14175번: The Cow-Signal

문제 출처 : https://www.acmicpc.net/problem/14175

 

언어 : Kotlin

 

문제 설명 :

Bessie and her cow friends are playing as their favorite cow superheroes. Of course, everyone knows that any self-respecting superhero needs a signal to call them to action. Bessie has drawn a special signal on a sheet of M×N paper (1≤M≤10,1≤N≤10), but this is too small, much too small! Bessie wants to amplify the signal so it is exactly K times bigger (1≤K≤10) in each direction.

The signal will consist only of the '.' and 'X' characters.

 

입력 :

The first line of input contains M, N, and K, separated by spaces.
The next M lines each contain a length-N string, collectively describing the picture of the signal.

 

출력 :

You should output KM lines, each with KN characters, giving a picture of the enlarged signal.

 

제한 사항 :

• 시간 제한 : 2초
• 메모리 제한 : 512MB

 

입출력 예 :

입력	출력
5 4 2 XXX. X..X XXX. X..X XXX.	XXXXXX.. XXXXXX.. XX....XX XX....XX XXXXXX.. XXXXXX.. XX....XX XX....XX XXXXXX.. XXXXXX..

 

풀이 : 

import java.io.BufferedReader
import java.io.BufferedWriter
import java.io.InputStreamReader
import java.io.OutputStreamWriter

fun main() = with(BufferedReader(InputStreamReader(System.`in`))) {
    val bw = BufferedWriter(OutputStreamWriter(System.out))

    val (m, n, k) = readLine().split(" ").map { it.toInt() }
    repeat(m) {
        val line = readLine()
        val mid = line.length / 2

        if (n % 2 == 0) {
            val left = line.slice(0 until mid).map { "$it".repeat(k) }.joinToString("")
            val right = line.slice(mid .. line.lastIndex).map { "$it".repeat(k) }.joinToString("")

            repeat(k) {
                bw.appendLine(left + right)
            }
        } else {
            val left = line.slice(0 until mid).map { "$it".repeat(k) }.joinToString("")
            val right = line.slice(mid + 1 .. line.lastIndex).map { "$it".repeat(k) }.joinToString("")
            repeat(k) {
                bw.appendLine(left + "${line[mid]}".repeat(k) + right)
            }
        }
    }

    bw.flush()
    bw.close()
}