B. Text Document Analysis

题目连接:

http://codeforces.com/contest/723/problem/B

Description

Modern text editors usually show some information regarding the document being edited. For example, the number of words, the number of pages, or the number of characters.

In this problem you should implement the similar functionality.

You are given a string which only consists of:

uppercase and lowercase English letters,
underscore symbols (they are used as separators),
parentheses (both opening and closing).

It is guaranteed that each opening parenthesis has a succeeding closing parenthesis. Similarly, each closing parentheses has a preceding opening parentheses matching it. For each pair of matching parentheses there are no other parenthesis between them. In other words, each parenthesis in the string belongs to a matching "opening-closing" pair, and such pairs can't be nested.

For example, the following string is valid: "_Hello_Vasya(and_Petya)_bye(and_OK)".

Word is a maximal sequence of consecutive letters, i.e. such sequence that the first character to the left and the first character to the right of it is an underscore, a parenthesis, or it just does not exist. For example, the string above consists of seven words: "Hello", "Vasya", "and", "Petya", "bye", "and" and "OK". Write a program that finds:

the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
the number of words inside the parentheses (print 0, if there is no word inside the parentheses).

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 255) — the length of the given string. The second line contains the string consisting of only lowercase and uppercase English letters, parentheses and underscore symbols.

Output

Print two space-separated integers:

the length of the longest word outside the parentheses (print 0, if there is no word outside the parentheses),
the number of words inside the parentheses (print 0, if there is no word inside the parentheses).

Sample Input

37

_Hello_Vasya(and_Petya)_bye(and_OK)

Sample Output

5 4

Hint

题意

问你括号外面的单词,最长的是多长。

括号内一共有多少个单词。

题解:

模拟一下就好了嘛,写的比较烦,可能。

代码

#include<bits/stdc++.h>
using namespace std; int n;
string s;
int solve(int l,int r)
{
int tmp = 0;
for(int i=l;i<=r;i++)
{
if(s[i]!='('&&s[i]!='_'&&s[i]!=')')
{
if(i==0||s[i-1]=='('||s[i-1]==')'||s[i-1]=='_')
tmp++;
}
}
return tmp;
}
int solve2(int l,int r)
{
int tmp = 0;
int ans = 0;
for(int i=l;i<=r;i++)
{
if(s[i]!='('&&s[i]!='_'&&s[i]!=')')tmp++;
else tmp=0;
ans=max(tmp,ans);
}
return ans;
}
int main()
{
cin>>n>>s;
int ans = 0;
int flag = 0;
for(int i=0;i<s.size();i++)
{
if(s[i]=='(')flag=i+1;
if(s[i]==')')ans+=solve(flag,i-1),flag=0;
}
int tmp = 0,ans2=0;
flag = 0;
for(int i=0;i<s.size();i++)
{
if(s[i]==')')flag=i+1;
else if(s[i]=='(')ans2=max(solve2(flag,i-1),ans2);
}
ans2=max(ans2,solve2(flag,s.size()-1));
cout<<ans2<<" "<<ans<<endl;
}

Codeforces Round #375 (Div. 2) B. Text Document Analysis 模拟的更多相关文章

  1. Codeforces Round #375 (Div. 2) A B C 水 模拟 贪心

    A. The New Year: Meeting Friends time limit per test 1 second memory limit per test 256 megabytes in ...

  2. Codeforces Round #375 (Div. 2)

    A. The New Year: Meeting Friends 水 #include <set> #include <map> #include <stack> ...

  3. Codeforces Round #375 (Div. 2) ABCDE

    A - The New Year: Meeting Friends 水 #include<iostream> #include<algorithm> using namespa ...

  4. Codeforces Round #367 (Div. 2) B. Interesting drink (模拟)

    Interesting drink 题目链接: http://codeforces.com/contest/706/problem/B Description Vasiliy likes to res ...

  5. Codeforces Round #375 (Div. 2) - D

    题目链接:http://codeforces.com/contest/723/problem/D 题意:给定n*m小大的字符矩阵.'*'表示陆地,'.'表示水域.然后湖的定义是:如果水域完全被陆地包围 ...

  6. Codeforces Round #375 (Div. 2) - C

    题目链接:http://codeforces.com/contest/723/problem/C 题意:给定长度为n的一个序列.还有一个m.现在可以改变序列的一些数.使得序列里面数字[1,m]出现次数 ...

  7. Codeforces Round #375 (Div. 2) - B

    题目链接:http://codeforces.com/contest/723/problem/B 题意:给定一个字符串.只包含_,大小写字母,左右括号(保证不会出现括号里面套括号的情况),_分隔开单词 ...

  8. Codeforces Round #375 (Div. 2) - A

    题目链接:http://codeforces.com/contest/723/problem/A 题意:在一维坐标下有3个人(坐标点).他们想选一个点使得他们3个到这个点的距离之和最小. 思路:水题. ...

  9. Codeforces Round #375 (Div. 2) F. st-Spanning Tree 生成树

    F. st-Spanning Tree 题目连接: http://codeforces.com/contest/723/problem/F Description You are given an u ...

随机推荐

  1. git push --set-upstream

    我在本地建了一个分支wangxiao,开发完之后,提交代码 git add .git commit -m '注释'git push 出现下面的问题,这个意思是:当前分支没有与远程分支关联. 因此导致了 ...

  2. pandas 实现通达信里的MFI

    pandas 实现通达信里的MFI 算法里的关键点: combine()和rolling().sum()方法 combine -- 综合运算, rolling().sum() -- 滚动求和 利用pd ...

  3. html5 canvas用图案填充形状

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...

  4. 第11月第20天 sqlite3_open xcode mysql connector

    1. sqlite3_open 死锁 * thread #1, queue = 'com.apple.main-thread', stop reason = signal SIGSTOP frame ...

  5. C - 青蛙的约会 POJ - 1061 (扩展欧几里得)

    题目链接:https://cn.vjudge.net/contest/276376#problem/C 题目大意:中文题目. 具体思路:扩展gcd,具体证明过程看图片(就这麽个题我搞了一天,,,). ...

  6. sync 解释

    sync命令用于强制被改变的内容立刻写入磁盘,更新超块信息. 在Linux/Unix系统中,在文件或数据处理过程中一般先放到内存缓冲区中,等到适当的时候再写入磁盘,以提高系统的运行效率.sync命令则 ...

  7. MAC连接安卓手机通过adb指令安装apk

    Android的apk可以通过adb命令来安装.在MAC电脑上,如果想通过命令行的方式给安卓手机安装apk,需要做以下操作: 一句话概括就是:将安卓SDK的adb命令添加到环境变量中,然后通过adb ...

  8. 重温CSS之文档结构

    我们来看看几个基本的HTML页面 <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "h ...

  9. android 8.0变更

    Android 8.0 行为变更 Android 8.0 除了提供诸多新特性和功能外,还对系统和 API 行为做出了各种变更.本文重点介绍您应该了解并在开发应用时加以考虑的一些主要变更. 其中大部分变 ...

  10. VS Code折腾记 - (4) 常用必备插件推荐【前端】

    前言 这篇文章只要让你做一些基础的配置,把vscode变得更加顺手: 插件的需求不是一成不变,有些插件我已经移除了..在最新的VSCODE 1.9.1中, 部分以前用插件实现的功能已经集成了,那就没有 ...