maya普及组的dp都要想很长时间,我真是越来越菜了qwq
设$f[i][j]$表示当前到第$i$个位置,剩下$j$个左括号没被匹配
转移的时候判断一下即可
/*
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<set>
#include<queue>
#include<cmath>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#include<set>
#include<vector>
//#define int long long
#define LL long long
//#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<22, stdin), p1 == p2) ? EOF : *p1++)
//char buf[(1 << 22)], *p1 = buf, *p2 = buf;
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N;
char s[MAXN];
int f[2][MAXN];
int main() {
// freopen("a.in", "r", stdin);
// freopen("c.out", "w", stdout);
N = read();
scanf("%s", s + 1);
int o = 1; f[0][0] = 1;
for(int i = 1; i <= N; i++, o ^= 1) {
//(f[i][j] += f[i - 1][j]) %= mod;
memset(f[o], 0, sizeof(f[o]));
if(s[i] == '(') {
for(int j = 0; j <= i; j++)
(f[o][j] += f[o ^ 1][j]) %= mod, (f[o][j] += f[o ^ 1][j - 1]) %= mod;
}
else {
for(int j = 0; j <= i; j++)
(f[o][j] += f[o ^ 1][j]) %= mod, (f[o][j] += f[o ^ 1][j + 1]) %= mod;
}
}
printf("%d", (f[o ^ 1][0] - 1 + mod) % mod);
return 0;
}
/*
2
()
3
())
8
)(()(())
*/