)
  }
  def map[@specialized A: ClassTag](n: Int, offset: Int = 0)(f: Int => A): Array[A] = {
    val res = Array.ofDim[A](n)
    REP(n)(i => res(i) = f(i + offset))
    res
  }

  def sumL(as: Array[Int]): Long = {
    var s = 0L
    REP(as.length)(i => s += as(i))
    s
  }
  def cumSum(as: Array[Int]): Array[Long] = {
    val cum = Array.ofDim[Long](as.length + 1)
    REP(as.length) { i =>
      cum(i + 1) = cum(i) + as(i)
    }
    cum
  }
}

object Workspace {
  import Main._
  import java.util.Arrays.sort

  import scala.collection.mutable
  import math.{abs, max, min}
  import mutable.ArrayBuffer
  def divisors(x: Long): Array[Long] = {
    val pre = ArrayBuffer[Long]()
    val post = ArrayBuffer[Long]()
    REP(math.sqrt(x).toInt, 1) { d =>
      if (x % d == 0) {
        pre += d
        if (d * d != x) post += x / d
      }
    }

    val res = Array.ofDim[Long](pre.length + post.length)
    REP(pre.length)(i => res(i) = pre(i))
    val preLen = pre.length
    val postLen = post.length
    REP(postLen)(i => res(i + preLen) = post(postLen - 1 - i))
    res
  }
}

class Main(out: java.io.PrintWriter, sc: Main.InputReader) {
  import sc._
  import Main._
  import java.util.Arrays.sort

  import scala.collection.mutable
  import math.{abs, max, min}
  import mutable.ArrayBuffer
  import Workspace._

  // toIntとか+7とかするならvalにしろ
  final private[this] val MOD = 1000000007

  def solve(): Unit = {
    val N = nl()
    val D = divisors(N)
    var ans = mutable.Set[Long]()

    // 約数から探す
    REP(D.length - 1, 1) { i =>
      val d = D(i)
      var x = N
      while(x % d == 0) {
        x /= d
      }
      if (x == 1 || x % d == 1) {
        debug(s"d:$d")
        ans += d
      }
    }

    // 約数以外から探す
    REP(math.sqrt(N).toInt, 1) { a =>
      if ((N - 1) % a == 0) {
        val x = (N - 1) / a
        ans += a
        ans += x
      }
    }

    out.println(ans.size - 1) // 1を除く
  }
}

Score : 600 points

Problem Statement

Given is a positive integer N.

We will choose an integer K between 2 and N (inclusive), then we will repeat the operation below until N becomes less than K.

Operation: if K divides N, replace N with N/K; otherwise, replace N with N-K.

In how many choices of K will N become 1 in the end?

Constraints

2 \leq N \leq 10^{12}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of choices of K in which N becomes 1 in the end.

Sample Input 1

6

Sample Output 1

3

There are three choices of K in which N becomes 1 in the end: 2, 5, and 6.

In each of these choices, N will change as follows:

When K=2: 6 \to 3 \to 1

When K=5: 6 \to 1

When K=6: 6 \to 1

Sample Input 2

3141

Sample Output 2

13

Sample Input 3

314159265358

Sample Output 3

9


object Main extends App {
  val S = io.StdIn.readLine()
  if (S(2) == S(3) && S(4) == S(5)){
    println("Yes")
  }else{
    println("No")
  }
}

Score : 100 points

Problem Statement

A string of length 6 consisting of lowercase English letters is said to be coffee-like if and only if its 3-rd and 4-th characters are equal and its 5-th and 6-th characters are also equal.

Given a string S, determine whether it is coffee-like.

Constraints

S is a string of length 6 consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If S is coffee-like, print Yes; otherwise, print No.

Sample Input 1

sippuu

Sample Output 1

Yes

In sippuu, the 3-rd and 4-th characters are equal, and the 5-th and 6-th characters are also equal.

