(hdu step 8.1.1)ACboy needs your help again!(STL中栈和队列的基本使用)

题目：

ACboy needs your help again!

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 73 Accepted Submission(s): 57

Problem Description 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

Source 2007省赛集训队练习赛（1）

Recommend lcy

题目分析：

栈和队列的基本使用，简单题。

事实上出题人的意思可能是让我们自己手写一个栈和队列。可是，作为一个早就知道STL的渣渣来说，是没有耐心再去写stack和queue了。。

。哎哎。。

代码例如以下：

/*
 * a.cpp
 * 栈和队列的模拟
 *
 *  Created on: 2015年3月19日
 *      Author: Administrator
 */
#include <iostream>
#include <cstdio>
#include <stack>
#include <queue>

using namespace std;

int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		int n;
		string type;
		cin >> n >> type;

		if(type == "FIFO"){
			queue<int> q;
			string cmd;
			int num;

			int i;
			for(i = 0 ; i < n ; ++i){
				cin >> cmd;

				if(cmd == "IN"){
					cin >> num;
					q.push(num);
				}else{
					if(q.empty() == true){
						printf("None\n");
					}else{
						int ans = q.front();
						q.pop();
						printf("%d\n",ans);
					}
				}
			}

		}else{
			stack<int> st;
			string cmd;
			int  num;

			int i;
			for(i = 0 ; i < n ; ++i){
				cin >> cmd;

				if(cmd == "IN"){
					cin >> num;
					st.push(num);
				}else{
					if(st.empty() == true){
						printf("None\n");
					}else{
						int ans = st.top();
						st.pop();
						printf("%d\n",ans);
					}
				}
			}
		}
	}

	return 0;
}