//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 A, B, T = sc.nextLong()
    println(T / A * B)
  }

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

  def calc(N: Long, R: Long, M: Long): Double = {
    math.sqrt(R * R - (R * 2 * ((M * 1.0 / N) - 0.5)) * (R * 2 * ((M * 1.0 / N) - 0.5))) * 2
  }

  var memo: Map[(Long, Long), Long] = Map[(Long, Long), Long]()

  def recursive(input: Long, maxV: Long): Long = {
    if (!memo.contains((input, maxV))) {
      val ans = if (input == 0) 1 else (1L to Math.min(input, maxV)).map(i => recursive(input - i, i)).sum
      memo += ((input, maxV) -> ans)
    }
    memo((input, maxV))
  }

  def recursive2(X: Set[Long], Y: Stream[Long]): Long = if (X.contains(Y.head)) X.size else recursive2(X + Y.head, Y.tail)

  def check(s: String): Boolean = {
    if (s == "0") true
    else s(0) != '0' && 0 <= s.toInt && s.toInt <= 255
  }

  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
  }

  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): Stream[Long] =
    Stream.cons(begin, getPermutation(begin + 1))

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

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

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

object ArabicRoman {

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

  val codeTable = 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 =>

      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
    }
    } =>

      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())
  }
}

Score : 100 points

Problem Statement

A biscuit making machine produces B biscuits at the following moments: A seconds, 2A seconds, 3A seconds and each subsequent multiple of A seconds after activation.

Find the total number of biscuits produced within T + 0.5 seconds after activation.

Constraints

All values in input are integers.

1 \leq A, B, T \leq 20

Input

Input is given from Standard Input in the following format:

A B T

Output

Print the total number of biscuits produced within T + 0.5 seconds after activation.

Sample Input 1

3 5 7

Sample Output 1

10

Five biscuits will be produced three seconds after activation.

Another five biscuits will be produced six seconds after activation.

Thus, a total of ten biscuits will be produced within 7.5 seconds after activation.

Sample Input 2

3 2 9

Sample Output 2

6

Sample Input 3

20 20 19

Sample Output 3

0


import java.util.Scanner

object Main extends App {
  val sc = new Scanner(System.in)

  val a, b, t = sc.nextInt
  println(t / a * b)
}

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

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 io.StdIn._
  object Main extends App {
    val diamonds = Input(readInt(), readLine(), readLine()).diamonds().foldLeft[Diamonds](Diamonds(Nil)) { (purchased: Diamonds, diamond: Diamond) =>
      val next = purchased.add(diamond)
      if (next.value() > purchased.value() && next.cost() < purchased.cost()) next
      else purchased
    }

    println(diamonds.value())
  }

  case class Input(numberOfDiamond: Int, values: Seq[Int], costs: Seq[Int]) {
    def diamonds(): Seq[Diamond] = {
      values.zip(costs).map { case (value, cost) => Diamond.apply(value, cost) }
    }
  }

  object Input {
    def apply(n: Int, values: String, costs: String): Input = {
      Input(n.toInt, values.split(' ').map(_.toInt), costs.split(' ').map(_.toInt))
    }
  }

case class Diamond(value:Int, cost: Int)

case class Diamonds(values: Seq[Diamond]) {
  def value(): Int =
    values.foldLeft(0)(_ + _.value)

  def cost(): Int =
    values.foldLeft(0)(_ + _.cost)

  def add(diamond: Diamond): Diamonds = Diamonds(values :+ diamond)
}

Score : 200 points

Problem Statement

There are N gems. The value of the i-th gem is V_i.

You will choose some of these gems, possibly all or none, and get them.

However, you need to pay a cost of C_i to get the i-th gem.

Let X be the sum of the values of the gems obtained, and Y be the sum of the costs paid.

Find the maximum possible value of X-Y.

Constraints

All values in input are integers.

1 \leq N \leq 20

1 \leq C_i, V_i \leq 50

Input

Input is given from Standard Input in the following format:

N
V_1 V_2 ... V_N
C_1 C_2 ... C_N

Output

Print the maximum possible value of X-Y.

Sample Input 1

3
10 2 5
6 3 4

Sample Output 1

5

If we choose the first and third gems, X = 10 + 5 = 15 and Y = 6 + 4 = 10.
We have X-Y = 5 here, which is the maximum possible value.

Sample Input 2

4
13 21 6 19
11 30 6 15

Sample Output 2

6

Sample Input 3

1
1
50

Sample Output 3

0


object Main {
  def main(args: Array[String]): Unit = {
    val jewelry_nums   = io.StdIn.readInt
    val jewelry_values = io.StdIn.readLine.split(" ").map{_.toInt}
    var jewelry_costs  = io.StdIn.readLine.split(" ").map{_.toInt}
    var max_profits    = 0

    (0 until jewelry_nums).foreach { i =>
      val profit = jewelry_values(i) - jewelry_costs(i)
      if(profit > 0) {
        max_profits += profit
      }
    }
    println(max_profits)

  }
}


object Main{
	def main(args:Array[String]){
		val s = new java.util.Scanner(System.in)
		val n = s.nextInt
		val a = Array.fill(n)(s.nextLong)
		for(i <- 0 to n - 2) {
			if(a(i) < 0) {
				a(i) *= -1
				a(i + 1) *= -1
			}
		}

		val sum=a.sum

		println(math.max(sum,sum-(a(0)+a(n-1))*2))
	}
}

Score : 400 points

Problem Statement

There are N integers, A_1, A_2, ..., A_N, arranged in a row in this order.

You can perform the following operation on this integer sequence any number of times:

Operation: Choose an integer i satisfying 1 \leq i \leq N-1. Multiply both A_i and A_{i+1} by -1.

Let B_1, B_2, ..., B_N be the integer sequence after your operations.

Find the maximum possible value of B_1 + B_2 + ... + B_N.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

-10^9 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible value of B_1 + B_2 + ... + B_N.

Sample Input 1

3
-10 5 -4

Sample Output 1

19

If we perform the operation as follows:

Choose 1 as i, which changes the sequence to 10, -5, -4.

Choose 2 as i, which changes the sequence to 10, 5, 4.

we have B_1 = 10, B_2 = 5, B_3 = 4. The sum here, B_1 + B_2 + B_3 = 10 + 5 + 4 = 19, is the maximum possible result.

Sample Input 2

5
10 -4 -8 -11 3

Sample Output 2

30

Sample Input 3

11
-1000000000 1000000000 -1000000000 1000000000 -1000000000 0 1000000000 -1000000000 1000000000 -1000000000 1000000000

Sample Output 3

10000000000

The output may not fit into a 32-bit integer type.


object Main {
  def main(args: Array[String]): Unit = {
    val s = new Main()
    s.solve()
    s.out.flush()
  }
}

class Main {
  import java.io._
  import java.util.StringTokenizer
  import java.util.Arrays.sort

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

  val MOD = 1000000007
  val out = new PrintWriter(System.out)

  def solve(): Unit = {
    val N = ni()
    val A = nal(N)
    val dp = Array.fill[Long](N + 1, 2)(-1e18.toLong)
    dp(0)(0) = 0
    REP(N) { i =>
      REP(2) { j =>
        REP(2) { k =>
          val sig = if (((j + k) & 1) == 0) 1 else -1
          debug(s"i:$i j:$j k:$k sig:$sig")
          dp(i + 1)(j) = max(dp(i + 1)(j), dp(i)(k) + sig * A(i))
        }
      }
    }

    debugDim(dp)

    val ans = dp(N)(0)
    out.println(ans)
  }

  private val oj = System.getenv("ATCODER_DEBUG") == null

  def DEBUG(f: => Unit): Unit = {
    if (!oj){ f }
  }

  def debug(as: Array[Boolean]): Unit = DEBUG {
    debug(as.map(x => if(x) "1" else "0").mkString)
  }

  def debug(as: Array[Int]): Unit = DEBUG {
    debug(as.mkString(" "))
  }

  def debug(as: Array[Long]): Unit = DEBUG {
    debug(as.mkString(" "))
  }

  def debugDim(m: Array[Array[Long]]): Unit = DEBUG {
    REP(m.length) { i =>
      debug(m(i))
    }
  }

  def debugDimFlip(m: Array[Array[Long]]): Unit = DEBUG {
    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 = DEBUG {
    System.err.println(s)
  }

  def debugL(num: => Long): Unit = DEBUG {
    System.err.println(num)
  }

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

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

    def nextInt(): Int = next().toInt
    def nextLong(): Long = next().toLong
    def nextChar(): Char = next().charAt(0)
  }
  val sc = new InputReader(System.in)
  def ni(): Int = sc.nextInt()
  def nl(): Long = sc.nextLong()
  def nc(): Char = sc.nextChar()
  def ns(): String = sc.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) { i =>
      f(i)
    }
  }

  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
  }
}


import scala.io.StdIn._

object Main extends App {
  def foo(n: Int, s: String): Unit = {
    if (s.count(_ == '#') == 0) {
      println(0)
    }
    else {
      val ns = s.substring(s.indexOf('#'), s.length)
      println(
        math.min(ns.count(_ == '.'),
          ns.count(_ == '#')))
    }
  }
  foo(readInt(), readLine.toString)
}

Score : 300 points

Problem Statement

There are N stones arranged in a row. Every stone is painted white or black.
A string S represents the color of the stones. The i-th stone from the left is white if the i-th character of S is ., and the stone is black if the character is #.

Takahashi wants to change the colors of some stones to black or white so that there will be no white stone immediately to the right of a black stone.
Find the minimum number of stones that needs to be recolored.

Constraints

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

S is a string of length N consisting of . and #.

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the minimum number of stones that needs to be recolored.

Sample Input 1

3
#.#

Sample Output 1

1

It is enough to change the color of the first stone to white.

