CF 266B Queue at the School （人生第一道B题）-优快云博客

探讨n个同学在餐厅排队过程中，男同学给紧挨着的女同学让位的问题。每次操作只能同时处理t次交换，求解t秒后的队列状态。

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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

题意（借助了有道词典）：有n个同学在餐厅排队，如果男同学再前面就会给后面紧挨着的女同学让个位置，但是对于一列而言只能同时操作t次；问t次后的队列应该什么样子的；

题解：这个题就和指针思想一样，再交换后要及时对后面的同学进行处理，跳过当前位置放置再次被交换，就没坑了；

AC代码：

#include <bits/stdc++.h>
using namespace std ;
char a[1000] ;
int ans , n ;
int main()
{
    cin>>n>>ans ;
    cin>>a;
    int len = strlen(a);
    while(ans--)
    {
        for(int i = 0 ; i < len-1 ; i++)
        {
            if(a[i]=='B'&&a[i+1]=='G')
            {
                char temp ;
                temp = a[i] ;
                a[i] = a[i+1];
                a[i+1] = temp ;
                i++;
            }
        }
    }

    cout<<a<<endl;
    return 0  ;
}