Sample Input 2

iphone

Sample Output 2

No

Sample Input 3

coffee

Sample Output 3

Yes


object Main extends App {
  val in = new java.util.Scanner(System.in)
  val S = in.next
  val coffee = S.charAt(2) == S.charAt(3) && S.charAt(4) == S.charAt(5)
  println(if (coffee) "Yes" else "No")
}


object Main extends App {
  import scala.io.StdIn.readLine

  	val k = readLine.split(" ").head.toInt;
  	val homes =readLine.split(" ").map(_.toInt).toList;
  	println((k - (homes zip (homes.tail :+ (k+homes.head))).map{case (x, y) => scala.math.abs(y - x)}.max).toString)
}

Score : 300 points

Problem Statement

There is a circular pond with a perimeter of K meters, and N houses around them.

The i-th house is built at a distance of A_i meters from the northmost point of the pond, measured clockwise around the pond.

When traveling between these houses, you can only go around the pond.

Find the minimum distance that needs to be traveled when you start at one of the houses and visit all the N houses.

Constraints

2 \leq K \leq 10^6

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

0 \leq A_1 < ... < A_N < K

All values in input are integers.

Input

Input is given from Standard Input in the following format:

K N
A_1 A_2 ... A_N

Output

Print the minimum distance that needs to be traveled when you start at one of the houses and visit all the N houses.

Sample Input 1

20 3
5 10 15

Sample Output 1

10

If you start at the 1-st house and go to the 2-nd and 3-rd houses in this order, the total distance traveled will be 10.

Sample Input 2

20 3
0 5 15

Sample Output 2

10

If you start at the 2-nd house and go to the 1-st and 3-rd houses in this order, the total distance traveled will be 10.


object Main extends App {
	val Array(k, n) = scala.io.StdIn.readLine.split(" ").map(_.toInt)
	val a = scala.io.StdIn.readLine.split(" ").map(_.toInt)
	val b = new Array[Int](n)

	for(i<-0 until n-1){
		b(i) = k - (a(i+1) - a(i))
	}

	b(n-1) = a(n-1) - a(0)

	println(b.min)
}


import scala.math.abs

object Main extends App {
  val Array(n, x, y) = io.StdIn.readLine().split(' ').map(_.toInt)

  val count = Array.fill(n)(0)
  for (k <- 1 until n) {
    for (i <- 1 to n; j = i+k; if j <= n) {
      val len = Seq(abs(i-j), abs(x-i)+1+abs(y-j), abs(y-i)+1+abs(x-j)).min
      count(len) += 1
    }
  }
  for (a <- count.drop(1)) println(a)
}

Score : 400 points

Problem Statement

We have an undirected graph G with N vertices numbered 1 to N and N edges as follows:

For each i=1,2,...,N-1, there is an edge between Vertex i and Vertex i+1.

There is an edge between Vertex X and Vertex Y.

For each k=1,2,...,N-1, solve the problem below:

Find the number of pairs of integers (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j in G is k.

Constraints

3 \leq N \leq 2 \times 10^3

1 \leq X,Y \leq N

X+1 < Y

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N X Y

Output

For each k=1, 2, ..., N-1 in this order, print a line containing the answer to the problem.

Sample Input 1

5 2 4

Sample Output 1

5
4
1
0

The graph in this input is as follows:

There are five pairs (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j is 1: (1,2)\,,(2,3)\,,(2,4)\,,(3,4)\,,(4,5).

There are four pairs (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j is 2: (1,3)\,,(1,4)\,,(2,5)\,,(3,5).

There is one pair (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j is 3: (1,5).

There are no pairs (i,j) (1 \leq i < j \leq N) such that the shortest distance between Vertex i and Vertex j is 4.

Sample Input 2

3 1 3

Sample Output 2

3
0

The graph in this input is as follows:

Sample Input 3

7 3 7

Sample Output 3

7
8
4
2
0
0

Sample Input 4

10 4 8

Sample Output 4

10
12
10
8
4
1
0
0
0


import java.io.{BufferedOutputStream, BufferedReader, InputStreamReader, PrintWriter}

object Main {
  def main(args: Array[String]): Unit = {
    val in = new BufferedReader(new InputStreamReader(System.in))
    val out = new PrintWriter(new BufferedOutputStream(System.out))
    val Array(n, m) = in.readLine().split("\\s+").map(_.toInt)

    out.println(n * (n - 1) / 2 + m * (m - 1) / 2)
    out.flush()
  }
}

Score : 100 points

Problem Statement

We have N+M balls, each of which has an integer written on it.