Sample Input 2

5
#.##.

Sample Output 2

2

Sample Input 3

9
.........

Sample Output 3

0


object Main {
  def main(args: Array[String]): Unit = {
    val s = new Main()
    s.solve()
    s.out.flush()
  }
}

class Main {
  import java.io._
  import java.util.StringTokenizer
  import java.util.Arrays.sort

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

  val MOD = 1000000007
  val out = new PrintWriter(System.out)

  def solve(): Unit = {
    val N = ni()
    val S = ns(N)
    var cnt = 0

    REP_r(N) { i =>
      if (i > 0) {
        if (S(i - 1) == '#' && S(i) == '.') {
          S(i) = '#'
          cnt += 1
        }
      }
    }

    out.println(cnt)
  }

  private val oj = System.getenv("ATCODER_DEBUG") == null

  def DEBUG(f: => Unit): Unit = {
    if (!oj){ f }
  }

  def debug(as: Array[Boolean]): Unit = DEBUG {
    debug(as.map(x => if(x) "1" else "0").mkString)
  }

  def debug(as: Array[Int]): Unit = DEBUG {
    debug(as.mkString(" "))
  }

  def debug(as: Array[Long]): Unit = DEBUG {
    debug(as.mkString(" "))
  }

  def debug(s: => String): Unit = DEBUG {
    System.err.println(s)
  }

  def debugL(num: => Long): Unit = DEBUG {
    System.err.println(num)
  }

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

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

    def nextInt(): Int = next().toInt
    def nextLong(): Long = next().toLong
    def nextChar(): Char = next().charAt(0)
  }
  val sc = new InputReader(System.in)
  def ni(): Int = sc.nextInt()
  def nl(): Long = sc.nextLong()
  def nc(): Char = sc.nextChar()
  def ns(): String = sc.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) { i =>
      f(i)
    }
  }

  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[Long]) = {
    val cum = Array.ofDim[Long](as.length + 1)
    REP(as.length) { i =>
      cum(i + 1) = cum(i) + as(i)
    }
    cum
  }
}


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

import scala.annotation.tailrec
import scala.collection.mutable.ArrayBuffer
import scala.reflect.ClassTag

object Main {
  val DEBUG = false

  lazy val debugOut: PrintWriter = {
    new PrintWriter(new BufferedOutputStream(new FileOutputStream("log.txt", true)))
  }

  def log(message: => String): Unit = {
    if (DEBUG) {
      Console.err.println(message)
    }
  }

  def debug(closure: => Unit): Unit = {
    if (DEBUG) {
      closure
    }
  }

  def get[T: ClassTag](implicit in: BufferedReader, converter: String => T): T = {
    converter(in.readLine)
  }

  def getArray[T: ClassTag](implicit in: BufferedReader, converter: String => T): Array[T] = {
    get[String].split("\\s+").map[T](converter)
  }

  implicit val idConverter: String => String = s => s
  implicit val intConverter: String => Int = s => Integer.parseInt(s)
  implicit val longConverter: String => Long = s => lang.Long.parseLong(s)

  def get[T1, T2](
                   implicit in: BufferedReader,
                   converter1: String => T1,
                   converter2: String => T2
                 ): (T1, T2) = {
    getArray[String] match {
      case Array(v1, v2) => (converter1(v1), converter2(v2))
    }
  }

  def get[T1, T2, T3](
                       implicit in: BufferedReader,
                       converter1: String => T1,
                       converter2: String => T2,
                       converter3: String => T3
                     ): (T1, T2, T3) = {
    getArray[String] match {
      case Array(v1, v2, v3) => (converter1(v1), converter2(v2), converter3(v3))
    }
  }

  def get[T1, T2, T3, T4](
                           implicit in: BufferedReader,
                           converter1: String => T1,
                           converter2: String => T2,
                           converter3: String => T3,
                           converter4: String => T4
                         ): (T1, T2, T3, T4) = {
    getArray[String] match {
      case Array(v1, v2, v3, v4) => (converter1(v1), converter2(v2), converter3(v3), converter4(v4))
    }
  }

  implicit class TimesInt(val n: Int) {
    def times(f: Int => Unit): Unit = {
      for (i <- 0 until n) {
        f(i)
      }
    }
  }

  def primes(n: Long): ArrayBuffer[Long] = {
    val buf = ArrayBuffer.empty[Long]
    val s = Math.sqrt(n).toInt

    var v = n
    for (i <- 2 to s) {
      if (v == 1 || v == 0) {
        return buf
      } else {
        while (v % i == 0) {
          buf.append(i)
          v /= i
        }
      }
    }
    if (v > s) {
      buf.append(v)
    }
    buf
  }

  def divisors(n: Int): ArrayBuffer[Int] = {
    val buf = ArrayBuffer.empty[Int]
    val s = Math.sqrt(n).toInt

    for (i <- 1 to s) {
      if (n % i == 0) {
        buf.append(i)
        if (i != n / i) {
          buf.append(n / i)
        }
      }
    }
    buf
  }

  @tailrec
  def gcd(a: Long, b: Long): Long = {
    val reminder = a % b
    if (reminder == 0) {
      b
    } else {
      gcd(b, reminder)
    }
  }

  @tailrec
  def gcd(a: Int, b: Int): Int = {
    val reminder = a % b
    if (reminder == 0) {
      b
    } else {
      gcd(b, reminder)
    }
  }

  def lcm(a: Long, b: Long): Long = {
    a / gcd(a, b) * b
  }

  def toBitArray(n: Int, width: Int): Array[Boolean] = {
    val buf = ArrayBuffer.fill(width)(false)

    @tailrec
    def inner(n: Int, i: Int): Array[Boolean] = {
      buf(i) = (n & 1) == 0
      if (i == 0) {
        buf.toArray
      } else {
        inner(n / 2, i - 1)
      }
    }

    inner(n, width - 1)
  }

  def modpow(n: Long, p: Long, m: Long): Long = {
    @tailrec
    def inner(n: Long, p: Long, acc: Long): Long = {
      if (p == 0) {
        acc
      } else if ((p & 1) == 0) {
        inner(n * n % m, p >> 1, acc)
      } else {
        inner(n * n % m, p >> 1, acc * n % m)
      }
    }

    inner(n % m, p, 1)
  }

  def main(args: Array[String]): Unit = {
    log("=== start")

    implicit val in: BufferedReader = new BufferedReader(new InputStreamReader(System.in))
    val out = new PrintWriter(new BufferedOutputStream(System.out))
    val n = get[Int]
    val s = get[String]

    def count: ArrayBuffer[Int] = {
      val counts = ArrayBuffer.empty[Int]
      var current = 'x'
      var count = 0

      for (c <- s) {
        if (current == c) {
          count += 1
        } else {
          if (count != 0) {
            counts += count
          }
          current = c
          count = 1
        }
      }

      if (count > 0) {
        counts += count
      }

      counts
    }

    val allCounts =
      s.head match {
        case '.' =>
          s.last match {
            case '.' => count.tail
            case '#' => count.tail.init
          }
        case '#' =>
          s.last match {
            case '.' => count
            case '#' => count.init
          }
      }

    val counts = Array.ofDim[Int](2)
    for (i <- allCounts.indices) {
      counts(i % 2) += allCounts(i)
    }

    out.println(counts.min)

    out.flush()
    log("=== end")
    debug {
      debugOut.flush()
    }
  }
}


import scala.io._
object Main extends App {
  val n = StdIn.readLine.toInt
  val s = StdIn.readLine

  def checkMin(i: Int, changes: Int): Int = {
    if (i == n) changes
    else {
      checkMin(i + 1, changes + (if (s.charAt(i) == '#') 1 else -1))
    }
  }
  println(checkMin(0, s.count(_ == '.')))
}


import scala.math._

object Main {
  def main(args: Array[String]): Unit = {
    val lines = io.Source.stdin.getLines().toSeq
    println(solve(lines))
  }

  def solve(lines: Seq[String]): String = {
    val Array(a, b) = lines.head.split(" ").map(_.toInt)
    var result = 0
    if (a == b) {
      result = a + b
    } else {
      result = max(a, b) + (max(a, b) - 1)
    }
    result.toString
  }
}

Score : 100 points

Problem Statement

There are two buttons, one of size A and one of size B.

When you press a button of size X, you get X coins and the size of that button decreases by 1.

You will press a button twice. Here, you can press the same button twice, or press both buttons once.

At most how many coins can you get?

Constraints

All values in input are integers.

3 \leq A, B \leq 20

Input

Input is given from Standard Input in the following format:

A B

Output

Print the maximum number of coins you can get.

Sample Input 1

5 3

Sample Output 1

9

You can get 5 + 4 = 9 coins by pressing the button of size 5 twice, and this is the maximum result.

Sample Input 2

3 4

Sample Output 2

7

Sample Input 3

6 6

Sample Output 3

12


import scala.io.StdIn

object Main extends App {
  val input = StdIn.readLine().split(" ").map(_.toInt)
  var a = input(0)
  var b = input(1)
  var ret = 0
  ret = exec
  ret += exec
  println(ret)

  def exec() = {
    var ret = 0
    if (a > b) {
      ret = a
      a -= 1
    } else {
      ret = b
      b -= 1
    }
    ret
  }
}


import java.util.Scanner

object Main {
  def main(args: Array[String]) = {
    val sc = new Scanner(System.in)
    val A, B = sc.nextInt()
    val c = A.max(B) + A.max(B) - 1

    def foo(a: Int, b: Int, sum: Int = 0, deps: Int = 1): Int = {
      if (deps == 0) sum else {
        if (a < b) {
          foo(a, b - 1, b, deps - 1)
        } else {
          foo(a - 1, b, a, deps - 1)
        }
      }
    }

    foo(A, B)
    println(c)
  }
}


object Main extends App {
  val in = scala.io.Source.stdin.getLines.toSeq
  val H = in.map(_.split(' ').map(_.toInt)).tail.head

  var ret = 0
  H.foldLeft(0){ (a,e) =>
    if(a<=e) ret+=1
    Math.max(a,e)
  }
  println(ret)
}

Score : 200 points

Problem Statement

There are N mountains ranging from east to west, and an ocean to the west.

