Book Pile （双端队列）

There is a pile of N books on the table. Two types of operations are performed over this pile:
- a book is added to the top of the pile,
- top K books are rotated. If there are less than K books on the table, the whole pile is rotated.
First operation is denoted as ADD(S) where S is the name of the book, and the second operations is denoted as ROTATE.
The maximum number of books is no more than 40000. All book names are non-empty sequences of no more than 3 capital Latin letters. The names of the books can be non-unique.

Input
The first line of input file contains 3 integer numbers N, M, K (0 <= N <= 40000; 0 <= M <= 100000; 0 <= K <= 40000). The following N lines are the names of the books in the pile before performing any operations. The book names are given in order from top book to bottom. Each of the following M lines contains the operation description.

Output
Output the sequence of books names in the pile after performing all operations. First line corresponds to the top book.

Sample test(s)

Input

2 3 2
A
B
ADD(C)
ROTATE
ADD(D)

Output

D
A
C
B

按照题目要求大模拟就一定会超时，由于只动了前面的字符，所以可以选择双端队列。
*注意，字符可能有多个。

#include <iostream>
#include <cstring>
#include <deque>
#include <cstdlib>
#include <cstdio>
#define INF 0x3f3f3f3f
using namespace std;
deque<string>ans1, ans2;
int main()
{
    int n, m, k;
    ios::sync_with_stdio(false);
    cin>>n>>m>>k;
    for(int i=0; i<n; i++)
    {
        string x;
        cin>>x;
        if(i<k)
            ans1.push_back(x);
        else
            ans2.push_back(x);//hou
    }
    int flag = 0;
    while(m--)
    {
        string c;
        cin>>c;
        if(c[0]=='A')
        {
            string ka=c.substr(4, c.size()-5);//注意这里，名字可能不是一个字符
            if(flag==0)
            {
                ans1.push_front(ka);
                if(ans1.size()>k)
                {
                    ans2.push_front(ans1.back());
                    ans1.pop_back();
                }
            }
            else
            {
                ans1.push_back(ka);
                if(ans1.size()>k)
                {
                    ans2.push_front(ans1.front());
                    ans1.pop_front();
                }
            }
        }
        else
        {
            if(!flag)
                flag = 1;
            else
                flag = 0;
        }
    }
    if(flag==0)
    {
        while(!ans1.empty())
        {
            cout<<ans1.front()<<endl;
            ans1.pop_front();
        }
    }
    else
    {
        while(!ans1.empty())
        {
            cout<<ans1.back()<<endl;
            ans1.pop_back();
        }
    }
    while(!ans2.empty())
    {
        cout<<ans2.front()<<endl;
        ans2.pop_front();
    }
    return 0;
}