It is known that:

The numbers written on N of the balls are even.

The numbers written on M of the balls are odd.

Find the number of ways to choose two of the N+M balls (disregarding order) so that the sum of the numbers written on them is even.

It can be shown that this count does not depend on the actual values written on the balls.

Constraints

0 \leq N,M \leq 100

2 \leq N+M

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the answer.

Sample Input 1

2 1

Sample Output 1

1

For example, let us assume that the numbers written on the three balls are 1,2,4.

If we choose the two balls with 1 and 2, the sum is odd;

If we choose the two balls with 1 and 4, the sum is odd;

If we choose the two balls with 2 and 4, the sum is even.

Thus, the answer is 1.

Sample Input 2

4 3

Sample Output 2

9

Sample Input 3

1 1

Sample Output 3

0

Sample Input 4

13 3

Sample Output 4

81

Sample Input 5

0 3

Sample Output 5

3


object Main extends App {
  val in = new java.util.Scanner(System.in)
  val S = in.next
  def isPalindrome(s: String) = s.reverse.zip(s).forall{case (l, r) => l == r}
  val result = isPalindrome(S) && isPalindrome(S.substring(0, (S.size - 1) / 2))
  println(if (result) "Yes" else "No")
}

Score : 200 points

Problem Statement

A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:

S is a palindrome.

Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.

The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.

Determine whether S is a strong palindrome.

Constraints

S consists of lowercase English letters.

The length of S is an odd number between 3 and 99 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

If S is a strong palindrome, print Yes;
otherwise, print No.

Sample Input 1

akasaka

Sample Output 1

Yes

S is akasaka.

The string formed by the 1-st through the 3-rd characters is aka.

The string formed by the 5-th through the 7-th characters is aka.
All of these are palindromes, so S is a strong palindrome.

Sample Input 2

level

Sample Output 2

No

Sample Input 3

atcoder

Sample Output 3

No


import java.util._

object Main {
  def main(args: Array[String]): Unit = {
    val scanner = new Scanner(System.in)
    val string = scanner.nextLine
    val revString = string.reverse
    if (string != revString) {
      println("No")
      return
    }

    val firstHalf = string.substring((string.length - 1) / 2 + 1)
    if (firstHalf != firstHalf.reverse) {
      println("No")
      return
    }

    val secondHalf = string.substring((string.length - 1) / 2 + 1, string.length)
    if (secondHalf != secondHalf.reverse) {
      println("No")
      return
    }

    println("Yes")
  }
}


import java.io.{BufferedOutputStream, BufferedReader, InputStreamReader, PrintWriter}

object Main {
  def main(args: Array[String]): Unit = {
    val in = new BufferedReader(new InputStreamReader(System.in))
    val out = new PrintWriter(new BufferedOutputStream(System.out))
    val n = in.readLine().toInt
    val as = in.readLine().split("\\s+").map(_.toInt)

    val map = Array.ofDim[Int](n + 1)
    for (a <- as) {
      map(a) += 1
    }

    val cns = Array.ofDim[Int](n + 1)
    val dcns = Array.ofDim[Int](n + 1)
    for (i <- as.indices) {
      val c = map(i)
      cns(i) = if (c > 1) {
        map(i) * (map(i) - 1) / 2
      } else {
        0
      }
      dcns(i) = if (c > 2) {
        (map(i) - 1) * (map(i) - 2) / 2
      } else {
        0
      }
    }

    val sumOfCns = cns.sum

    for (a <- as) {
      out.println(sumOfCns - cns(a) + dcns(a))
    }

    out.flush()
  }
}

Score : 400 points

Problem Statement

We have N balls. The i-th ball has an integer A_i written on it.

For each k=1, 2, ..., N, solve the following problem and print the answer.

Find the number of ways to choose two distinct balls (disregarding order) from the N-1 balls other than the k-th ball so that the integers written on them are equal.