At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.

The height of the i-th mountain from the west is H_i.

You can certainly see the ocean from the inn at the top of the westmost mountain.

For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.

From how many of these N inns can you see the ocean?

Constraints

All values in input are integers.

1 \leq N \leq 20

1 \leq H_i \leq 100

Input

Input is given from Standard Input in the following format:

N
H_1 H_2 ... H_N

Output

Print the number of inns from which you can see the ocean.

Sample Input 1

4
6 5 6 8

Sample Output 1

3

You can see the ocean from the first, third and fourth inns from the west.

Sample Input 2

5
4 5 3 5 4

Sample Output 2

3

Sample Input 3

5
9 5 6 8 4

Sample Output 3

1


object Main extends App {

  val line1 = io.StdIn.readLine.split(' ').map(_.toInt)
  val s = io.StdIn.readLine

  val N = line1(0)
  val K = line1(1)

  var previous = '-'

  var seq = Seq.empty[Int]
  var accum = 0

  for (i <- 0 until s.size) {
    s(i) match {
      case ss if ss == previous => accum += 1
      case ss if ss != previous && accum > 0=> {
        seq = accum +: seq
        accum = 1
      }
      case _ => accum += 1
    }
    previous = s(i)
  }

  if (accum > 0)
    seq = accum +: seq

  val num = if (s.head == '0') 2*K else 2*K+1
  val ans = seq.sliding(num).map(_.sum).max

  println(ans)

}

Score : 400 points

Problem Statement

N people are arranged in a row from left to right.

You are given a string S of length N consisting of 0 and 1, and a positive integer K.

The i-th person from the left is standing on feet if the i-th character of S is 0, and standing on hands if that character is 1.

You will give the following direction at most K times (possibly zero):

Direction: Choose integers l and r satisfying 1 \leq l \leq r \leq N, and flip the l-th, (l+1)-th, ..., and r-th persons. That is, for each i = l, l+1, ..., r, the i-th person from the left now stands on hands if he/she was standing on feet, and stands on feet if he/she was standing on hands.

Find the maximum possible number of consecutive people standing on hands after at most K directions.

Constraints

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

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

The length of the string S is N.

Each character of the string S is 0 or 1.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of consecutive people standing on hands after at most K directions.

Sample Input 1

5 1
00010

Sample Output 1

4

We can have four consecutive people standing on hands, which is the maximum result, by giving the following direction:

Give the direction with l = 1, r = 3, which flips the first, second and third persons from the left.

Sample Input 2

14 2
11101010110011

Sample Output 2

8

Sample Input 3

1 1
1

Sample Output 3

1

No directions are necessary.


import java.util.Scanner

import scala.collection.mutable.ArrayBuffer

object Main {
  def read(): (Int, Int, String) = {
    val sc = new Scanner(System.in)
    val n, k = sc.nextInt()
    val s = sc.next()
    (n, k, s)
  }

  def compressRunlength(s: String): IndexedSeq[(Int, Char)] = {

    var currentChar = s(0)
    var currentCount = 1
    var buf = ArrayBuffer.empty[(Int, Char)]
    for (i <- 1 until s.length) {
      if (currentChar == s(i)) {
        currentCount += 1
      } else {
        buf.append((currentCount, currentChar))
        currentChar = s(i)
        currentCount = 1
      }
    }
    buf.append((currentCount, currentChar))
    buf.toIndexedSeq
  }

  def solve(n: Int, k: Int, s: String): Int = {
    // s.length == n
    val runLength = compressRunlength(s)
    val cumSum = runLength.map(_._1).scanLeft(0)(_ + _)

    // (i, ..., j-1)の和: cumSum(j)-cumSum(i)
    def sum(min: Int, max: Int): Int = cumSum(max + 1) - cumSum(min)

    def cut(i: Int) = Math.min(i, runLength.length - 1)

    // i番目が0: i, i+1,...,i+2k-1
    // i番目が1: i, i+1,...,i+2k
    runLength.indices
      .map(i =>
        if (runLength(i)._2 == '0') (i, cut(i + 2 * k - 1)) else (i, cut(i + 2 * k)))
      .map { case (min, max) =>
        sum(min, max)
      }.max
  }

  def main(args: Array[String]): Unit = {
    val (n, k, s) = read()
    println(solve(n, k, s))
  }
}


object Main {
  def main(args: Array[String]): Unit = {
    val a = io.StdIn.readInt()
    val b = io.StdIn.readInt()
    val c = io.StdIn.readInt()
    val d = io.StdIn.readInt()
    val e = io.StdIn.readInt()
    val k = io.StdIn.readInt()

    if(e - a <= k) println("Yay!")
    else           println(":(")
  }
}

Score: 100 points

Problem Statement

In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.

Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.

Determine if there exists a pair of antennas that cannot communicate directly.

Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.

Constraints

a, b, c, d, e and k are integers between 0 and 123 (inclusive).

a < b < c < d < e

Input

Input is given from Standard Input in the following format:

a
b
c
d
e
k

Output

Print :( if there exists a pair of antennas that cannot communicate directly, and print Yay! if there is no such pair.

Sample Input 1

1
2
4
8
9
15

Sample Output 1

Yay!

In this case, there is no pair of antennas that cannot communicate directly, because:

the distance between A and B is 2 - 1 = 1

the distance between A and C is 4 - 1 = 3

the distance between A and D is 8 - 1 = 7

the distance between A and E is 9 - 1 = 8

the distance between B and C is 4 - 2 = 2

the distance between B and D is 8 - 2 = 6

the distance between B and E is 9 - 2 = 7

the distance between C and D is 8 - 4 = 4

the distance between C and E is 9 - 4 = 5

the distance between D and E is 9 - 8 = 1

and none of them is greater than 15. Thus, the correct output is Yay!.

Sample Input 2

15
18
26
35
36
18

Sample Output 2

:(

In this case, the distance between antennas A and D is 35 - 15 = 20 and exceeds 18, so they cannot communicate directly.
Thus, the correct output is :(.


object Main extends App {
	val a = new Array[Int](5)

	for(i<-0 until 5){
		a(i) = scala.io.StdIn.readInt
	}

	val k = scala.io.StdIn.readInt

	var flag = true

	for(i<-0 to 3){
		for(j<-i to 4){
			if((a(j) - a(i)).abs > k) flag = false
		}
	}

	if(flag) println("Yay!")
	else println(":(")

}


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

  val antennas = Array.tabulate(5){_ ⇒ readInt()}
  val k = readInt()
  println(
    if (antennas.max - antennas.min <= k) "Yay!" else ":("
  )

}


object Main{
	def main(args:Array[String]){
		val s = new java.util.Scanner(System.in)

		val a=Array.fill(5)(s.nextInt).sortBy(i=>(i+9)%10)

		for(i<-1 to 4)
			a(i)=(a(i) + 9) / 10 * 10

		println(a.sum)
	}
}

Score: 200 points

Problem Statement

The restaurant AtCoder serves the following five dishes:

ABC Don (rice bowl): takes A minutes to serve.

ARC Curry: takes B minutes to serve.

AGC Pasta: takes C minutes to serve.

APC Ramen: takes D minutes to serve.

ATC Hanbagu (hamburger patty): takes E minutes to serve.

Here, the time to serve a dish is the time between when an order is placed and when the dish is delivered.

This restaurant has the following rules on orders:

An order can only be placed at a time that is a multiple of 10 (time 0, 10, 20, ...).

Only one dish can be ordered at a time.

No new order can be placed when an order is already placed and the dish is still not delivered, but a new order can be placed at the exact time when the dish is delivered.

E869120 arrives at this restaurant at time 0. He will order all five dishes. Find the earliest possible time for the last dish to be delivered.

Here, he can order the dishes in any order he likes, and he can place an order already at time 0.

Constraints

A, B, C, D and E are integers between 1 and 123 (inclusive).

Input

Input is given from Standard Input in the following format:

A
B
C
D
E

Output

Print the earliest possible time for the last dish to be delivered, as an integer.

Sample Input 1

29
20
7
35
120

Sample Output 1

215

If we decide to order the dishes in the order ABC Don, ARC Curry, AGC Pasta, ATC Hanbagu, APC Ramen, the earliest possible time for each order is as follows:

Order ABC Don at time 0, which will be delivered at time 29.

Order ARC Curry at time 30, which will be delivered at time 50.

Order AGC Pasta at time 50, which will be delivered at time 57.

Order ATC Hanbagu at time 60, which will be delivered at time 180.

Order APC Ramen at time 180, which will be delivered at time 215.

There is no way to order the dishes in which the last dish will be delivered earlier than this.

Sample Input 2

101
86
119
108
57

Sample Output 2

481

If we decide to order the dishes in the order AGC Pasta, ARC Curry, ATC Hanbagu, APC Ramen, ABC Don, the earliest possible time for each order is as follows:

Order AGC Pasta at time 0, which will be delivered at time 119.

Order ARC Curry at time 120, which will be delivered at time 206.

Order ATC Hanbagu at time 210, which will be delivered at time 267.

Order APC Ramen at time 270, which will be delivered at time 378.

Order ABC Don at time 380, which will be delivered at time 481.

There is no way to order the dishes in which the last dish will be delivered earlier than this.

