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 程式師世界 >> 編程語言 >> C語言 >> C++ >> 關於C++ >> Uva11988 Broken Keyboard (a.k.a. Beiju Text)(就是先輸出括號的字符)

Uva11988 Broken Keyboard (a.k.a. Beiju Text)(就是先輸出括號的字符)

編輯:關於C++

Description

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Problem B

Broken Keyboard (a.k.a. Beiju Text)

You're typing a long text with a broken keyboard. Well it's not so badly broken. The only problem with the keyboard is that sometimes the "home" key or the "end" key gets automatically pressed (internally).

You're not aware of this issue, since you're focusing on the text and did not even turn on the monitor! After you finished typing, you can see a text on the screen (if you turn on the monitor).

In Chinese, we can call it Beiju. Your task is to find the Beiju text.

Input

There are several test cases. Each test case is a single line containing at least one and at most 100,000 letters, underscores and two special characters '[' and ']'. '[' means the "Home" key is pressed internally, and ']' means the "End" key is pressed internally. The input is terminated by end-of-file (EOF). The size of input file does not exceed 5MB.

Output

For each case, print the Beiju text on the screen.

Sample Input

This_is_a_[Beiju]_text
[[]][][]Happy_Birthday_to_Tsinghua_University

Output for the Sample Input

BeijuThis_is_a__text
Happy_Birthday_to_Tsinghua_University

Rujia Liu's Present 3: A Data Structure Contest Celebrating the 100th Anniversary of Tsinghua University
Special Thanks: Yiming Li
Note: Please make sure to test your program with the gift I/O files before submitting!



看到網上一種指針的做法,感覺太厲害了,每次cur指向當前的點,然後把需要輸出的字符插到cur的後面,當然這樣後面的[]裡面的東西會先輸出來

根據是 '[' 還是']‘改變 cur

具體自己好好想想


#include
#include
#include
#include
using namespace std;
#define N 100005

char a[N];
int nest[N];
char ans[N];

int main()
{
   int i,j,len;
   while(gets(a))
   {
   	  int k,cur,last;
   	  k=cur=last=0;
   	  nest[0]=-1;
   	  for(i=0;a[i];i++)
	  {
	  	if(a[i]=='[')
			cur=0;
		else
		  if(a[i]==']')
			cur=last;
		else
		{
			k++;
			ans[k]=a[i];
			nest[k]=nest[cur];
			nest[cur]=k;
			cur=k;
			
			if(nest[k]==-1) last=k;
		}
	  }
	  int t=nest[0];
	  while(t!=-1)
	  {
	  	printf("%c",ans[t]);
	  	t=nest[t];
	  }
	  printf("\n");
   }
   return 0;
}








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