Constraints

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

1 \leq A_i \leq N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

For each k=1,2,...,N, print a line containing the answer.

Sample Input 1

5
1 1 2 1 2

Sample Output 1

2
2
3
2
3

Consider the case k=1 for example. The numbers written on the remaining balls are 1,2,1,2.

From these balls, there are two ways to choose two distinct balls so that the integers written on them are equal.

Thus, the answer for k=1 is 2.

Sample Input 2

4
1 2 3 4

Sample Output 2

0
0
0
0

No two balls have equal numbers written on them.

Sample Input 3

5
3 3 3 3 3

Sample Output 3

6
6
6
6
6

Any two balls have equal numbers written on them.

Sample Input 4

8
1 2 1 4 2 1 4 1

Sample Output 4

5
7
5
7
7
5
7
5


import scala.math.round

object Main extends App {

  val sc = new java.util.Scanner(System.in)
  val H = sc.nextInt()
  val W = sc.nextInt()

  println(round(H * W / 2))

}

Score : 200 points

Problem Statement

We have a board with H horizontal rows and W vertical columns of squares.
There is a bishop at the top-left square on this board.
How many squares can this bishop reach by zero or more movements?

Here the bishop can only move diagonally.
More formally, the bishop can move from the square at the r_1-th row (from the top) and the c_1-th column (from the left) to the square at the r_2-th row and the c_2-th column if and only if exactly one of the following holds:

r_1 + c_1 = r_2 + c_2

r_1 - c_1 = r_2 - c_2

For example, in the following figure, the bishop can move to any of the red squares in one move:

Constraints

1 \leq H, W \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H \ W

Output

Print the number of squares the bishop can reach.

Sample Input 1

4 5

Sample Output 1

10

The bishop can reach the cyan squares in the following figure:

Sample Input 2

7 3

Sample Output 2

11

The bishop can reach the cyan squares in the following figure:

Sample Input 3

1000000000 1000000000

Sample Output 3

500000000000000000


object Main extends App {
  val Array(h, w) = scala.io.StdIn.readLine.split(" ").map(_.toInt)

  val ans = {
    val min = List(h, w).min
    val max = List(h, w).max
    if (min == 1) {
      1
    } else {
      // 偶数, 偶数
      if (min % 2 == 0 && max % 2 == 0) {
        (BigInt(h) * w) / 2
      } else if (min % 2 != 0 && max % 2 != 0) {
        // 奇数, 奇数
        (((BigInt(h) - 1) * w) / 2) + (BigInt(h) / 2) + 1
      } else {
        // 奇数, 偶数
        (BigInt(h) * w) / 2
      }
    }
  }

  println(ans)
}


object Main {
  import java.io.{BufferedReader, InputStream, InputStreamReader}
  import java.util.StringTokenizer
  import scala.reflect.ClassTag

  def main(args: Array[String]): Unit = {
    val out = new java.io.PrintWriter(System.out)
    new Main(out, new InputReader(System.in)).solve()
    out.flush()
  }

  private[this] val oj = System.getenv("MY_DEBUG") == null
  def DEBUG(f: => Unit): Unit = {
    if (!oj){ f }
  }
  def debug(as: Array[Boolean]): Unit = if (!oj){ debug(as.map(x => if(x) "1" else "0").mkString) }
  def debug(as: Array[Int]): Unit = if (!oj){ debug(as.mkString(" ")) }
  def debug(as: Array[Long]): Unit =if (!oj){ debug(as.mkString(" ")) }
  def debugDim(m: Array[Array[Int]]): Unit = if (!oj){
    REP(m.length) { i =>
      debug(m(i))
    }
  }
  def debugDimFlip(m: Array[Array[Long]]): Unit = if (!oj){
    REP(m(0).length) { j =>
      REP(m.length) { i =>
        System.err.print(m(i)(j))
        System.err.print(" ")
      }
      System.err.println()
    }
  }
  def debug(s: => String): Unit = {
    if (!oj){ System.err.println(s) }
  }
  def isDebug[A](debug: => A, online: => A): A = {
    if (oj) online else debug
  }