Sample Input 3

123
123
123
123
123

Sample Output 3

643

This is the largest valid case.


import scala.io.StdIn._

object Main extends App {

  val n = readLong()
  val A = Array.fill(5)(readLong())

  def solve() = {
    val bottleneck = A.min
    if(n <= bottleneck) 5L
    else 4 + (n + bottleneck - 1)/bottleneck
  }
  println(solve())

}

Score: 300 points

Problem Statement

In 2028 and after a continuous growth, AtCoder Inc. finally built an empire with six cities (City 1, 2, 3, 4, 5, 6)!

There are five means of transport in this empire:

Train: travels from City 1 to 2 in one minute. A train can occupy at most A people.

Bus: travels from City 2 to 3 in one minute. A bus can occupy at most B people.

Taxi: travels from City 3 to 4 in one minute. A taxi can occupy at most C people.

Airplane: travels from City 4 to 5 in one minute. An airplane can occupy at most D people.

Ship: travels from City 5 to 6 in one minute. A ship can occupy at most E people.

For each of them, one vehicle leaves the city at each integer time (time 0, 1, 2, ...).

There is a group of N people at City 1, and they all want to go to City 6.

At least how long does it take for all of them to reach there?
You can ignore the time needed to transfer.

Constraints

1 \leq N, A, B, C, D, E \leq 10^{15}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A
B
C
D
E

Output

Print the minimum time required for all of the people to reach City 6, in minutes.

Sample Input 1

5
3
2
4
3
5

Sample Output 1

7

One possible way to travel is as follows.
First, there are N = 5 people at City 1, as shown in the following image:

In the first minute, three people travels from City 1 to City 2 by train. Note that a train can only occupy at most three people.

In the second minute, the remaining two people travels from City 1 to City 2 by train, and two of the three people who were already at City 2 travels to City 3 by bus. Note that a bus can only occupy at most two people.

In the third minute, two people travels from City 2 to City 3 by train, and another two people travels from City 3 to City 4 by taxi.

From then on, if they continue traveling without stopping until they reach City 6, all of them can reach there in seven minutes.

There is no way for them to reach City 6 in 6 minutes or less.

Sample Input 2

10
123
123
123
123
123

Sample Output 2

5

All kinds of vehicles can occupy N = 10 people at a time.
Thus, if they continue traveling without stopping until they reach City 6, all of them can reach there in five minutes.

Sample Input 3

10000000007
2
3
5
7
11

Sample Output 3

5000000008

Note that the input or output may not fit into a 32-bit integer type.


object Main extends App {
  val (us, limits) = {
    val s = new java.util.Scanner(System.in)
    (s.nextLong(), (1 to 5).map(_ => s.nextInt).toList)
  }

  // ボトルネックで切り上げ + 残りの移動回数
  val ans = (us + limits.min - 1) / 2 + 4

  println(ans.toLong)
}


//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 N = sc.nextLong()
    println((N - 1) / Array.fill(5)(sc.nextLong()).min + 5)

  }

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

  def calc(N: Long, R: Long, M: Long): Double = {
    math.sqrt(R * R - (R * 2 * ((M * 1.0 / N) - 0.5)) * (R * 2 * ((M * 1.0 / N) - 0.5))) * 2
  }

  def recursive(now: Long, count: Long): Long = {
    if (count == 1000) now else {
      if (now % 2 == 1) recursive(now * 3 + 1, count + 1) else recursive(now / 2, count + 1)
    }
  }

  def recursive2(X: Set[Long], Y: Stream[Long]): Long = if (X.contains(Y.head)) X.size else recursive2(X + Y.head, Y.tail)

  def check(s: String): Boolean = {
    if (s == "0") true
    else s(0) != '0' && 0 <= s.toInt && s.toInt <= 255
  }

  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
  }

  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): Stream[Long] =
    Stream.cons(begin, getPermutation(begin + 1))

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

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

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

object ArabicRoman {

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

  val codeTable = 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 =>

      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
    }
    } =>

      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 Main extends App {
  var count = readLine.split(" ")
  var k = count(3).toInt
  var a, b , c = readLine.split(" ").map(_.toLong)
  var sum = for(v1 <- a;v2 <- b) yield {
  				v1 + v2
			   }
  sum = sum.take(k)
  sum = sum.sorted(Ordering[Long].reverse)
  var sum2 =  for(v1 <- sum;v2 <- c) yield {
  				v1 + v2
			   }
  sum2 = sum2.take(k)
  sum2 = sum2.sorted(Ordering[Long].reverse)

  val out = new java.io.PrintWriter(System.out)
  for(i <- 0 to k - 1){
  out.println(sum2(i))
  }
  out.flush
}

Score: 400 points

Problem Statement

The Patisserie AtCoder sells cakes with number-shaped candles.
There are X, Y and Z kinds of cakes with 1-shaped, 2-shaped and 3-shaped candles, respectively.
Each cake has an integer value called deliciousness, as follows:

The deliciousness of the cakes with 1-shaped candles are A_1, A_2, ..., A_X.

The deliciousness of the cakes with 2-shaped candles are B_1, B_2, ..., B_Y.

The deliciousness of the cakes with 3-shaped candles are C_1, C_2, ..., C_Z.

Takahashi decides to buy three cakes, one for each of the three shapes of the candles, to celebrate ABC 123.

There are X \times Y \times Z such ways to choose three cakes.

We will arrange these X \times Y \times Z ways in descending order of the sum of the deliciousness of the cakes.

Print the sums of the deliciousness of the cakes for the first, second, ..., K-th ways in this list.

Constraints

1 \leq X \leq 1 \ 000

1 \leq Y \leq 1 \ 000

1 \leq Z \leq 1 \ 000

1 \leq K \leq \min(3 \ 000, X \times Y \times Z)

1 \leq A_i \leq 10 \ 000 \ 000 \ 000

1 \leq B_i \leq 10 \ 000 \ 000 \ 000

1 \leq C_i \leq 10 \ 000 \ 000 \ 000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X Y Z K
A_1 \ A_2 \ A_3 \ ... \ A_X
B_1 \ B_2 \ B_3 \ ... \ B_Y
C_1 \ C_2 \ C_3 \ ... \ C_Z

Output

Print K lines. The i-th line should contain the i-th value stated in the problem statement.

Sample Input 1

2 2 2 8
4 6
1 5
3 8

Sample Output 1

19
17
15
14
13
12
10
8

There are 2 \times 2 \times 2 = 8 ways to choose three cakes, as shown below in descending order of the sum of the deliciousness of the cakes:

(A_2, B_2, C_2): 6 + 5 + 8 = 19

(A_1, B_2, C_2): 4 + 5 + 8 = 17

(A_2, B_1, C_2): 6 + 1 + 8 = 15

(A_2, B_2, C_1): 6 + 5 + 3 = 14

(A_1, B_1, C_2): 4 + 1 + 8 = 13

(A_1, B_2, C_1): 4 + 5 + 3 = 12

(A_2, B_1, C_1): 6 + 1 + 3 = 10

(A_1, B_1, C_1): 4 + 1 + 3 = 8

Sample Input 2

3 3 3 5
1 10 100
2 20 200
1 10 100

Sample Output 2

400
310
310
301
301

There may be multiple combinations of cakes with the same sum of the deliciousness. For example, in this test case, the sum of A_1, B_3, C_3 and the sum of A_3, B_3, C_1 are both 301.
However, they are different ways of choosing cakes, so 301 occurs twice in the output.

Sample Input 3

10 10 10 20
7467038376 5724769290 292794712 2843504496 3381970101 8402252870 249131806 6310293640 6690322794 6082257488
1873977926 2576529623 1144842195 1379118507 6003234687 4925540914 3902539811 3326692703 484657758 2877436338
4975681328 8974383988 2882263257 7690203955 514305523 6679823484 4263279310 585966808 3752282379 620585736

Sample Output 3

23379871545
22444657051
22302177772
22095691512
21667941469
21366963278
21287912315
21279176669
21160477018
21085311041
21059876163
21017997739
20703329561
20702387965
20590247696
20383761436
20343962175
20254073196
20210218542
20150096547

Note that the input or output may not fit into a 32-bit integer type.


import java.util.Scanner

// [x1 + x2 | x1<- l1, x2<- l2]をソート (TLE)
object Main {
  def read(): (Int, Int, Int, Int, IndexedSeq[Long], IndexedSeq[Long], IndexedSeq[Long]) = {
    val sc = new Scanner(System.in)
    val n1, n2, n3, k = sc.nextInt()
    val l1 = for (_ <- 0 until n1) yield sc.nextLong()
    val l2 = for (_ <- 0 until n2) yield sc.nextLong()
    val l3 = for (_ <- 0 until n3) yield sc.nextLong()
    (n1, n2, n3, k, l1, l2, l3)
  }

  def solve(n1: Int, n2: Int, n3: Int, k: Int, l1: IndexedSeq[Long], l2: IndexedSeq[Long], l3: IndexedSeq[Long]): IndexedSeq[Long] = {
    val l1_plus_l2 = for (x1 <- l1; x2 <- l2) yield x1 + x2
    // O(n1*n2)
    // l1+l2の時点でtop kにないものはl3を足してもtop kにならない。
    val sorted_l1_plus_l2_top_k = l1_plus_l2.sorted.reverse.take(k)
    //O((n1*n2)log(n1*n2))
    // k個取れない場合もエラーは出ない。取れるだけ取る
    val sorted_l1_plus_l2_top_k_plus_l3 = for (x12 <- sorted_l1_plus_l2_top_k; x3 <- l3) yield x12 + x3
    // O(k*n3)
    val sorted_sorted_l1_plus_l2_top_k_plus_l3_topk = sorted_l1_plus_l2_top_k_plus_l3.sorted.reverse.take(k) //O((k*n3)log(k*n3))
    sorted_sorted_l1_plus_l2_top_k_plus_l3_topk
  }

