题干：
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（一趟下来所有符合条件的都交换一次），求t 秒后，队列分布情况。
解题报告：
就直接模拟就行了。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
char s[55];
int main() {
int n,t;
scanf("%d%d\n",&n,&t);
scanf("%s",s);
while(t--) {
for (int i=0;i<n;) {
if(s[i]=='B') {
if(s[i+1]=='G') {
swap(s[i],s[i+1]);
i+=2;
}
else i++;
}
else i++;
}
}
printf("%s\n",s);
return 0;
}