  class InputReader(val stream: InputStream) {
    private[this] val reader = new BufferedReader(new InputStreamReader(stream), 32768)
    private[this] var tokenizer: StringTokenizer = _

    private[this] def next(): String = {
      while (tokenizer == null || !tokenizer.hasMoreTokens)
        tokenizer = new StringTokenizer(reader.readLine)
      tokenizer.nextToken
    }

    def nextInt(): Int = Integer.parseInt(next())
    def nextLong(): Long = java.lang.Long.parseLong(next())
    def nextChar(): Char = next().charAt(0)

    def ni(): Int = nextInt()
    def nl(): Long = nextLong()
    def nc(): Char = nextChar()
    def ns(): String = next()
    def ns(n: Int): Array[Char] = ns().toCharArray
    def na(n: Int, offset: Int = 0): Array[Int] = map(n)(_ => ni() + offset)
    def na2(n: Int, offset: Int = 0): (Array[Int], Array[Int]) = {
      val A1, A2 = Array.ofDim[Int](n)
      REP(n) { i =>
        A1(i) = ni() + offset
        A2(i) = ni() + offset
      }
      (A1, A2)
    }
    def nm(n: Int, m: Int): Array[Array[Int]] = {
      val A = Array.ofDim[Int](n, m)
      REP(n) { i =>
        REP(m) { j =>
          A(i)(j) = ni()
        }
      }
      A
    }
    def nal(n: Int): Array[Long] = map(n)(_ => nl())
    def nm_c(n: Int, m: Int): Array[Array[Char]] = map(n) (_ => ns(m))
  }

  def REP(n: Int, offset: Int = 0)(f: Int => Unit): Unit = {
    var i = offset
    val N = n + offset
    while(i < N) { f(i); i += 1 }
  }
  def REP_r(n: Int, offset: Int = 0)(f: Int => Unit): Unit = {
    var i = n - 1 + offset
    while(i >= offset) { f(i); i -= 1 }
  }
  def TO(from: Int, to: Int)(f: Int => Unit): Unit = {
    REP(to - from + 1, from)(f)
  }
  def map[@specialized A: ClassTag](n: Int, offset: Int = 0)(f: Int => A): Array[A] = {
    val res = Array.ofDim[A](n)
    REP(n)(i => res(i) = f(i + offset))
    res
  }

  def sumL(as: Array[Int]): Long = {
    var s = 0L
    REP(as.length)(i => s += as(i))
    s
  }
  def cumSum(as: Array[Int]): Array[Long] = {
    val cum = Array.ofDim[Long](as.length + 1)
    REP(as.length) { i =>
      cum(i + 1) = cum(i) + as(i)
    }
    cum
  }
}

object Workspace {
  import Main._
  import java.util.Arrays.sort

  import scala.collection.mutable
  import math.{abs, max, min}
  import mutable.ArrayBuffer
}

class Main(out: java.io.PrintWriter, sc: Main.InputReader) {
  import sc._
  import Main._
  import java.util.Arrays.sort

  import scala.collection.mutable
  import math.{abs, max, min}
  import mutable.ArrayBuffer
  import Workspace._

  // toIntとか+7とかするならvalにしろ
  final private[this] val MOD = 1000000007

  def solve(): Unit = {
    val N = ni()
    val chars = Array.ofDim[Char](N)
    chars(0) = 'a'
    def dfs(i: Int, d: Int): Unit = {
      if (i == N) {
        out.println(chars.mkString)
      } else {
        chars(i) = (d+'a').toChar
        dfs(i + 1, d)
        chars(i) = (d + 1+'a').toChar
        dfs(i + 1, d + 1)
      }
    }

    dfs(1, 0)
  }
}

Score : 400 points

Problem Statement

In this problem, we only consider strings consisting of lowercase English letters.

Strings s and t are said to be isomorphic when the following conditions are satisfied:

|s| = |t| holds.

For every pair i, j, one of the following holds:

s_i = s_j and t_i = t_j.

s_i \neq s_j and t_i \neq t_j.

For example, abcac and zyxzx are isomorphic, while abcac and ppppp are not.