  def printResult(result: IndexedSeq[Long]) = {
    result.foreach(println)
  }

  def main(args: Array[String]): Unit = {
    val (n1, n2, n3, k, l1, l2, l3) = read()
    printResult(solve(n1, n2, n3, k, l1, l2, l3))
  }
}


import java.util.Scanner

import scala.collection.mutable.ArrayBuffer

object Main {
  def main(args: Array[String]) {
    val result = solve(new Scanner(System.in))
    if (result.length > 0) println(result)
  }

  def solve(in: String): String = solve(new Scanner(in))

  def solve(sc: Scanner): String = {
    val X = sc.nextInt
    val Y = sc.nextInt
    val Z = sc.nextInt
    val K = sc.nextInt
    val A = Array.fill(X)(sc.nextLong).sorted.reverse
    val B = Array.fill(X)(sc.nextLong).sorted.reverse
    val C = Array.fill(X)(sc.nextLong).sorted.reverse

    val ar = new ArrayBuffer[Long]()
    var ai = 0
    while (ai < A.length) {
      var bi = 0
      while (bi < B.length && (ai + 1) * (bi * 1) <= K) {
        var ci = 0
        while (ci < C.length && (ai + 1) * (bi + 1) * (ci + 1) <= K) {
          ar.append(A(ai) + B(bi) + C(ci))
          ci += 1
        }
        bi += 1
      }
      ai += 1
    }
    ar.sorted.reverse.take(K).mkString("\n")
  }
}


object Main {
  def main(args: Array[String]): Unit = {
    val s = new Main()
    s.solve()
    s.out.flush()
  }
}

class Main {
  import java.io._
  import java.util.StringTokenizer
  import java.util.Arrays

  import scala.collection.mutable
  import scala.util.Sorting
  import math.{abs, max, min}
  import mutable.{ArrayBuffer, ListBuffer}
  import scala.reflect.ClassTag

  val MOD = 1000000007
  val out = new PrintWriter(System.out)

  def solve(): Unit = {
    val A, B, C = ni()
    if (A == B && B == C) out.println("Yes")
    else out.println("No")
  }

  private val oj = System.getenv("ATCODER_DEBUG") == null

  def DEBUG(f: => Unit): Unit = {
    if (!oj){ f }
  }

  def debug(as: Array[Boolean]): Unit = DEBUG {
    debug(as.map(x => if(x) "1" else "0").mkString)
  }

  def debug(as: Array[Int]): Unit = DEBUG {
    debug(as.mkString(" "))
  }

  def debug(as: Array[Long]): Unit = DEBUG {
    debug(as.mkString(" "))
  }

  def debug(s: => String): Unit = DEBUG {
    System.err.println(s)
  }

  def debugL(num: => Long): Unit = DEBUG {
    System.err.println(num)
  }

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

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

    def nextInt(): Int = next().toInt
    def nextLong(): Long = next().toLong
    def nextChar(): Char = next().charAt(0)
  }
  val sc = new InputReader(System.in)
  def ni(): Int = sc.nextInt()
  def nl(): Long = sc.nextLong()
  def nc(): Char = sc.nextChar()
  def ns(): String = sc.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) { i =>
      f(i)
    }
  }

  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]) = {
    val cum = Array.ofDim[Long](as.length + 1)
    REP(as.length) { i =>
      cum(i + 1) = cum(i) + as(i)
    }
    cum
  }
}

Score : 100 points

Problem Statement

You are given three integers A, B and C.

Determine if there exists an equilateral triangle whose sides have lengths A, B and C.

Constraints

All values in input are integers.

1 \leq A,B,C \leq 100

Input

Input is given from Standard Input in the following format:

A B C

Output

If there exists an equilateral triangle whose sides have lengths A, B and C, print Yes; otherwise, print No.

Sample Input 1

2 2 2

Sample Output 1

Yes

There exists an equilateral triangle whose sides have lengths 2, 2 and 2.

Sample Input 2

3 4 5

Sample Output 2

No

There is no equilateral triangle whose sides have lengths 3, 4 and 5.


import java.util.Scanner

object Main {
  def read() = {
    val sc = new Scanner(System.in)
    val a, b, c = sc.nextInt()
    (a,b,c)
  }

  def solve(a: Int, b:Int, c:Int) : String = {
    val isOK :Boolean= a==b && b == c
    if(isOK) "Yes" else "No"
  }

  def main(args: Array[String]): Unit = {
    val (a,b,c) = read()
    println(solve(a,b,c))
  }
}


object Main {
  import java.util.Scanner

  implicit class RichScanner(sc: Scanner) {
    def readStrings(n: Int): Array[String] = Array.fill(n) { sc.next() }
    def readInts(n: Int): Array[Int] = Array.fill(n) { sc.nextInt() }
    def readLongs(n: Int): Array[Long] = Array.fill(n) { sc.nextLong() }
    def readLineStrings: Array[String] = sc.nextLine().split(' ')
    def readLineInts: Array[Int] = readLineStrings.map(_.toInt)
    def readLineLongs: Array[Long] = readLineStrings.map(_.toLong)
  }

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

  def solve(sc: Scanner): Unit = {
    val Array(a, b, c) = sc.readLineInts
    if (a == b && b == c) {
      println("Yes")
    } else {
      println("No")
    }
  }
}


import scala.io.{StdIn => in}

object Main extends App {
  val n = in.readInt
  val s = in.readLine.toCharArray
  val ans = if(s.count(_ == 'R') > s.count(_ == 'B')) "Yes" else "No"
  println(ans)
}

Score : 200 points

Problem Statement

There are N people numbered 1 to N. Each person wears a red hat or a blue hat.

You are given a string s representing the colors of the people. Person i wears a red hat if s_i is R, and a blue hat if s_i is B.

Determine if there are more people wearing a red hat than people wearing a blue hat.

Constraints

1 \leq N \leq 100

|s| = N

s_i is R or B.

Input

Input is given from Standard Input in the following format:

N
s

Output

If there are more people wearing a red hat than there are people wearing a blue hat, print Yes; otherwise, print No.

Sample Input 1

4
RRBR

Sample Output 1

Yes

There are three people wearing a red hat, and one person wearing a blue hat.

Since there are more people wearing a red hat than people wearing a blue hat, the answer is Yes.

Sample Input 2

4
BRBR

Sample Output 2

No

There are two people wearing a red hat, and two people wearing a blue hat.

Since there are as many people wearing a red hat as people wearing a blue hat, the answer is No.


object Main extends App {
  var num = readInt
    var get = readLine
  var r = 0
  var b = 0
  for(i <- 0  to num-1) {
    if(get.charAt(i)=='R')
    r+=1
    else
    b+=1
  }
  if(r>b)
      println("Yes")
  else
  println("No")
    }


object Main {
  def main(args: Array[String]): Unit = {
    val s = new Main()
    s.solve()
    s.out.flush()
  }
}

class Main {
  import java.io._
  import java.util.StringTokenizer
  import java.util.Arrays

  import scala.collection.mutable
  import scala.util.Sorting
  import math.{abs, max, min}
  import mutable.{ArrayBuffer, ListBuffer}
  import scala.reflect.ClassTag

  val MOD = 1000000007
  val out = new PrintWriter(System.out)

  def solve(): Unit = {
    val N = ni()
    val S = ns(N)
    val R = S.count(_ == 'R')
    val B = N - R
    if (R > B) out.println("Yes")
    else out.println("No")
  }

  private val oj = System.getenv("ATCODER_DEBUG") == null

  def DEBUG(f: => Unit): Unit = {
    if (!oj){ f }
  }

  def debug(as: Array[Boolean]): Unit = DEBUG {
    debug(as.map(x => if(x) "1" else "0").mkString)
  }

  def debug(as: Array[Int]): Unit = DEBUG {
    debug(as.mkString(" "))
  }

  def debug(as: Array[Long]): Unit = DEBUG {
    debug(as.mkString(" "))
  }

  def debug(s: => String): Unit = DEBUG {
    System.err.println(s)
  }

  def debugL(num: => Long): Unit = DEBUG {
    System.err.println(num)
  }

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

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

    def nextInt(): Int = next().toInt
    def nextLong(): Long = next().toLong
    def nextChar(): Char = next().charAt(0)
  }
  val sc = new InputReader(System.in)
  def ni(): Int = sc.nextInt()
  def nl(): Long = sc.nextLong()
  def nc(): Char = sc.nextChar()
  def ns(): String = sc.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) { i =>
      f(i)
    }
  }

  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]) = {
    val cum = Array.ofDim[Long](as.length + 1)
    REP(as.length) { i =>
      cum(i + 1) = cum(i) + as(i)
    }
    cum
  }
}


import java.io._
import java.util.StringTokenizer
import scala.util.control.Breaks.{breakable,break}

object Main extends App {
  val in = new InputReader(System.in)
  val pw = new PrintWriter(System.out)

  val MOD = 1e9.toInt + 7
  // (x ^ n) % mod
  def repeatSquares(x: Long, n: Int, mod: Int): Long = {
    if(n == 0)
      1
    else if(n % 2 == 0)
      repeatSquares(x*x % mod, n/2, mod)
    else
      (x * repeatSquares(x*x % mod, n/2, mod)) % mod
  }

  val N = in.next().toInt
  val X = in.next().toInt
  val _S = new Array[Int](N)

  for (i <- 0 until N)
    _S(i) = in.next().toInt

  //
  val S = 0 +: _S.sorted.reverse
  val dp = Array.ofDim[Long](N+1, X+1)
  for (n <- 0 to N; x <- 0 to X)
    dp(n)(x) = -1L

  def calc(i: Int, X: Int): Long = {
    if(i == N)
      X % S(i)
    else if(dp(i)(X) != -1L)
      dp(i)(X)
    else {
      val denom = repeatSquares(N+1-i, MOD-2, MOD)
      val r1 = (denom * calc(i+1, X % S(i))) % MOD
      val r2 = ((N-i) * denom % MOD) * calc(i+1, X) % MOD
      dp(i)(X) = (r1 + r2) % MOD
      dp(i)(X)
    }
  }

  def fact(n: Long, acc: Long): Long =
    if(n == 1)
      acc
    else
      fact(n-1, (n*acc) % MOD)

  val expected = calc(1, X) % MOD
  val ans = expected * fact(N, 1L)
  println(ans % MOD)
}

class InputReader(stream: InputStream) {
  private val reader = new BufferedReader(new InputStreamReader(stream))
  private var tokenizer: StringTokenizer = new StringTokenizer(reader.readLine())
  def next(): String = {
    while (!tokenizer.hasMoreTokens()) {
      tokenizer = new StringTokenizer(reader.readLine())
    }
    tokenizer.nextToken()
  }
}

Score : 600 points

Problem Statement

Snuke has a blackboard and a set S consisting of N integers.
The i-th element in S is S_i.

