STL练习-ACboy needs your help again!

 

ACboy was kidnapped!!
he miss his mother very much and is very scare now.You can't image how dark the room he was put into is, so poor :(.

As a smart ACMer, you want to get ACboy out of the monster's labyrinth.But when you arrive at the gate of the maze, the monste say :" I have heard that you are very clever, but if can't solve my problems,

you will die with ACboy."

The problems of the monster is shown on the wall:

Each problem's first line is a integer N(the number of commands), and a word "FIFO" or "FILO".(you are very happy because you know "FIFO" stands for "First In First Out", and "FILO" means "First In Last Out").

and the following N lines, each line is "IN M" or "OUT", (M represent a integer).

and the answer of a problem is a passowrd of a door, so if you want to rescue ACboy, answer the problem carefully!

 

Input

The input contains multiple test cases. The first line has one integer,represent the number oftest cases.And the input of each subproblem are described above . 

 

Output

For each command "OUT", you should output a integer depend on the word is "FIFO" or "FILO", or a word "None" if you don't have any integer.

 

Sample Input

4
4 FIFO
IN 1
IN 2
OUT
OUT
4 FILO
IN 1
IN 2
OUT
OUT
5 FIFO
IN 1
IN 2
OUT
OUT
OUT
5 FILO
IN 1
IN 2
OUT
IN 3
OUT

Sample Output

1
2
2
1
1
2
None
2
3
分别定义一个栈和一个队列，输入FIFO即先入先出，操作队列，输入FILO即先入后出，操作栈。

 1 #include<iostream>
 2 #include<stack>
 3 #include<queue>
 4 using namespace std;
 5 int main()
 6 {
 7 stack<int> s;
 8 queue<int> q;
 9 int n;
10 cin>>n;
11   while(n--)
12   {
13    int m;
14    string s1;
15    cin>>m>>s1;
16   while(!s.empty())s.pop();//循环开始清空栈和队列
17   while(!q.empty())q.pop();
18    if(s1=="FIFO")//先入先出操作队列
19    {
20      for(int i=0;i<m;i++)
21      {
22       string s2;int num;
23       cin>>s2;
24       if(s2=="IN"){cin>>num;q.push(num);}//入队
25       if(s2=="OUT"){
26                    if(q.empty())cout<<"None"<<endl;//队列为空不能"OUT"，输出None
27                       else
28                       {
29                        cout<<q.front()<<endl;//q.front()取队头值
30                        q.pop();//无返回值，借助q.front()获得队头值
31                       }
32                    }
33      }
34    }
35    if(s1=="FILO")
36    {
37      for(int i=0;i<m;i++)
38      {
39       string s2;int num;
40       cin>>s2;
41       if(s2=="IN"){cin>>num;s.push(num);}//入栈
42       if(s2=="OUT"){
43                    if(s.empty())cout<<"None"<<endl;//栈为空时
44                       else
45                       {
46                        cout<<s.top()<<endl;s.pop();//s.pop()无返回值，借助s.top()获得栈顶元素
47                       }
48
49                    }
50      }
51    }
52
53   }
54   return 0;
55 }