Codeforces 629C Famil Door and Brackets【dp】-优快云博客

本文介绍了一种算法，用于解决给定一个长度为M的括号串，如何通过在其前后添加括号来构造长度为N的有效括号序列的问题。讨论了动态规划方法并给出了AC代码。

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As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length n more than any other strings!

The sequence of round brackets is called valid if and only if:

the total number of opening brackets is equal to the total number of closing brackets;
for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets.

Gabi bought a string s of length m (m ≤ n) and want to complete it to obtain a valid sequence of brackets of length n. He is going to pick some strings p and q consisting of round brackets and merge them in a string p + s + q, that is add the string p at the beginning of the string s and string q at the end of the string s.

Now he wonders, how many pairs of strings p and q exists, such that the string p + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 10⁹ + 7.

Input

First line contains n and m (1 ≤ m ≤ n ≤ 100 000, n - m ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.

The second line contains string s of length m consisting of characters '(' and ')' only.

Output

Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 10⁹ + 7.

Examples

Input

4 1
(

Output

Input

4 4
(())

Output

Input

4 3
(((

Output

Note

In the first sample there are four different valid pairs:

p = "(", q = "))"
p = "()", q = ")"
p = "", q = "())"
p = "", q = ")()"

In the second sample the only way to obtain a desired string is choose empty p and q.

In the third sample there is no way to get a valid sequence of brackets.

题目大意：

给你一个长为M的括号串（只有左括号和右括号），需要在当前串的前边加一个p串，在当前串后边加一个q串，使得整个串长度为N，并且是一个匹配的串，问一共有多少种方案。

思路：

1、因为给出的字符串有一个左括号数量和右括号数量的比较问题，那么我们考虑dp其括号数量：

①设定dp【i】【j】表示长度为i的串，其中左括号比右括号多j个的方案数。

②那么不难想到其状态转移方程：
dp【i】【j】=dp【i-1】【j-1】+dp【i-1】【j+1】；

2、预处理dp数组之后，我们再考虑统计问题：

①首先我们处理一下给出的串S中，至少需要p串中含有左括号的个数，记做minn；

②然后我们设定一个变量：sumzuo，在扫s串的过程中，其遇到左括号，sumzuo++，否则sumzuo--，扫完串s之后，保留sumzuo的值，表示左括号比右括号多的个数（可能为负）。

3、

①然后我们考虑枚举p串的起点记做i，以及p串中左括号比右括号多的个数记做j。然后将p串和s串合并得到一个新的串，其中这个新的串的左括号比右括号多的个数为：j+sumzuo，而且此时我们能够得知，q串的长度为n-m-i；

②然后我们再考虑这样一点，我们想要得到的串是左右括号个数平衡的一个状态，那么考虑到dp【i】【j】还可以表示长度为i的串，其中右括号比左括号多j个的方案数。那么我们想要使得这个串是平衡的，那么我们只要保证q串中右括号比左括号多j+sumzuo个即可。

③那么我们枚举出的i，j，能够增加的方案数为：dp【i】【j】*dp【n-m-i】【j+sumzuo】；

④根据之前处理，保证i，j，n-m-i，j+sumzuo均合法即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define ll __int64
char a[100400];
ll dp[2010][2010];
ll mod=1e9+7;
int n,m;
void init()
{
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    for(int i=1;i<=2005;i++)
    {
        for(int j=0;j<=i;j++)
        {
            if(j==0)dp[i][j]=dp[i-1][j+1];
            dp[i][j]=(dp[i-1][j-1]+dp[i-1][j+1])%mod;
        }
    }
}
int main()
{
    init();
    while(~scanf("%d%d",&n,&m))
    {
        scanf("%s",a);
        int sumzuo=0,minn=0;
        for(int i=0;i<m;i++)
        {
            if(a[i]=='(')sumzuo++;
            else sumzuo--;
            minn=min(minn,sumzuo);
        }
        ll ans=0;
        for(int i=0;i<n-m+1;i++)
        {
            for(int j=0;j<=i;j++)
            {
                if(j+minn>=0&&j+sumzuo<=n-m-i)
                ans+=(dp[i][j]*dp[n-m-i][j+sumzuo]);
                ans%=mod;
            }
        }
        printf("%I64d\n",ans);
    }
}