Codeforces 문제 푸는 길 - 266B Queue at the School

B. Queue at the School
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen. Initially the children stood in the order they entered the canteen. However, after a while the boys started feeling awkward for standing in front of the girls in the queue and they started letting the girls move forward each second.
Let's describe the process more precisely. Let's say that the positions in the queue are sequentially numbered by integers from 1 to n, at that the person in the position number 1 is served first. Then, if at time x a boy stands on the i-th position and a girl stands on the(i + 1)-th position, then at time x + 1 the i-th position will have a girl and the (i + 1)-th position will have a boy. The time is given in seconds.
You've got the initial position of the children, at the initial moment of time. Determine the way the queue is going to look after t seconds.
Input
The first line contains two integers n and t (1 ≤ n, t ≤ 50), which represent the number of children in the queue and the time after which the queue will transform into the arrangement you need to find.
The next line contains string s, which represents the schoolchildren's initial arrangement. If the i-th position in the queue contains a boy, then the i-th character of string s equals "B", otherwise the i-th character equals "G".
Output
Print string a, which describes the arrangement after t seconds. If the i-th position has a boy after the needed time, then the i-th character a must equal "B", otherwise it must equal "G".
Examples
input

5 1
BGGBG

output

GBGGB

input

5 2
BGGBG

output

GGBGB

input

4 1
GGGB

output

GGGB

제목 대의: n(학생 수), t(시간)를 입력하고 매초에 한 번씩 위치를 전환한다. 여자가 우선하는 신사 스타일을 나타내기 위해 대열에 서 있는 남학생과 뒤에 인접한 여학생이 한 번 위치를 교환하고 매초에 한 번 교환한다. t초 후에 대열의 분포 상황을 구한다.
문제풀이 사고방식: 두 층이 순환하고 바깥쪽은 시간 t이며 안쪽은 반복 대기열이다. 조건이 충족되면 위치를 교환한다.
주의 사항: 1초 동안 한 쌍만 교환할 수 있는 것이 아니라 만족하면 위치를 동기화합니다.
다음은 문제풀이 코드(java 코드)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scanner = new Scanner(System.in);
		int n = scanner.nextInt();
		int t = scanner.nextInt();
		scanner.nextLine();
		String string = scanner.nextLine();
		char[] ch = string.toCharArray();
		
		for(int i = 0;i < t;i++){
			for(int j = 0;j < ch.length;j++){
				if((j+1) < ch.length && (ch[j] < ch[j+1])  ){
					char temp = ch[j];
					ch[j] = ch[j+1];
					ch[j+1] = temp;
					j++;
				}
			}
		}
		System.out.println(ch);
		scanner.close();		
	}
}