A string s is said to be in normal form when the following condition is satisfied:

For every string t that is isomorphic to s, s \leq t holds. Here \leq denotes lexicographic comparison.

For example, abcac is in normal form, but zyxzx is not since it is isomorphic to abcac, which is lexicographically smaller than zyxzx.

You are given an integer N.
Print all strings of length N that are in normal form, in lexicographically ascending order.

Constraints

1 \leq N \leq 10

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N

Output

Assume that there are K strings of length N that are in normal form: w_1, \ldots, w_K in lexicographical order.
Output should be in the following format:

w_1
:
w_K

Sample Input 1

1

Sample Output 1

a

Sample Input 2

2

Sample Output 2

aa
ab


object Main {
  import java.io.{BufferedReader, InputStream, InputStreamReader}
  import java.util.StringTokenizer
  import scala.reflect.ClassTag

  def main(args: Array[String]): Unit = {
    val out = new java.io.PrintWriter(System.out)
    new Main(out, new InputReader(System.in)).solve()
    out.flush()
  }

  private[this] val oj = System.getenv("MY_DEBUG") == null
  def DEBUG(f: => Unit): Unit = {
    if (!oj){ f }
  }
  def debug(as: Array[Boolean]): Unit = if (!oj){ debug(as.map(x => if(x) "1" else "0").mkString) }
  def debug(as: Array[Int]): Unit = if (!oj){ debug(as.mkString(" ")) }
  def debug(as: Array[Long]): Unit =if (!oj){ debug(as.mkString(" ")) }
  def debugDim(m: Array[Array[Int]]): Unit = if (!oj){
    REP(m.length) { i =>
      debug(m(i))
    }
  }
  def debugDimFlip(m: Array[Array[Long]]): Unit = if (!oj){
    REP(m(0).length) { j =>
      REP(m.length) { i =>
        System.err.print(m(i)(j))
        System.err.print(" ")
      }
      System.err.println()
    }
  }
  def debug(s: => String): Unit = {
    if (!oj){ System.err.println(s) }
  }
  def isDebug[A](debug: => A, online: => A): A = {
    if (oj) online else debug
  }

  class InputReader(val stream: InputStream) {
    private[this] val reader = new BufferedReader(new InputStreamReader(stream), 32768)
    private[this] var tokenizer: StringTokenizer = _

    private[this] def next(): String = {
      while (tokenizer == null || !tokenizer.hasMoreTokens)
        tokenizer = new StringTokenizer(reader.readLine)
      tokenizer.nextToken
    }

    def nextInt(): Int = Integer.parseInt(next())
    def nextLong(): Long = java.lang.Long.parseLong(next())
    def nextChar(): Char = next().charAt(0)

    def ni(): Int = nextInt()
    def nl(): Long = nextLong()
    def nc(): Char = nextChar()
    def ns(): String = next()
    def ns(n: Int): Array[Char] = ns().toCharArray
    def na(n: Int, offset: Int = 0): Array[Int] = map(n)(_ => ni() + offset)
    def na2(n: Int, offset: Int = 0): (Array[Int], Array[Int]) = {
      val A1, A2 = Array.ofDim[Int](n)
      REP(n) { i =>
        A1(i) = ni() + offset
        A2(i) = ni() + offset
      }
      (A1, A2)
    }
    def nm(n: Int, m: Int): Array[Array[Int]] = {
      val A = Array.ofDim[Int](n, m)
      REP(n) { i =>
        REP(m) { j =>
          A(i)(j) = ni()
        }
      }
      A
    }
    def nal(n: Int): Array[Long] = map(n)(_ => nl())
    def nm_c(n: Int, m: Int): Array[Array[Char]] = map(n) (_ => ns(m))
  }

  def REP(n: Int, offset: Int = 0)(f: Int => Unit): Unit = {
    var i = offset
    val N = n + offset
    while(i < N) { f(i); i += 1 }
  }
  def REP_r(n: Int, offset: Int = 0)(f: Int => Unit): Unit = {
    var i = n - 1 + offset
    while(i >= offset) { f(i); i -= 1 }
  }
  def TO(from: Int, to: Int)(f: Int => Unit): Unit = {
    REP(to - from + 1, from)(f)
  }
  def map[@specialized A: ClassTag](n: Int, offset: Int = 0)(f: Int => A): Array[A] = {
    val res = Array.ofDim[A](n)
    REP(n)(i => res(i) = f(i + offset))
    res
  }

