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).

Example

Input
37
_Hello_Vasya(and_Petya)__bye_(and_OK)
Output
5 4
Input
37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__
Output
2 6
Input
27
(LoooonG)__shOrt__(LoooonG)
Output
5 2
Input
5
(___)
Output
0 0

Note

In the first sample, the words "Hello", "Vasya" and "bye" are outside any of the parentheses, and the words "and", "Petya", "and" and "OK" are inside. Note, that the word "and" is given twice and you should count it twice in the answer.

 
给你一个长度为n的字符串,求出括号里面单词的数量和括号外面最长单词的长度 
即便是暴力模拟 还是很需要掌握技巧的
 
 #include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cctype>
using namespace std; int main() {
int n;
char a[];
while(scanf("%d%s",&n,a)!=EOF){
int sum=,maxn=,len=,flag=;
for (int i= ;i<n ;i++){
if (a[i]=='(') flag=;
if (a[i]==')') flag=;
if (isalpha(a[i])) len++;
else len=;
if (flag==) maxn=max(maxn,len);
else if (len==) sum++;
}
printf("%d %d\n",maxn,sum);
}
return ;
}
 

Text Document Analysis CodeForces - 723B的更多相关文章

  1. Codefoces 723B Text Document Analysis

    B. Text Document Analysis time limit per test:1 second memory limit per test:256 megabytes input:sta ...

  2. Codeforces Round #375 (Div. 2) B. Text Document Analysis 模拟

    B. Text Document Analysis 题目连接: http://codeforces.com/contest/723/problem/B Description Modern text ...

  3. codeforces 723B:Text Document Analysis

    Description Modern text editors usually show some information regarding the document being edited. F ...

  4. 【44.10%】【codeforces 723B】Text Document Analysis

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

  5. 【Codeforces 723B】Text Document Analysis 模拟

    求括号外最长单词长度,和括号里单词个数. 有限状态自动机处理一下. http://codeforces.com/problemset/problem/723/B Examples input 37_H ...

  6. codeforces 723B Text Document Analysis(字符串模拟,)

    题目链接:http://codeforces.com/problemset/problem/723/B 题目大意: 输入n,给出n个字符的字符串,字符串由 英文字母(大小写都包括). 下划线'_' . ...

  7. CodeForces 723B Text Document Analysis (水题模拟)

    题意:给定一行字符串,让你统计在括号外最长的单词和在括号内的单词数. 析:直接模拟,注意一下在左右括号的时候有没有单词.碰到下划线或者括号表示单词结束了. 代码如下: #pragma comment( ...

  8. cf723b Text Document Analysis

    Modern text editors usually show some information regarding the document being edited. For example, ...

  9. 【Codeforces】Round #375 (Div. 2)

    Position:http://codeforces.com/contest/723 我的情况 啊哈哈,这次raiting肯定要涨,接受过上次的教训,先用小号送肉,大号都是一发切,重回蓝咯 结果... ...

随机推荐

  1. CF 2015 ICL, Finals, Div. 1 J. Ceizenpok’s formula [Lucas定理]

    http://codeforces.com/gym/100633/problem/J Lucas定理P不是质数裸题 #include <iostream> #include <cst ...

  2. HDU 4333 Revolving Digits [扩展KMP]【学习笔记】

    题意:给一个数字,每一次把它的最后一位拿到最前面,一直那样下去,分别求形成的数字小于,等于和大于原来数的个数. SAM乱搞失败 当然要先变SS了 然后考虑每个后缀前长为n个字符,把它跟S比较就行了 如 ...

  3. 《Web Scraping With Python》Chapter 2的学习笔记

    You Don't Always Need a Hammer When Michelangelo was asked how he could sculpt a work of art as mast ...

  4. 树莓派上运行.net core 2.0程序

    记录中 参考: https://www.cnblogs.com/songxingzhu/p/7399991.html https://www.cnblogs.com/goodfulcom/p/7624 ...

  5. Angular20 nginx安装,angular项目部署

    1 nginx安装(Windows版本) 1.1 下载安装包 到官网下载Windows版本的nginx安装包 技巧01:下载好的压缩包解压即可,无需安装 1.2 启动nginx 进入到解压目录,点击 ...

  6. [Python Study Notes]匿名函数

    Python 使用 lambda 来创建匿名函数. lambda这个名称来自于LISP,而LISP则是从lambda calculus(一种符号逻辑形式)取这个名称的.在Python中,lambda作 ...

  7. 如何在Centos 7上用Logrotate管理日志文件

    何为Logrotate? Logrotate是一个实用的日志管理工具,旨在简化对系统上生成大量的日志文件进行管理. Logrotate允许自动旋转压缩,删除和邮寄日志文件,从而节省宝贵的磁盘空间. L ...

  8. 使用scp从远程服务器下载文件到本地

    [下载远程文件到本地] scp -P 6008 root@192.168.1.123:/usr/data/1.zip   /Users/abc/www [上传本地文件到远程] scp -P 6008  ...

  9. 原生js总结(干货)

    1.js基本数据类型 number string boolean underfined null 2.查找文档中的特定元素 document.getElementById("id" ...

  10. java程序员理解js中的闭包

    1.闭包概念: 就是函数内部通过某种方式访问一个函数内部的局部变量 再次理解: 闭包产生原因: 1.内部函数引用了外部函数的变量 作用:延长局部变量的生命周期 让函数外部可以调用到函数内部的数据 利用 ...