CodeForces - 266B （水题）
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
G代表女孩 B代表男孩，如果有男孩在女孩前面（挨着），那么下一秒他俩就交换位置，女士优先嘛。。然后问T秒后的队列是什么样的。 - 解题思路：
这还要啥思路啊？？50的数据量。。。暴力出奇迹啊，，直接模拟一下这个过程就ok了，贼水的题。 - AC代码：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn=1e6+10;
int l,r;
char a[maxn];
int ans[maxn];
int main()
{
scanf("%s",a);
int len=strlen(a);
l=1;r=len;
for(int i=0;i<len;i++)
{
if(a[i]=='l')
ans[r--]=i+1;
else
ans[l++]=i+1;
}
for(int i=1;i<=len;i++)
//cout<<ans[i]<<endl;
printf("%d\n",ans[i]);
return 0;
}