  def sumL(as: Array[Int]): Long = {
    var s = 0L
    REP(as.length)(i => s += as(i))
    s
  }
  def cumSum(as: Array[Int]): Array[Long] = {
    val cum = Array.ofDim[Long](as.length + 1)
    REP(as.length) { i =>
      cum(i + 1) = cum(i) + as(i)
    }
    cum
  }
}

object Workspace {
  import Main._
  import java.util.Arrays.sort

  import scala.collection.mutable
  import math.{abs, max, min}
  import mutable.ArrayBuffer
}

class Main(out: java.io.PrintWriter, sc: Main.InputReader) {
  import sc._
  import Main._
  import java.util.Arrays.sort

  import scala.collection.mutable
  import math.{abs, max, min}
  import mutable.ArrayBuffer
  import Workspace._

  // toIntとか+7とかするならvalにしろ
  final private[this] val MOD = 1000000007

  def solve(): Unit = {
    val H, W = nl()
    if (H == 1 || W == 1) out.println(1)
    else {
      val ans = if (H * W % 2 == 1) H * W / 2 + 1 else H * W / 2
      out.println(ans)
    }
  }
}

Score : 500 points

Problem Statement

Snuke has a string s.
From this string, Anuke, Bnuke, and Cnuke obtained strings a, b, and c, respectively, as follows:

Choose a non-empty (contiguous) substring of s (possibly s itself). Then, replace some characters (possibly all or none) in it with ?s.

For example, if s is mississippi, we can choose the substring ssissip and replace its 1-st and 3-rd characters with ? to obtain ?s?ssip.

You are given the strings a, b, and c.
Find the minimum possible length of s.

Constraints

1 \leq |a|, |b|, |c| \leq 2000

a, b, and c consists of lowercase English letters and ?s.

Input

Input is given from Standard Input in the following format:

a
b
c

Output

Print the minimum possible length of s.

Sample Input 1

a?c
der
cod

Sample Output 1

7

For example, s could be atcoder.

Sample Input 2

atcoder
atcoder
???????

Sample Output 2

7

a, b, and c may not be distinct.


object Main extends App {
  val abmInput  = scala.io.StdIn.readLine()
  val abmArray = abmInput.split(" ").map(_.toInt).toSeq
  val aLength = abmArray(0)
  val bLength = abmArray(1)
  val mLength = abmArray(2)

  val aInput = scala.io.StdIn.readLine()
  val aArray = aInput.split(" ").map(_.toInt).toVector
  val bInput = scala.io.StdIn.readLine()
  val bArray = bInput.split(" ").map(_.toInt).toVector

  var mMap = scala.collection.mutable.Map.empty[(Int, Int), Int]
  (1 to mLength).foreach { _ =>
    val mInput  = scala.io.StdIn.readLine()
    val mInputArray = mInput.split(" ").map(_.toInt).toSeq
    val key = (mInputArray(0), mInputArray(1))
    if (mMap.contains(key)) {
      val existingValue = mMap.get(key).get
      if(existingValue < mInputArray(2)) {
        mMap.put(key, mInputArray(2))
      }
    } else {
      mMap.put(key, mInputArray(2))
    }
  }

  var min = aArray.min + bArray.min
  mMap.keys.foreach { key =>
    val mCalculate = aArray(key._1 - 1) + bArray(key._2 - 1) - mMap.get(key).get
    if(mCalculate < min) {
      min = mCalculate
    }
  }
  println(min)
}

Score : 200 points

Problem Statement

You are visiting a large electronics store to buy a refrigerator and a microwave.

The store sells A kinds of refrigerators and B kinds of microwaves. The i-th refrigerator ( 1 \le i \le A ) is sold at a_i yen (the currency of Japan), and the j-th microwave ( 1 \le j \le B ) is sold at b_j yen.

