C. Bracket Subsequence

最新推荐文章于 2022-01-16 17:52:18 发布

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本文介绍了一种算法，用于从给定的合法括号序列中构建特定长度的子序列，该子序列同样为合法括号序列。文章通过示例说明了如何选择适当的左括号和右括号来构建所需的子序列。

A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not.

Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

You are given a regular bracket sequence s

and an integer number k. Your task is to find a regular bracket sequence of length exactly k such that it is also a subsequence of s

It is guaranteed that such sequence always exists.

Input

The first line contains two integers n

and k (2≤k≤n≤2⋅105, both n and k are even) — the length of s

and the length of the sequence you are asked to find.

The second line is a string s

— regular bracket sequence of length n

Output

Print a single string — a regular bracket sequence of length exactly k

such that it is also a subsequence of s

It is guaranteed that such sequence always exists.

Examples

Input

Copy

6 4
()(())

Output

Copy

()()

Input

Copy

8 8
(()(()))

Output

Copy

(()(()))

题意：

给你2个长度，一个字符串，一个是字符串的长度，一个是要你构造的长度，一个字符串合法当且仅当每一位的左括号数大于等于右括号数，且最后一位的左括号数等于右括号数。

思路：

观察：首先我读懂了题意提出假设：贪心就行

code:

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<vector>
#include<queue>
#include<string>
#include<stack>
#include<string.h>
# include <cstdio>
# include <iostream>
#include <algorithm>
#include <iostream>
#include <iterator>
#include <ostream>
#include <string>
#include <utility>
#include <vector>
using namespace std;
int n,m,tl,tr,book[200100];
char s[200100];
int main()
{
	memset(book,0,sizeof(book));
	cin>>n>>m;
	scanf("%s",s+1);
	tl=tr=(n-m)/2;
	for(int i=1;i<=n;i++)
	{
		if(tr>0)
		{
			if(s[i]==')')
			{
				book[i]=1;
				tr--;
			}
		}
		else
			break;
	}
	for(int i=1;i<=n;i++)
	{
		if(tl>0)
		{
			if(s[i]=='(')
			{
				book[i]=1;
				tl--;
			}
		}
		else
			break;
	}
	for(int i=1;i<=n;i++)
	{
		if(!book[i])
			cout<<s[i];
	}
	cout<<endl;
	return 0;
}