266b.md

题目地址(B. Queue at the School)

https://codeforces.com/problemset/problem/266/B

题目描述

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

前置知识

• 模拟

公司

• 暂无

思路

按照题目的思路进行模拟即可。

具体来说，遍历字符串 s，如果满足题目描述的当前位置是男孩并且后面紧跟着女孩就交换位置，重复这个过程 t 次即可得到结果。

关键点

代码

• 语言支持：Python3

Python3 Code:

n, t = map(int, input().split(" "))
s = list(input())
d = 0

while d < t:
    i = 0
    while i < len(s) - 1:
        if s[i] == "B" and s[i + 1] == "G":
            s[i], s[i + 1] = s[i + 1], s[i]
            i += 1
        i += 1
    d += 1
print("".join(s))

复杂度分析

令 n 为数组长度。

• 时间复杂度：$O(n)$
• 空间复杂度：$O(1)$