10.28 test9 T1

dp[i][j][k] 表示加了i个，给定串匹配到第j个，sum1-sum0=k；

转移：
若放左括号：
dp[i][j][k] -> dp[i+1][j][k+1] , dp[i][j+1][k+1]
方右括号：
dp[i][j][k] -> dp[i+1][j][k-1] ,dp[i][j+1][k-1]

能放给的括号优先放给的

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
	char ch=' ';
	int f=1;int x=0;
	while(ch<'0'||ch>'9')
	{
		if(ch=='-') f=-1;ch=getchar();
	}
	while(ch>='0'&&ch<='9')
	{
	    x=x*10+ch-'0';ch=getchar();
	}
	return x*f;
}
const int N=110;
const int p=1e9+7;
char s[N];
int dp[N<<1][N<<1][N<<1];
int main()
{	
	int n;
	n=read();
	int i,j,k;
	cin>>s;
	int len=strlen(s);
	dp[0][0][0]=1;
	for(i=0;i<=2*n-len;i++)
	{
		for(j=0;j<=len;j++)
		{
			for(k=0;k<=n;k++)
			{
				if(dp[i][j][k])
				{
					if(j<len&&s[j]=='(') 
					{
						dp[i][j+1][k+1]+=dp[i][j][k];
						dp[i][j+1][k+1]%=p;
					}
					else
					{
						dp[i+1][j][k+1]+=dp[i][j][k];
						dp[i+1][j][k+1]%=p;
					}
					if(k>0)
					{
						if(j<len&&s[j]==')')
						{
							dp[i][j+1][k-1]+=dp[i][j][k];
							dp[i][j+1][k-1]%=p;
						}
						else
						{
							dp[i+1][j][k-1]+=dp[i][j][k];
							dp[i+1][j][k-1]%=p;
						}
					}
				}
			}
		}
	}
	printf("%d\n",dp[n*2-len][len][0]);
	return 0; 
}