题目连接：

http://www.codeforces.com/contest/670/problem/E

Description

Recently Polycarp started to develop a text editor that works only with correct bracket sequences (abbreviated as CBS).

Note that a bracket sequence is correct if it is possible to get a correct mathematical expression by adding "+"-s and "1"-s to it. For example, sequences "(())()", "()" and "(()(()))" are correct, while ")(", "(()" and "(()))(" are not. Each bracket in CBS has a pair. For example, in "(()(()))":

1st bracket is paired with 8th,
2d bracket is paired with 3d,
3d bracket is paired with 2d,
4th bracket is paired with 7th,
5th bracket is paired with 6th,
6th bracket is paired with 5th,
7th bracket is paired with 4th,
8th bracket is paired with 1st.
Polycarp's editor currently supports only three operations during the use of CBS. The cursor in the editor takes the whole position of one of the brackets (not the position between the brackets!). There are three operations being supported:

«L» — move the cursor one position to the left,
«R» — move the cursor one position to the right,
«D» — delete the bracket in which the cursor is located, delete the bracket it's paired to and all brackets between them (that is, delete a substring between the bracket in which the cursor is located and the one it's paired to).
After the operation "D" the cursor moves to the nearest bracket to the right (of course, among the non-deleted). If there is no such bracket (that is, the suffix of the CBS was deleted), then the cursor moves to the nearest bracket to the left (of course, among the non-deleted).

There are pictures illustrated several usages of operation "D" below.

All incorrect operations (shift cursor over the end of CBS, delete the whole CBS, etc.) are not supported by Polycarp's editor.

Polycarp is very proud of his development, can you implement the functionality of his editor?

Input

The first line contains three positive integers n, m and p (2 ≤ n ≤ 500 000, 1 ≤ m ≤ 500 000, 1 ≤ p ≤ n) — the number of brackets in the correct bracket sequence, the number of operations and the initial position of cursor. Positions in the sequence are numbered from left to right, starting from one. It is guaranteed that n is even.

It is followed by the string of n characters "(" and ")" forming the correct bracket sequence.

Then follow a string of m characters "L", "R" and "D" — a sequence of the operations. Operations are carried out one by one from the first to the last. It is guaranteed that the given operations never move the cursor outside the bracket sequence, as well as the fact that after all operations a bracket sequence will be non-empty.

Output

Print the correct bracket sequence, obtained as a result of applying all operations to the initial sequence.

Sample Input

8 4 5
(())()()
RDLD

Sample Output

()

题意

你有长度为n的括号序列，有m个操作，光标现在在pos位置

操作有3个

d,l,r

删除，左移动，右移动。

删除就是，你每次可以删除光标所在的位置以及他所对应的括号，和这括号内包括的东西

然后这个光标就跑到这个地方的右边去了

问你操作完之后是什么样子了？

题解：

对应的括号，这个用栈去找就好了

然后操作，我们就模拟一下链表就好了

然后就完了……

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e5+7;
struct node{int l,r;}p[maxn];
int n,m,pos,a[maxn],d[maxn],len,now=0;
char s1[maxn],s2[maxn];
int main()
{scanf("%d%d%d",&n,&m,&pos);scanf("%s",s1+1);len=strlen(s1+1);for(int i=0;i<=n+1;i++)p[i].l=i-1,p[i].r=i+1;for(int i=1;i<=len;i++){if(s1[i]=='(')a[now++]=i;else d[i]=a[--now],d[a[now]]=i;}scanf("%s",s2+1);for(int i=1;i<=m;i++){if(s2[i]=='R')pos=p[pos].r;else if(s2[i]=='L')pos=p[pos].l;else{int l=min(d[pos],pos);int r=max(d[pos],pos);p[p[l].l].r=p[r].r;p[p[r].r].l=p[l].l;pos=p[r].r;if(pos==n+1)pos=p[n+1].l;if(pos==0)pos=p[0].r;}}int pos=p[0].r;while(pos!=n+1){cout<<s1[pos];pos=p[pos].r;}cout<<endl;return 0;
}