B - Ada and Queue

本篇介绍了一个关于AdaLadybug的问题,她需要管理一个双向队列,执行多种操作如添加元素到队首或队尾、反转队列、从队首或队尾移除并打印元素等。通过一个标记来跟踪队列的方向变化。

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Ada the Ladybug has many things to do. She puts them into her queue. Anyway she is very indecisive, so sometime she uses the top, sometime the back and sometime she decides to reverses it.

Input

The first line consists of 1 ≤ Q ≤ 106, number of queries. Each of them contains one of following commands

back - Print number from back and then erase it

front - Print number from front and then erase it

reverse - Reverses all elements in queue

push_back N - Add element N to back

toFront N - Put element N to front

All numbers will be 0 ≤ N ≤ 100

Output

For each back/front query print appropriate number.

If you would get this type of query and the queue would be empty, print "No job for Ada?" instead.

Example Input

15
toFront 93
front
back
reverse
back
reverse
toFront 80
push_back 53
push_back 50
front
front
reverse
push_back 66
reverse
front

Example Output

93
No job for Ada?
No job for Ada?
80
53

66

解析:双向队列模拟;主要是reverse,定义一个flag标记,对头队尾即可;

#include<bits/stdc++.h>
using namespace std;
int Q;

char s[15];
int num;
deque<int>q;
deque<int>qq;

deque<int>::iterator it;

int main()
{
    int ans=0;
    scanf("%d",&Q);
    for(int i=0;i<Q;i++)
    {
        scanf("%s",s);
        if(s[0]=='t')
        {
            scanf("%d",&num);
            if(ans%2==0)
            q.push_front(num);
            else
                q.push_back(num);
        }
        else if(s[0]=='p')
        {
            scanf("%d",&num);
            if(ans%2==0)
            q.push_back(num);
            else
                q.push_front(num);
        }
        else if(s[0]=='f')
        {
            if(q.empty())
            {
                puts("No job for Ada?");
            }
            else
            {
                if(ans%2==0)
                {
                     int top=q.front();
                q.pop_front();
                printf("%d\n",top);
                }
                else
                {
                     int tail=q.back();
                q.pop_back();
                printf("%d\n",tail);
                }

            }
        }
        else if(s[0]=='b')
        {
             if(q.empty())
            {
                puts("No job for Ada?");
            }
            else
            {
                if(ans%2==0)
                {
                    int tail=q.back();
                q.pop_back();
                printf("%d\n",tail);
                }
                else
                {
                     int top=q.front();
                q.pop_front();
                printf("%d\n",top);
                }
            }
        }
        else
        {
            ans++;
        }

    }
}

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