Codeforces629 C. Famil Door and Brackets
2 seconds
256 megabytes
standard input
standard 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.
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.
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.
4 1
(
4
4 4
(())
1
4 3
(((
0
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.
思路:dp;
dp[i][j]表示前i个括号,开口向右减开口向左的值SS。方程:dp[i][j]=dp[i-1][j-1]+dp[i-1][j+1].在i的时候当前选向左开口或向右开口。
可以知道dp[0][0]=1;由于n-m<=2000;所以枚举p,因为最后要平衡所以q也可以知道了,并且q的取值也就是前面的dp,因为要和前面匹配,前面p+s结束时必定有向右的
括号数大于向左的括号数,所以在剩下的q只要取向左的括号数数-向右的括号数,与前面的匹配中和就行了,也就是将前面的dp看成在i时开口向右减开口向左的值SS,所以套用前面的dp就行了。 最后sum=((sum)%N+(dp[i][j]*dp[cc][ans])%N)%N;
1 #include<stdio.h>
2 #include<algorithm>
3 #include<iostream>
4 #include<queue>
5 #include<stdlib.h>
6 #include<string.h>
7 using namespace std;
8 typedef long long LL;
9 LL dp[2205][2205];
10 char str[100005];
11 int QZ[100005];
12 const LL N=1e9+7;
13 int main(void)
14 {
15 LL i,j,k,p,q;
16 dp[0][0]=1;
17 for(i=1;i<=2200;i++)
18 {
19 for(j=0;j<=i;j++)
20 {
21 if(j>0)
22 {
23 dp[i][j]=(dp[i][j]+dp[i-1][j-1])%N;
24 }
25 dp[i][j]=(dp[i][j]+dp[i-1][j+1])%N;
26 }
27 }
28 while(scanf("%I64d %I64d",&p,&q)!=EOF)
29 {memset(QZ,0,sizeof(QZ));
30 scanf("%s",str);LL meq=1e8;LL maxx;
31 for(i=0;i<q;i++)
32 {
33 if(str[i]=='(')
34 {
35 QZ[i+1]=QZ[i]+1;
36 }
37 else
38 {
39 QZ[i+1]=QZ[i]-1;
40 }
41 if(QZ[i+1]<meq)
42 {
43 meq=QZ[i+1];
44 }
45 }LL sum=0;if(meq<=0){maxx=-meq;}
46 else maxx=0;
47 for(i=maxx;i<=2000;i++)
48 {
49 for(j=maxx;j<=i;j++)
50 {
51 LL cc=p-q-i;
52 LL ans=j+QZ[q];
53 if(cc>=0&&ans<=cc)
54 {
55 sum=((sum)%N+(dp[i][j]*dp[cc][ans])%N)%N;
56 }
57 }
58 }
59 printf("%I64d\n",sum);
60 }
61 return 0;
62 }
Codeforces629 C. Famil Door and Brackets的更多相关文章
- 【Codeforces629C】Famil Door and Brackets [DP]
Famil Door and Brackets Time Limit: 20 Sec Memory Limit: 512 MB Description Input Output Sample Inp ...
- Codeforces Round #343 (Div. 2) C. Famil Door and Brackets dp
C. Famil Door and Brackets 题目连接: http://www.codeforces.com/contest/629/problem/C Description As Fami ...
- codeforces 629C Famil Door and Brackets (dp + 枚举)
题目链接: codeforces 629C Famil Door and Brackets 题目描述: 给出完整的括号序列长度n,现在给出一个序列s长度为m.枚举串p,q,使得p+s+q是合法的括号串 ...
- 【23.24%】【codeforces 629C】Famil Door and Brackets
time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...
- codeforces629C Famil Door and Brackets (dp)
As Famil Door's birthday is coming, some of his friends (like Gabi) decided to buy a present for him ...
- Codeforces 629C Famil Door and Brackets(DP)
题目大概说给一个长m的括号序列s,要在其前面和后面添加括号使其变为合法的长度n的括号序列,p+s+q,问有几种方式.(合法的括号序列当且仅当左括号总数等于右括号总数且任何一个前缀左括号数大于等于右括号 ...
- 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 ...
- Codeforces 629C Famil Door and Brackets DP
题意:给你一个由括号组成的字符串,长度为m,现在希望获得一个长度为n(全由括号组成)的字符串,0<=n-m<=2000 这个长度为n的字符串要求有两个性质:1:就是任意前缀,左括号数量大于 ...
- codeforces629C Famil Door and Brackets (dp)
题意:给你一个长度为n的括号匹配串(不一定恰好匹配),让你在这个串的前面加p串和后面加上q串,使得这个括号串平衡(平衡的含义是对于任意位置的括号前缀和大于等于0,且最后的前缀和为0). 思路:枚举这个 ...
随机推荐
- 11 — springboot集成swagger — 更新完毕
1.前言 理论知识滤过,自行百度百科swagger是什么 2.导入依赖 <!-- swagger所需要的依赖--> <dependency> <groupId>io ...
- 日常Java 2021/11/2
Java提供了一种对象序列化的机制,该机制中,一个对象可以被表示为一个字节序列,该字节序列包括该对象的数据.有关对象的类型的信息和存储在对象中数据的类型.将序列化对象写入文件之后,可以从文件中读取出来 ...
- day01 前端bootstrap框架
day01 django框架之bootstrap框架 今日内容概要 前端框架之bootstrap 该框架支持cv编写前端页面 利用socket模块编写一个简易版本的web框架 利用wsgiref模块编 ...
- 转 android开发笔记之handler+Runnable的一个巧妙应用
本文链接:https://blog.csdn.net/hfreeman2008/article/details/12118817 版权 1. 一个有趣Demo: (1)定义一个handler变量 pr ...
- ORACEL 创建DIRECTORY
oracle要直接对文件进行读写必须先创建一个DIRECTORY. 语法如下: CREATE DIRECTORY UTL_FILE_DIR AS '/home/oracle/oradir'; 可以通过 ...
- Linux安装软件出错
1.Delta RPMs disabled because /usr/bin/applydeltarpm not installed. yum provides '*/applydeltarpm' # ...
- IDE搬进浏览器里——JetBrains Projector
发展 提起 JetBrains,你会想到什么?各路强大的 IDE,比如 Android Studio.IDEA.WebStorm--这些对于开发者来说耳熟能详的产品都出自这家公司,这些 IDE 的功能 ...
- hash 模式与 history 模式小记
hash 模式 这里的 hash 就是指 url 后的 # 号以及后面的字符.比如说 "www.baidu.com/#hashhash" ,其中 "#hashhash&q ...
- 使用Lock接口来解决线程安全的问题
package cn.itcast.demo16.Demo09.Lock;import java.util.concurrent.locks.Lock;import java.util.concurr ...
- 使用Spring JDBC连接数据库(以SQL Server为例)
一.配置Spring JDBC 1.导入相关jar包 (略) 2.配置文件applicationContext.xml <?xml version="1.0" encodin ...