He wrote an integer X on the blackboard, then performed the following operation N times:

Choose one element from S and remove it.

Let x be the number written on the blackboard now, and y be the integer removed from S. Replace the number on the blackboard with x \bmod {y}.

There are N! possible orders in which the elements are removed from S.
For each of them, find the number that would be written on the blackboard after the N operations, and compute the sum of all those N! numbers modulo 10^{9}+7.

Constraints

All values in input are integers.

1 \leq N \leq 200

1 \leq S_i, X \leq 10^{5}

S_i are pairwise distinct.

Input

Input is given from Standard Input in the following format:

N X
S_1 S_2 \ldots S_{N}

Output

Print the answer.

Sample Input 1

2 19
3 7

Sample Output 1

3

There are two possible orders in which we remove the numbers from S.

If we remove 3 and 7 in this order, the number on the blackboard changes as follows: 19 \rightarrow 1 \rightarrow 1.

If we remove 7 and 3 in this order, the number on the blackboard changes as follows: 19 \rightarrow 5 \rightarrow 2.

The output should be the sum of these: 3.

Sample Input 2

5 82
22 11 6 5 13

Sample Output 2

288

Sample Input 3

10 100000
50000 50001 50002 50003 50004 50005 50006 50007 50008 50009

Sample Output 3

279669259

Be sure to compute the sum modulo 10^{9}+7.


object Main extends App {
  println(new java.util.Scanner(System.in).next match {
	case "A" => "T"
    case "T" => "A"
    case "G" => "C"
    case _ => "G"
  })
}

Score : 100 points

Problem Statement

On the Planet AtCoder, there are four types of bases: A, C, G and T. A bonds with T, and C bonds with G.

You are given a letter b as input, which is A, C, G or T. Write a program that prints the letter representing the base that bonds with the base b.

Constraints

b is one of the letters A, C, G and T.

Input

Input is given from Standard Input in the following format:

b

Output

Print the letter representing the base that bonds with the base b.

Sample Input 1

A

Sample Output 1

T

Sample Input 2

G

Sample Output 2

C


import java.util.Scanner

object Main {
  def main(args: Array[String]): Unit = {
    val sc = new Scanner(System.in)
    val b = sc.next()

    if (b == "A") {
      println("C")
    }
    if (b == "C") {
      println("A")
    }
    if (b == "G") {
      println("T")
    }
    println("G")
  }
}


import java.util.Scanner

object Main extends App {

  val sc = new Scanner(System.in)

  val s = sc.next.toCharArray

  s.map(c => c match {
    case 'A' => 'T'
    case 'T' => 'A'
    case 'C' => 'G'
    case 'G' => 'C'
  }
  ).foreach(print)
  println()
}


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

  val tg   = "ATCG"
  val line = readLine()

  val result =
    line
      .foldLeft((0, 0)) {
        case ((maxCount, count), c) =>
          if (tg.contains(c)) (Math.max(maxCount, count + 1), count + 1)
          else (maxCount, 0)
      }
      ._1

  println(result)
}

Score : 200 points

Problem Statement

You are given a string S consisting of uppercase English letters. Find the length of the longest ACGT string that is a substring (see Notes) of S.

Here, a ACGT string is a string that contains no characters other than A, C, G and T.

Notes

A substring of a string T is a string obtained by removing zero or more characters from the beginning and the end of T.

For example, the substrings of ATCODER include TCO, AT, CODER, ATCODER and (the empty string), but not AC.

Constraints

S is a string of length between 1 and 10 (inclusive).

Each character in S is an uppercase English letter.

Input

Input is given from Standard Input in the following format:

S

Output

Print the length of the longest ACGT string that is a substring of S.

Sample Input 1

ATCODER

Sample Output 1

3

Among the ACGT strings that are substrings of ATCODER, the longest one is ATC.

Sample Input 2

HATAGAYA

Sample Output 2

5

Among the ACGT strings that are substrings of HATAGAYA, the longest one is ATAGA.

Sample Input 3

SHINJUKU

Sample Output 3

0

Among the ACGT strings that are substrings of SHINJUKU, the longest one is (the empty string).


import java.util.Scanner

object Main {
  val sc = new Scanner(System.in)
  def main(args: Array[String]): Unit = {
    val s = sc.next

    val bl = s.split("(?<=.)").map{"ATGC".contains(_)}
    val res =
      bl.foldLeft(List.empty[Int]){ case (l, b) =>
        if( l.isEmpty) if(b) List(1) else List(0)
        else if(b) l :+ (l.last + 1) else l :+ 0
      }
    println(res.max)
  }
}


import scala.io.StdIn._

object Main extends App {

  val Array(n,q) = readLine().split(" ").map(_.toInt)
  val s = readLine()
  val lr = Array.fill(q)(readLine().split(" ").map(_.toInt -1))

  def solve() = {
    val cSum = Array.fill(n)(0)
    for(i <- 0 until n) {
      if(i == 0) cSum(i) = 0
      else if(s(i-1) == 'A' && s(i) == 'C') cSum(i) = cSum(i-1) + 1
      else cSum(i) = cSum(i-1)
    }
    lr.map{ list =>
      cSum(list(1)) - cSum(list(0))
    }.mkString("\n")
  }
  println(solve())

}

Score : 300 points

Problem Statement

You are given a string S of length N consisting of A, C, G and T. Answer the following Q queries:

Query i (1 \leq i \leq Q): You will be given integers l_i and r_i (1 \leq l_i < r_i \leq N). Consider the substring of S starting at index l_i and ending at index r_i (both inclusive). In this string, how many times does AC occurs as a substring?

Notes

A substring of a string T is a string obtained by removing zero or more characters from the beginning and the end of T.

For example, the substrings of ATCODER include TCO, AT, CODER, ATCODER and (the empty string), but not AC.

Constraints

2 \leq N \leq 10^5

1 \leq Q \leq 10^5

S is a string of length N.

Each character in S is A, C, G or T.

1 \leq l_i < r_i \leq N

Input

Input is given from Standard Input in the following format:

N Q
S
l_1 r_1
:
l_Q r_Q

Output

Print Q lines. The i-th line should contain the answer to the i-th query.

Sample Input 1

8 3
ACACTACG
3 7
2 3
1 8

Sample Output 1

2
0
3

Query 1: the substring of S starting at index 3 and ending at index 7 is ACTAC. In this string, AC occurs twice as a substring.

Query 2: the substring of S starting at index 2 and ending at index 3 is CA. In this string, AC occurs zero times as a substring.

Query 3: the substring of S starting at index 1 and ending at index 8 is ACACTACG. In this string, AC occurs three times as a substring.


import java.util.Scanner

object Main extends App {

  val sc = new Scanner(System.in)

  val n = sc.nextInt
  val q = sc.nextInt
  val s = sc.next

  // (li, ri) の組 として扱う
  val lrs = Seq.fill(q)((sc.nextInt, sc.nextInt))

  // Vector[Int]は略せるが、わかりやすさのため。。。
  val commutativeCount: Vector[Int] = {
    s.tail.foldLeft((Vector(0), s.head)){case ((vec, pre), ch) =>
      (pre, ch) match {
        case ('A', 'C') => ( vec :+ (vec.last + 1), ch )
        case _          => ( vec :+  vec.last     , ch )
      }
    }._1
  }

  val res = lrs.map{ case (l,r) =>
    commutativeCount(r - 1) - commutativeCount(l - 1)
  }

  res.foreach(println)

  //文字列Sに含まれるACの個数を求めるメソッド countAC を実装する
//  def countAC(str: String): Int = {
//    str.tail.foldLeft((0, str.head)){ case ( (cnt, pre), ch) =>
//      (pre, ch) match {
//        case ('A','C') => (cnt + 1, ch)
//        case _         => (cnt    , ch)
//      }
//    }._1
//  }

  //各iに対して文字列Sを左からli,右からriだけ切り出してcountACを施す
//  val res =
//    lrs.map{ case (l, r) =>
//      val subStr = s.substring(l - 1, r)
//      countAC(subStr)
//    }
//  res.foreach(println)
}


object Main extends App {
var get1 = readLine.split(" ")
var loop = get1(1).toInt
var word = readLine.split("")
var count : Array[Int] = new Array(word.size)
	for(i <- word.indexOf("A") to word.size - 1) {
    	if(word(i-1)=="A" && word(i)=="C")
      		count(i) = count(i - 1) + 1
        else
          count(i) = count(i - 1)
    }
  	for(i <- 0 to loop - 1) {
      var position = readLine.split(" ")
    	println(count(position(1).toInt) - count(position(0).toInt))
    }
}


import java.util.Scanner

import scala.collection.mutable.ArrayBuffer

object Main {
  def main(args: Array[String]) {
    val result = solve(new Scanner(System.in))
    if (result.length > 0) println(result)
  }

  def solve(in: String): String = solve(new Scanner(in))

  def solve(sc: Scanner): String = {
    val N = sc.nextInt
    val Q = sc.nextInt
    val S = sc.next
    val lr = Array.fill(Q)(sc.nextInt, sc.nextInt)

    val L = ArrayBuffer.fill(N + 2)(0)
    var n = 0
    for (i <- S.indices if i < S.length - 1) {
      if (S(i) == 'A' && S(i + 1) == 'C') {
        n += 1
      }
      L(i + 1) = n
    }
    L(S.length + 1) = n

    val R = ArrayBuffer.fill(N + 2)(0)
    n = 0
    for (i <- S.indices.reverse if i > 0) {
      if (S(i - 1) == 'A' && S(i) == 'C') {
        n += 1
      }
      R(i + 1) = n
    }
    R(S.length + 1) = 0
    R(1) = R(2)
    R(0) = R(1)
    val all = n

