Codeforces Round #343 (Div. 2) C. Famil Door and Brackets

题目链接：

http://codeforces.com/contest/629/problem/C

题意：

长度为n的括号，已经知道的部分的长度为m，现在其前面和后面补充‘(',或')'，使得其长度为n，且每个左括号都能找到右括号与它匹配。

题解：

dp[i][j]表示长度为i，平衡度为j的合法括号序列的总数，这里平衡度定义是‘('比')'多多少个，或')'比’('多多少个。

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 using namespace std;

 const int maxn = ;
 const int maxm = 1e5 + ;
 const int mod = 1e9 + ;
 const int INF = 0x3f3f3f3f;
 typedef __int64 LL;

 LL dp[maxn][maxn];
 void pre() {
     memset(dp, , sizeof(dp));
     dp[][] = ;
     for (int i = ; i < maxn; i++) {
         dp[i][] = dp[i - ][];
         for (int j = ; j <= i; j++) {
             dp[i][j] = (dp[i - ][j - ] + dp[i - ][j + ]) % mod;
         }
     }
 }

 int n, m;
 char str[maxm];

 int main() {
     pre();
     scanf("%d%d", &n, &m);
     scanf("%s", str);
     int l = , r = , mi = INF;
     for (int i = ; i < m; i++) {
         if (str[i] == '(') l++;
         else r++;
         mi = min(mi, l - r);
     }
     LL ans = ;
     for (int i = ; i <= n - m; i++) {
         for (int j = max(, -mi); j <= i; j++) {
             int t = j + l - r;
             if (t > n - m - i) continue;
             ans = (ans + dp[i][j] * dp[n - m - i][t] % mod) % mod;
         }
     }
     printf("%I64d\n", ans);
     return ;
 }