[JZOJ 5804] 简单的序列

思路：
似乎和某次培训的题很像啊。。。
将左括号记为1，右括号记为-1，那么最终一定加和为0，然后再求最小前缀和。
用dp解决即可。

#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn = 1000010;
const int mod = 1e9+7;
char s[maxn];
int n,m;
ll sum;
int mn;
ll dp[2010][2010];
ll ans;
int main () {
	#ifdef ONLINE_JUDGE
		freopen("bracket.in","r",stdin);
		freopen("bracket.out","w",stdout);
	#endif
	scanf("%d %d",&n,&m);
	scanf("%s",s+1);
	for(int i = 1;i <= m; ++i) {
		if(i == 1) mn = (s[i] == '(' ? 1 : -1);
		sum += (s[i] == '(' ? 1 : -1);
		mn = min(mn,(int)sum);
	}
	dp[0][0] = 1;
	for(int i = 1;i <= n - m; ++i) {
		for(int j = 0;j <= i; ++j) {
			if(!j) {
				dp[i][j] = dp[i - 1][j + 1];
			}
			else dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j + 1]) % mod;
		}
	}
	for(int i = 0;i <= n - m; ++i) {
		for(int j = 0;j <= i; ++j) {
			if(j + sum <= n - m && j + mn >= 0) {
				ans = (ans + dp[i][j] * dp[n - m - i][j + sum])%mod;
			}
		}
	}
	cout<<ans<<endl;
	return 0;
}