CodeForces - 723B - B. Text Document Analysis

怼烎@ 2022-06-12 07:20 72阅读 0赞

--------------------

题目连接：[http://codeforces.com/problemset/problem/723/B][http_codeforces.com_problemset_problem_723_B]

## 题目描述 ##

### 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).

现代文本编辑器通常具备一些文本统计功能，例如单词数、字符数、行数等等。

现在给你一段文本，请你完成一些简单的文本统计功能

文本是一行字符串，由下列元素组成：

*  英文字母，大小写均可
 *  下划线，表示分隔符
 *  小括号
 *  文本保证括号是完全匹配的，且不含有嵌套括号

例如这个字符串是一个合法的文本 “*Hello\_Vasya(and\_Petya)\_\_bye*(and\_OK)”.

可以在里面找到7个单词（单词即连续的字母）： “Hello”, “Vasya”, “and”, “Petya”, “bye”, “and” 和 “OK”.试统计:

*  括号外的最长单词长度 (若括号外没有单词输出0),
 *  括号内的单词总数.

### 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.

第一行一个整数 n (1 ≤ n ≤ 255).

第二行一个字符串s，长度为n。

### Output ###

Print two space-separated integers:

输出一行两个整数，用空格隔开，分别为括号外的最长单词长度和括号内单词的总数。

### Sample Input ###

37  
*Hello\_Vasya(and\_Petya)\_\_bye*(and\_OK)

### Sample Output ###

5 4

### Sample Input ###

37  
*a*(*b\_\_\_c)**de\_f(g***)\_\_h\_\_i(j\_k\_l)m

### Sample Output ###

2 6

### Sample Input ###

27  
(LoooonG)**shOrt**(LoooonG)

### Sample Output ###

5 2

### Sample Input ###

5  
(*\_*)

### Sample Output ###

0 0

## 解题思路 ##

主要是利用好的思路写好求位置的算法，首先过滤字母，利用一个参数记录过滤的个数，  
当到达不是字母的时候，判断前面是否是字母，如果是，那么肯定存在一个单词，然后比较单词的长度，记录最长的，其次是用一个 flag 标记是否在括号内，如果在括号内只需记录个数即可，  
要注意的是标记的 flag 要更新， 还有记录单词长度的 len 实时更新。

## AC代码 ##

#include<iostream>
    #include<algorithm>
    using namespace std;
    int main () {
        int max = 0;
        int len = 0;
        int num = 0;
        int n;
        int flag = 1;
        char str[260];
        cin >> n;
        scanf("%s", str);
        for(int i = 0; i <= n; i++) {
            if(!isalpha(str[i])) {    //判断是否为字母
                if(flag == 1 && max < len) max = len;
                else if(flag == 0 && len != 0) num++;
                len = 0;
                if(str[i] == '(') flag = 0;
                else if(str[i] == ')') flag = 1;
            }
            else len++;
        }
        cout << max << " " << num << endl;
        return 0;
    }

[http_codeforces.com_problemset_problem_723_B]: http://codeforces.com/problemset/problem/723/B