B. Text Document Analysis
time limit per test:1 second
memory limit per test:256 megabytes
input:standard input
output:standard output
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).
Examples
Input
代码语言：javascript复制37
_Hello_Vasya(and_Petya)__bye_(and_OK)Output
代码语言：javascript复制5 4Input
代码语言：javascript复制37
_a_(_b___c)__de_f(g_)__h__i(j_k_l)m__Output
代码语言：javascript复制2 6Input
代码语言：javascript复制27
(LoooonG)__shOrt__(LoooonG)Output
代码语言：javascript复制5 2Input
代码语言：javascript复制5
(___)Output
代码语言：javascript复制0 0Note
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.
题目链接：http://codeforces.com/problemset/problem/723/B
正解：模拟
解题报告：
直接模拟，注意处理字符串的计数器的诸多情况，有一点细节。
下面给出AC代码：
代码语言：javascript复制 1 #include <bits/stdc .h>
2 using namespace std;
3 int main()
4 {
5 char s[10000];
6 int n,len=0,s1 = 0,s2 = 0;
7 bool flag = 0;
8 while(scanf("%d%s",&n,&s)!=EOF)
9 {
10 //for(int i=0;i<n;i )
11 //scanf("%c",&s[i]);
12 for(int j = 0; j < n; j )
13 {
14 if(s[j] == '(') flag=1;
15 if(s[j] == ')') flag=0;
16 if((s[j] >= 'a' && s[j] <= 'z') || (s[j] >= 'A' && s[j] <= 'Z')) {
17 len ;
18 }
19 else len = 0;
20 if(!flag) s1=max(s1, len);
21 else if(len==1)s2 ;
22 }
23 printf("%d %dn", s1,s2);
24 }
25 return 0;
26 }