    //    println(L.map(_.toString).mkString(" "))
    //    println(R.map(_.toString).mkString(" "))

    lr.map { case (l, r) =>
      (all - L(l - 1) - R(r + 1)).toString
    }.mkString("\n")
  }
}


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

import scala.annotation.tailrec
import scala.collection.mutable.ArrayBuffer
import scala.reflect.ClassTag

object Main {
  def get[T: ClassTag](implicit in: BufferedReader, converter: String => T): T = {
    converter(in.readLine)
  }

  def getArray[T: ClassTag](implicit in: BufferedReader, converter: String => T): Array[T] = {
    get[String].split("\\s+").map[T](converter)
  }

  implicit val idConverter: String => String = s => s
  implicit val intConverter: String => Int = s => Integer.parseInt(s)
  implicit val longConverter: String => Long = s => lang.Long.parseLong(s)

  def get[T1, T2](
                   implicit in: BufferedReader,
                   converter1: String => T1,
                   converter2: String => T2
                 ): (T1, T2) = {
    getArray[String] match {
      case Array(v1, v2) => (converter1(v1), converter2(v2))
    }
  }

  def get[T1, T2, T3](
                       implicit in: BufferedReader,
                       converter1: String => T1,
                       converter2: String => T2,
                       converter3: String => T3
                     ): (T1, T2, T3) = {
    getArray[String] match {
      case Array(v1, v2, v3) => (converter1(v1), converter2(v2), converter3(v3))
    }
  }

  def get[T1, T2, T3, T4](
                           implicit in: BufferedReader,
                           converter1: String => T1,
                           converter2: String => T2,
                           converter3: String => T3,
                           converter4: String => T4
                         ): (T1, T2, T3, T4) = {
    getArray[String] match {
      case Array(v1, v2, v3, v4) => (converter1(v1), converter2(v2), converter3(v3), converter4(v4))
    }
  }

  implicit class TimesInt(val n: Int) {
    def times(f: Int => Unit): Unit = {
      for (i <- 0 until n) {
        f(i)
      }
    }
  }

  def primes(n: Long): ArrayBuffer[Long] = {
    val buf = ArrayBuffer.empty[Long]
    val s = Math.sqrt(n).toInt

    var v = n
    for (i <- 2 to s) {
      if (v == 1 || v == 0) {
        return buf
      } else {
        while (v % i == 0) {
          buf.append(i)
          v /= i
        }
      }
    }
    if (v > s) {
      buf.append(v)
    }
    buf
  }

  def divisors(n: Int): ArrayBuffer[Int] = {
    val buf = ArrayBuffer.empty[Int]
    val s = Math.sqrt(n).toInt

    for (i <- 1 to s) {
      if (n % i == 0) {
        buf.append(i)
        if (i != n / i) {
          buf.append(n / i)
        }
      }
    }
    buf
  }

  @tailrec
  def gcd(a: Long, b: Long): Long = {
    val reminder = a % b
    if (reminder == 0) {
      b
    } else {
      gcd(b, reminder)
    }
  }

  def lcm(a: Long, b: Long): Long = {
    a / gcd(a, b) * b
  }

  def toBitArray(n: Int, width: Int): Array[Boolean] = {
    val buf = ArrayBuffer.fill(width)(false)

    @tailrec
    def inner(n: Int, i: Int): Array[Boolean] = {
      buf(i) = (n & 1) == 0
      if (i == 0) {
        buf.toArray
      } else {
        inner(n / 2, i - 1)
      }
    }

    inner(n, width - 1)
  }

  def main(args: Array[String]): Unit = {
    implicit val in: BufferedReader = new BufferedReader(new InputStreamReader(System.in))
    val out = new PrintWriter(new BufferedOutputStream(System.out))
    val n = get[Int]
    val bs = getArray[Int]

    @tailrec
    def take(bs: ArrayBuffer[Int], operations: ArrayBuffer[Int]): Option[ArrayBuffer[Int]] = {
      if (bs.isEmpty) {
        Some(operations)
      } else {
        bs.zipWithIndex.findLast {
          case (b, index) => b == index + 1
        } match {
          case Some((b, index)) =>
            operations += b
            bs.remove(index)
            take(bs, operations)
          case None =>
            None
        }
      }
    }

    take(ArrayBuffer(bs: _*), ArrayBuffer.empty[Int]) match {
      case Some(operations) => operations.reverse.foreach(out.println)
      case None => out.println(-1)
    }

    out.flush()
  }
}

Score : 400 points

Problem Statement

Snuke has an empty sequence a.

He will perform N operations on this sequence.

In the i-th operation, he chooses an integer j satisfying 1 \leq j \leq i, and insert j at position j in a (the beginning is position 1).

You are given a sequence b of length N. Determine if it is possible that a is equal to b after N operations. If it is, show one possible sequence of operations that achieves it.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq b_i \leq N

Input

Input is given from Standard Input in the following format:

N
b_1 \dots b_N

Output

If there is no sequence of N operations after which a would be equal to b, print -1.
If there is, print N lines. In the i-th line, the integer chosen in the i-th operation should be printed. If there are multiple solutions, any of them is accepted.

Sample Input 1

3
1 2 1

Sample Output 1

1
1
2

In this sequence of operations, the sequence a changes as follows:

After the first operation: (1)

After the second operation: (1,1)

After the third operation: (1,2,1)

Sample Input 2

2
2 2

Sample Output 2

-1

2 cannot be inserted at the beginning of the sequence, so this is impossible.

Sample Input 3

9
1 1 1 2 2 1 2 3 2

Sample Output 3

1
2
2
3
1
2
2
1
1


import scala.collection.mutable.ArrayBuffer

object Main extends App {
  val n = readInt
  val arr = readLine.split(" ").map(_.toInt).toBuffer
  val outBuf = ArrayBuffer.fill(n)(0)
  var i = 0

  var arraySize = n
  while(i < n) {
    var j = arraySize
    var found = false
    while (found == false) {
      j -= 1
      val num = arr(j)

      if (num == j + 1) {
        outBuf(arraySize - 1) = num
        arr.remove(j)
        found = true
      }

      if (j == 0 && found == false) {
        println(-1)
        sys.exit(-1)
      }
    }
    arraySize -= 1
    i += 1
  }

  outBuf.foreach(println)
}


object Main extends App {
  val in = scala.io.Source.stdin.getLines.toSeq
  val B = in.tail.head.split(' ').map(_.toInt.toChar).toArray

  private def func(b:Array[Char]):Seq[Int] = {
    if(b.isEmpty) return Seq()
    //println(b)
    var i=b.size - 1
    val sz=b.size
    while(i >= 0){
      if((i+1) == b(i) && b(i) <= sz ){
        val sub = func(b.take(i) ++ b.drop(i+1) )
        //println(sub)
        if(sub != null) return sub :+ (i+1)
      }
      i -= 1
    }
    return null
  }
  val ret = func(B)
  if(ret == null) println(-1) else println(ret.mkString("\n"))
}


object Main extends App {
  var n = scala.io.StdIn.readInt()
  var vectorCList = Vector.empty[Vector[Int]]
  val divideVal:Long = 1000000007

  for(i <- 1 to n) {
    var newVectorCList = Vector.empty[Vector[Int]]
    val cInput = scala.io.StdIn.readLine()
    val c = cInput.toInt
    if (vectorCList.length == 0) {
      var list = Vector.empty[Int]
      list :+= c
      newVectorCList :+= list
    } else {
      for(k <- 0 to (vectorCList.length - 1)){
        var list = vectorCList(k)
        if(list.exists(_ == c) && list(list.length - 1) != c) {
          var newList = Vector.empty[Int]
          val findIndex = list.lastIndexOf(c)
          for(l <- 0 to list.length) {
            if(findIndex < l) {
              newList :+= c
            } else {
              newList :+= list(l)
            }
          }
          newVectorCList :+= newList
        }
        list :+= c
        newVectorCList :+= list
      }
    }
    vectorCList = newVectorCList.distinct
  }
  println(vectorCList.size % divideVal)
}

Score : 700 points

Problem Statement

There are N stones arranged in a row. The i-th stone from the left is painted in the color C_i.

Snuke will perform the following operation zero or more times:

Choose two stones painted in the same color. Repaint all the stones between them, with the color of the chosen stones.

Find the number of possible final sequences of colors of the stones, modulo 10^9+7.

Constraints

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

1 \leq C_i \leq 2\times 10^5(1\leq i\leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
C_1
:
C_N

Output

Print the number of possible final sequences of colors of the stones, modulo 10^9+7.

Sample Input 1

5
1
2
1
2
2

Sample Output 1

3

We can make three sequences of colors of stones, as follows:

(1,2,1,2,2), by doing nothing.

(1,1,1,2,2), by choosing the first and third stones to perform the operation.

(1,2,2,2,2), by choosing the second and fourth stones to perform the operation.

