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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
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1
2
5 1 BGGBG
output
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1
GBGGB
input
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1
2
5 2 BGGBG
output
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1
GGBGB
input
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1
2
4 1 GGGB
output
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1
GGGB

题目大意:输入n(学生人数),t(时间),每秒切换一次位置,为了体现女士优先的绅士风格,站在队列靠前的男生与紧邻靠后的女生交换一次位置,每秒交换一次,求t 秒后,队列分布情况。
解题思路:两层循环,外层为时间 t ,内层为 遍历队列,若满足条件则交换位置。
注意事项:1 秒内只要满足就同步交换位置,而不是只能交换一对。


以下为解题代码(java代码)
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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(); } }

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