既求从点(0,0)只能向上或者向右并且不穿越y=x到达点(a,b)有多少总走法...

有公式: C(a+b,min(a,b))-C(a+b,min(a,b)-1)  ///

折纸法证明卡特兰数: http://blog.sina.com.cn/s/blog_6917f47301010cno.html

Brackets

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

We give the following inductive definition of a “regular brackets” sequence:

● the empty sequence is a regular brackets sequence,

● if s is a regular brackets sequence, then (s) are regular brackets sequences, and

● if a and b are regular brackets sequences, then ab is a regular brackets sequence.

● no other sequence is a regular brackets sequence

For instance, all of the following character sequences are regular brackets sequences:

(), (()), ()(), ()(())

while the following character sequences are not:

(, ), )(, ((), ((()

Now we want to construct a regular brackets sequence of length n,
how many regular brackets sequences we can get when the front several brackets are given already.

Input

Multi test cases (about 2000),
every case occupies two lines.

The first line contains an integer n.

Then second line contains a string str which indicates the front several brackets.

Please process to the end of file.

[Technical Specification]

1≤n≤1000000

str contains only ‘(‘ and ‘)‘ and length of str is larger than 0 and no more than n.

Output

For each case，output answer % 1000000007 in
a single line.

Sample Input

4
()
4
(
6
()

Sample Output

1
2
2

Hint

For the first case the only regular sequence is ()().
For the second case regular sequences are (()) and ()().
For the third case regular sequences are ()()() and ()(()).

/* ***********************************************
Author        :CKboss
Created Time  :2015年03月18日 星期三 20时10分21秒
File Name     :HDOJ5184.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

const int maxn=1001000;
const LL mod=1000000007LL;

int n,len;
char str[maxn];

LL inv[maxn];
LL jc[maxn],jcv[maxn];

void init()
{
	inv[1]=1; jc[0]=1; jcv[0]=1;
	jc[1]=1; jcv[1]=1;

	for(int i=2;i<maxn;i++)
	{
		inv[i]=inv[mod%i]*(mod-mod/i)%mod;
		jc[i]=(jc[i-1]*i)%mod;
		jcv[i]=(jcv[i-1]*inv[i])%mod;
	}
}

LL COMB(LL n,LL m)
{
	if(m<0||m>n) return 0LL;
	if(m==0||m==n) return 1LL;
	LL ret=((jc[n]*jcv[n-m])%mod*jcv[m])%mod;
	return ret;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	init();
	while(scanf("%d",&n)!=EOF)
	{
		scanf("%s",str);
		len=strlen(str);

		bool flag=true;
		if(n%2==1) flag=false;
		int left=0,right=0;
		for(int i=0;i<len&&flag;i++)
		{
			if(str[i]=='(') left++;
			else if(str[i]==')') right++;
			if(left>=right) continue;
			else flag=false;
		}
		if(flag==false) { puts("0"); continue; }

		int a=n/2-left; /// remain left
		int b=n/2-right; /// remain right

		if(b>a) swap(a,b);
		LL ans = (COMB(a+b,b)-COMB(a+b,b-1)+mod)%mod;
		cout<<ans<<endl;
	}

    return 0;
}