Sample Input 2

6
4
2
5
4
2
4

Sample Output 2

5

Sample Input 3

7
1
3
1
2
3
3
2

Sample Output 3

5


object Main extends App {

    val Array(ttlRow, ttlCol) =  io.StdIn.readLine.split(" ").map(_.toInt)
    val Array(row, col) =  io.StdIn.readLine.split(" ").map(_.toInt)
    println(balanceCell(ttlRow, ttlCol, row, col))
    def balanceCell(ttlRow:Int, ttlCol:Int, row:Int, col:Int):Int = {
        val rowBal = ttlRow - row
        val colBal = ttlCol - col
        rowBal * colBal
    }
}

Score : 100 points

Problem Statement

There are H rows and W columns of white square cells.

You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.

How many white cells will remain?

It can be proved that this count does not depend on what rows and columns are chosen.

Constraints

All values in input are integers.

1 \leq H, W \leq 20

1 \leq h \leq H

1 \leq w \leq W

Input

Input is given from Standard Input in the following format:

H W
h w

Output

Print the number of white cells that will remain.

Sample Input 1

3 2
2 1

Sample Output 1

1

There are 3 rows and 2 columns of cells. When two rows and one column are chosen and painted in black, there is always one white cell that remains.

Sample Input 2

5 5
2 3

Sample Output 2

6

Sample Input 3

2 4
2 4

Sample Output 3

0


object Main extends App {
    var get = readLine
    var get2 = readLine
    var a = get.split(" ")(0).toInt - get2.split(" ")(0).toInt

    var c = get.split(" ")(1).toInt - get2.split(" ")(1).toInt
   println(a*c)
}


object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)
    val N, M = sc.nextInt()
    val ABs = Array.fill(N, 2)(sc.nextLong())
    val s = ABs.sortBy(_.head)
    println(s.foldLeft((0L, M.toLong)) {
      case ((acc, m), arr) if m == 0 =>
        (acc, 0)
      case ((acc, m), arr) =>
        val a = arr.head
        val b = arr.last
        val buy = Math.min(m, b)
        (acc + (a * buy), m - buy)
    }._1)
  }
}

Score : 300 points

Problem Statement

Hearing that energy drinks increase rating in those sites, Takahashi decides to buy up M cans of energy drinks.

There are N stores that sell energy drinks. In the i-th store, he can buy at most B_i cans of energy drinks for A_i yen (the currency of Japan) each.

What is the minimum amount of money with which he can buy M cans of energy drinks?

It is guaranteed that, in the given inputs, a sufficient amount of money can always buy M cans of energy drinks.

Constraints

All values in input are integers.

1 \leq N, M \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^5

B_1 + ... + B_N \geq M

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the minimum amount of money with which Takahashi can buy M cans of energy drinks.

Sample Input 1

2 5
4 9
2 4

Sample Output 1

12

With 12 yen, we can buy one drink at the first store and four drinks at the second store, for the total of five drinks. However, we cannot buy 5 drinks with 11 yen or less.

Sample Input 2

4 30
6 18
2 5
3 10
7 9

Sample Output 2

130

Sample Input 3

1 100000
1000000000 100000

Sample Output 3

100000000000000

The output may not fit into a 32-bit integer type.


object Main extends App {

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

  def solve(): Int = {
    (Math.min(a, b) to 1 by -1).filter { i =>
      a % i == 0 && b % i == 0
    }(k-1)
  }

  println(solve)

}

Score : 200 points

Problem Statement

You are given positive integers A and B.

Find the K-th largest positive integer that divides both A and B.

The input guarantees that there exists such a number.

Constraints

All values in input are integers.

1 \leq A, B \leq 100

The K-th largest positive integer that divides both A and B exists.

K \geq 1

Input

Input is given from Standard Input in the following format:

A B K

Output

Print the K-th largest positive integer that divides both A and B.

Sample Input 1

8 12 2

Sample Output 1

2

Three positive integers divides both 8 and 12: 1, 2 and 4.
Among them, the second largest is 2.

Sample Input 2

100 50 4

Sample Output 2

5

Sample Input 3

1 1 1

Sample Output 3

1


object Main extends App {
  val sc = new java.util.Scanner(System.in)
  val A = sc.nextInt
  val B = sc.nextInt
  val K = sc.nextInt
  val result = (List(A, B).min to 1 by -1).filter(x => A % x == 0 && B % x == 0)(K - 1)
  println(result)
}


import java.util.Scanner

object Main {
  def read() = {
    val sc = new Scanner(System.in)
    val a = sc.nextInt()
    val b = sc.nextInt()
    val k = sc.nextInt()
    (a, b, k)
  }

  def solve(a: Int, b: Int, k: Int)= {
    def isDevidedByAandB(n:Int) = a%n ==0 && b%n==0

    val cmList = (1 to 100).filter(isDevidedByAandB) // 小さい順で並んだaとbの公約数
    cmList.reverse(k-1)
  }

  def main(args: Array[String]): Unit = {
    val (a, b, k) = read()
    println(solve(a, b, k))
  }
}


import scala.io.StdIn._
object Main extends App {
  val s = readLine
  println(math.min(s.count(_ == '0'), s.count(_ == '1')) * 2)
}

Score : 300 points

Problem Statement

There are N cubes stacked vertically on a desk.

You are given a string S of length N. The color of the i-th cube from the bottom is red if the i-th character in S is 0, and blue if that character is 1.

You can perform the following operation any number of times: choose a red cube and a blue cube that are adjacent, and remove them. Here, the cubes that were stacked on the removed cubes will fall down onto the object below them.

At most how many cubes can be removed?

Constraints

1 \leq N \leq 10^5

|S| = N

Each character in S is 0 or 1.

Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum number of cubes that can be removed.

Sample Input 1

0011

Sample Output 1

4

All four cubes can be removed, by performing the operation as follows:

Remove the second and third cubes from the bottom. Then, the fourth cube drops onto the first cube.

Remove the first and second cubes from the bottom.

Sample Input 2

11011010001011

Sample Output 2

12

Sample Input 3

0

Sample Output 3

0


//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 = {
    println(sc.next().groupBy(c => c).toArray.map(x => x._2.length).min * 2)
  }

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

  def calc(N: Long, R: Long, M: Long): Double = {
    math.sqrt(R * R - (R * 2 * ((M * 1.0 / N) - 0.5)) * (R * 2 * ((M * 1.0 / N) - 0.5))) * 2
  }

  def recursive(sc: => Scanner, n: Long): Unit = {

  }

  def recursive2(X: Set[Long], Y: Stream[Long]): Long = if (X.contains(Y.head)) X.size else recursive2(X + Y.head, Y.tail)

  def check(s: String): Boolean = {
    if (s == "0") true
    else s(0) != '0' && 0 <= s.toInt && s.toInt <= 255
  }

  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
  }

  def gcd(i: Long, j: Long): Long = {
    if (i < j) gcd(j, i)
    else if (j == 0) i
    else gcd(j, i % j)
  }

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

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

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

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

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

object ArabicRoman {

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

  val codeTable = 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 =>

      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
    }
    } =>

      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 Main {
  def main(args: Array[String]): Unit = {
    val s = new Main()
    s.solve()
    s.out.flush()
  }
}

class Main {
  import java.io._
  import java.util.StringTokenizer

  import scala.collection.mutable
  import scala.util.Sorting
  import math.{abs, max, min}
  import mutable.{ArrayBuffer, ListBuffer}
  import scala.reflect.ClassTag

  val MOD = 1000000007
  val out = new PrintWriter(System.out)

  def solve(): Unit = {
    val S = ns()
    val N = S.length
    val zero = S.count(_ == '0')
    val one = S.count(_ == '1')
    val remain = abs(zero - one)
    val ans = N - remain
    out.println(ans)

  }

  def debug(as: Array[Boolean]): Unit = {
    System.err.println(as.map(x => if(x) "1" else "0").mkString)
  }

  def debug(as: Array[Int]): Unit = {
    System.err.println(as.mkString(" "))
  }

  def debug(as: Array[Long]): Unit = {
    System.err.println(as.mkString(" "))
  }

  def debug(s: String): Unit = {
    System.err.println(s)
  }

  def debug(num: Long): Unit = {
    System.err.println(num)
  }

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

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

    def nextInt(): Int = next().toInt
    def nextLong(): Long = next().toLong
    def nextChar(): Char = next().charAt(0)
  }
  val sc = new InputReader(System.in)
  def ni(): Int = sc.nextInt()
  def nl(): Long = sc.nextLong()
  def nc(): Char = sc.nextChar()
  def ns(): String = sc.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 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]) = {
    val cum = Array.ofDim[Long](as.length + 1)
    REP(as.length) { i =>
      cum(i + 1) = cum(i) + as(i)
    }
    cum
  }
}


import java.util.Scanner

object Main {
  def main(args: Array[String]) {
    val result = solve(new Scanner(System.in))
    if (result.length > 0) println(result)
  }

  def solve(in: String): String = solve(new Scanner(in))

  def solve(sc: Scanner): String = {
    val S = sc.next
    val ymd = S.split("/").map(_.toInt)

    if (ymd(0) <= 2019 && ymd(1) <=4 && ymd(2) <= 30) "Heisei" else "TBD"
  }
}

Score : 100 points

Problem Statement

You are given a string S as input. This represents a valid date in the year 2019 in the yyyy/mm/dd format. (For example, April 30, 2019 is represented as 2019/04/30.)

Write a program that prints Heisei if the date represented by S is not later than April 30, 2019, and prints TBD otherwise.

Constraints

S is a string that represents a valid date in the year 2019 in the yyyy/mm/dd format.

Input

Input is given from Standard Input in the following format:

S

Output

Print Heisei if the date represented by S is not later than April 30, 2019, and print TBD otherwise.

Sample Input 1

2019/04/30

Sample Output 1

Heisei

Sample Input 2

2019/11/01

Sample Output 2

TBD