zoukankan      html  css  js  c++  java
  • 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 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;
    }
  • 相关阅读:
    python3中模块初识
    Django 应用程序 + 模型 + 基本数据访问
    Axure文本框验证和外部url的调用
    MATLAB 实时脚本(live-script)使用
    Django MTV 开发模式 + 数据库配置
    Django 模板继承
    Django 修改视图文件(views.py)并加载Django模块 + 利用render_to_response()简化加载模块 +locals()
    Django Context对象 + 过滤器 + 标签
    Axure 页面内多组内容切换的实现 + 利用一个内联框架实现百度地图访问
    MATLAB绘图功能(2) 二维底层绘图修饰
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5930082.html
Copyright © 2011-2022 走看看