sgu 302. BHTML 1.0 栈

最新推荐文章于 2022-07-09 16:40:01 发布

原创最新推荐文章于 2022-07-09 16:40:01 发布 · 181 阅读

0 ·

CC 4.0 BY-SA版权

sgu !!!!!!!!!! 同时被 2 个专栏收录

50 篇文章

订阅专栏

栈

1 篇文章

订阅专栏

本文介绍了一个简单的BHTML1.0解析器实现，该解析器能够正确处理UP和DOWN标签，使文本中的字母按标签要求转换为大写或小写。通过使用C++标准库中的栈来跟踪标签状态，确保了文本处理的准确性。

302. BHTML 1.0

Time limit per test: 0.25 second(s)
Memory limit: 65536 kilobytes

input: standard
output: standard

The hypertext markup language BHTML 1.0 has only two paired tags. They are

/UP

and

DOWN

/DOWN

. The

/UP

tag capitalizes all letters inside its body (between an open tag and a close one), and

DOWN

/DOWN

makes all inside the body letters lowercase. You are given the text consisting of latin letters and tags. Your task is to write the text right as it will be shown in the Bernet Explorer browser window. Tags in the text are arranged correctly, i.e. they form correct bracket sequence. If a letter lays inside several tags, its case is defined by the most inner tag.

Input

The input contains the string S with the text. The length of the string is a natural number not exceeding 1000. Tags are always written in uppercase.

Output

Write to the output text after the processing.

Example(s)

sample input	sample output
Thi<UP>sIs<DOWN>EaSY</DOWN>Pr<UP>O</UP>ble</UP>m	ThiSISeasyPROBLEm

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
char up(char c){
    if('a'<=c&&c<='z')return c-'a'+'A';
    return c;
}
char down(char c){
    if('A'<=c&&c<='Z')return c-'A'+'a';
    return c;
}
char s[1005];
int main(){
    scanf("%s",s);
    stack<int> st;
    int len=strlen(s);
    for(int i=0;i<len;i++){
        if(s[i]=='<'){
            if(s[i+1]=='/'){
                st.pop();
                if(s[i+1+1]=='U'){
                    i+=4;
                }
                else{
                    i+=6;
                }
            }
            else{
                if(s[i+1]=='U'){
                    st.push(1);
                    i+=3;
                }
                else{
                    st.push(2);
                    i+=5;
                }
            }
        }
        else{
            if(st.empty()){
                printf("%c",s[i]);
            }
            else{
                if(st.top()==1){
                    printf("%c",up(s[i]));
                }
                else{
                    printf("%c",down(s[i]));
                }
            }
        }

    }
    return 0;
}