Queue at the School

文章目录

一、Queue at the School

总结

一、Queue at the School

本题链接：Queue at the School

题目：

B. Queue at the School

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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

本博客给出本题截图：

题意：每次交换一个'B'和'G'的位置，求交换t次后的结果

AC代码

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
    int n, t;
    cin >> n >> t;
    string a;
    cin >> a;
    for (int i = 0; i < t; i ++ )
        for (int j = 0; j < n - 1; j ++ )
            if (a[j] == 'B' && a[j + 1] == 'G')
            {
                swap(a[j], a[j + 1]);
                j ++;
            }
    cout << a;
    return 0;
}

总结

水题，不解释