time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
As Famil Door’s birthday is coming, some of his friends (like Gabi) decided to buy a present for him. His friends are going to buy a string consisted of round brackets since Famil Door loves string of brackets of length n more than any other strings!
The sequence of round brackets is called valid if and only if:
the total number of opening brackets is equal to the total number of closing brackets;
for any prefix of the sequence, the number of opening brackets is greater or equal than the number of closing brackets.
Gabi bought a string s of length m (m ≤ n) and want to complete it to obtain a valid sequence of brackets of length n. He is going to pick some strings p and q consisting of round brackets and merge them in a string p + s + q, that is add the string p at the beginning of the string s and string q at the end of the string s.
Now he wonders, how many pairs of strings p and q exists, such that the string p + s + q is a valid sequence of round brackets. As this number may be pretty large, he wants to calculate it modulo 109 + 7.
Input
First line contains n and m (1 ≤ m ≤ n ≤ 100 000, n - m ≤ 2000) — the desired length of the string and the length of the string bought by Gabi, respectively.
The second line contains string s of length m consisting of characters ‘(’ and ‘)’ only.
Output
Print the number of pairs of string p and q such that p + s + q is a valid sequence of round brackets modulo 109 + 7.
Examples
input
4 1
(
output
4
input
4 4
(())
output
1
input
4 3
(((
output
0
Note
In the first sample there are four different valid pairs:
p = “(“, q = “))”
p = “()”, q = “)”
p = “”, q = “())”
p = “”, q = “)()”
In the second sample the only way to obtain a desired string is choose empty p and q.
In the third sample there is no way to get a valid sequence of brackets.
【题解】
给你一个串；
只包含圆括号；
可以让你在最左边和最右边加上两个串p和q；
这两个串也只能包括括号；
要使得最后的括号是匹配的；
问你有多少对P和q,p和q可以为空
每时每刻左括号的数目都要大于等于右括号的数目;
最后那个限制起了很大的作用；
把左括号看做1，右括号看做-1；
则它的要求是每时每刻前缀和都大于0；
则先求出所给的s的前缀和，并记录前缀和的最小值mi；
预处理出dp[i][j]表示i个括号能够造出前缀和为j的方案数(满足每时每刻前缀和都大于等于0的方案)；
然后枚举q的长度i,前缀和为j
如果dp[i][j]+mi>=0;
则从左到中间那个串都是满足题意的了(前缀和始终大于等于0)
然后再获取P;
p的长度就是n-m-i;
它的前缀和该是啥?j+pres；
这里的j+pres指的是负数->加上dp[n-m-i][j+pres];
可以理解为从右到左进行DP;然后从右到左的过程中每时每刻右括号的数目都大于左括号的数目;(等价转化);
这样可以保证从最左边的q到最右边的p的过程中间都不会出现前缀和小于0的情况；
假设在p处出现了前缀和小于0;则在这个位置的右边右括号的数目大于左括号的数目;最后结果就不可能为0;只可能是负数;
这和j+pres-(j+pres)==0不符合；
所以在这个串中不会出现前缀和为负数的情况；
具体实现看代码；
（那个DP很简单的。。);
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <string>
#define LL long long
using namespace std;
const int MAXN = 2000;
const LL MOD = 1e9+7;
int n,m;
LL dp[MAXN+20][MAXN+20];
string s;
const int dx[5] = {0,1,-1,0,0};
const int dy[5] = {0,0,0,-1,1};
void input_LL(LL &r)
{
r = 0;
char t = getchar();
while (!isdigit(t)) t = getchar();
LL sign = 1;
if (t == '-')sign = -1;
while (!isdigit(t)) t = getchar();
while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
r = r*sign;
}
void input_int(int &r)
{
r = 0;
char t = getchar();
while (!isdigit(t)) t = getchar();
int sign = 1;
if (t == '-')sign = -1;
while (!isdigit(t)) t = getchar();
while (isdigit(t)) r = r * 10 + t - '0', t = getchar();
r = r*sign;
}
int main()
{
//freopen("F:\\rush.txt", "r", stdin);
input_int(n);input_int(m);
cin >> s;
int pres = 0,mi=0;
for (int i = 0;i<=m-1;i++)
{
if (s[i]=='(')
pres++;
else
pres--;
mi = min(mi,pres);
}
dp[0][0] = 1;
for (int i = 1;i <= 2000;i++)
for (int j = 0;j <= i;j++)
{
if (j)
dp[i][j]=(dp[i][j]+dp[i-1][j-1])%MOD;//add "("
dp[i][j] = (dp[i][j]+dp[i-1][j+1]) % MOD;//add ")"
}
LL ans = 0;
for (int p = 0;p <= n-m;p++)
for (int j = 0;j <= p;j++)
if (mi+j>=0 && j+pres<=(n-m))
ans = (ans + dp[p][j]*dp[n-m-p][pres+j])%MOD;
cout << ans << endl;
return 0;
}