You have M discount tickets. With the i-th ticket ( 1 \le i \le M ), you can get a discount of c_i yen from the total price when buying the x_i-th refrigerator and the y_i-th microwave together. Only one ticket can be used at a time.

You are planning to buy one refrigerator and one microwave. Find the minimum amount of money required.

Constraints

All values in input are integers.

1 \le A \le 10^5

1 \le B \le 10^5

1 \le M \le 10^5

1 \le a_i , b_i , c_i \le 10^5

1 \le x_i \le A

1 \le y_i \le B

c_i \le a_{x_i} + b_{y_i}

Input

Input is given from Standard Input in the following format:

A B M
a_1 a_2 ... a_A
b_1 b_2 ... b_B
x_1 y_1 c_1
\vdots
x_M y_M c_M

Output

Print the answer.

Sample Input 1

2 3 1
3 3
3 3 3
1 2 1

Sample Output 1

5

With the ticket, you can get the 1-st refrigerator and the 2-nd microwave for 3+3-1=5 yen.

Sample Input 2

1 1 2
10
10
1 1 5
1 1 10

Sample Output 2

10

Note that you cannot use more than one ticket at a time.

Sample Input 3

2 2 1
3 5
3 5
2 2 2

Sample Output 3

6

You can get the 1-st refrigerator and the 1-st microwave for 6 yen, which is the minimum amount to pay in this case.
Note that using a ticket is optional.


object Main extends App {

  val s = scala.io.StdIn.readLine().split("").toList
  s match {
    case List(a, b, c) if a == b && b == c =>
      println("No")
    case _ =>
      println("Yes")
  }
}

Score : 100 points

Problem Statement

In AtCoder City, there are three stations numbered 1, 2, and 3.

Each of these stations is operated by one of the two railway companies, A and B. A string S of length 3 represents which company operates each station. If S_i is A, Company A operates Station i; if S_i is B, Company B operates Station i.

To improve the transportation condition, for each pair of a station operated by Company A and one operated by Company B, there will be a bus service connecting them.

Determine if there is a pair of stations that will be connected by a bus service.

Constraints

Each character of S is A or B.

|S| = 3

Input

Input is given from Standard Input in the following format:

S

Output

If there is a pair of stations that will be connected by a bus service, print Yes; otherwise, print No.

Sample Input 1

ABA

Sample Output 1

Yes

Company A operates Station 1 and 3, while Company B operates Station 2.

There will be a bus service between Station 1 and 2, and between Station 2 and 3, so print Yes.

Sample Input 2

BBA

Sample Output 2

Yes

Company B operates Station 1 and 2, while Company A operates Station 3.

There will be a bus service between Station 1 and 3, and between Station 2 and 3, so print Yes.

Sample Input 3

BBB

Sample Output 3

No

Company B operates all the stations. Thus, there will be no bus service, so print No.


import java.util.Scanner

object Main extends App {
  val sc = new Scanner(System.in)
  val n = sc.nextLong()
  val a = sc.nextLong()
  val b = sc.nextLong()

  val x = n / (a + b) * a
  val y = n % (a + b)
  val z = List(y, a).min

  println(x + z)
}

Score : 200 points

Problem Statement

Takahashi has many red balls and blue balls. Now, he will place them in a row.

Initially, there is no ball placed.

Takahashi, who is very patient, will do the following operation 10^{100} times:

Place A blue balls at the end of the row of balls already placed. Then, place B red balls at the end of the row.

How many blue balls will be there among the first N balls in the row of balls made this way?

Constraints

1 \leq N \leq 10^{18}

A, B \geq 0

0 < A + B \leq 10^{18}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B

Output

Print the number of blue balls that will be there among the first N balls in the row of balls.

Sample Input 1

8 3 4

Sample Output 1

4

Let b denote a blue ball, and r denote a red ball. The first eight balls in the row will be bbbrrrrb, among which there are four blue balls.

Sample Input 2

8 0 4

Sample Output 2

0

He placed only red balls from the beginning.

Sample Input 3

6 2 4

Sample Output 3

2

Among bbrrrr, there are two blue balls.