ACboy needs your help again!

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16823    Accepted Submission(s): 8254

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
 
这个题的思路其实也很简单,就是模拟栈和队列,来实现对应的操作,主要是为了熟悉queue的使用
#include <iostream>
#include <queue>
#include <stack>
#include <string> using namespace std; int main()
{
int t,n,temp;
cin>>t;
while(t--){
string str,str1;
queue<int>Q;
stack<int>S;
cin>>n>>str;
for(int i = ;i < n; i++){
if(str=="FIFO"){
cin>>str1;
if(str1=="IN"){
cin>>temp;
Q.push(temp);
}
if(str1=="OUT"){
if(Q.empty())
cout<<"None"<<endl;
else{
cout<<Q.front()<<endl;
Q.pop();
}
}
}
else{
cin>>str1;
if(str1=="IN"){
cin>>temp;
S.push(temp);
}
if(str1=="OUT"){
if(S.empty())
cout<<"None"<<endl;
else{
cout<<S.top()<<endl;
S.pop();
}
}
}
}
}
return ;
}

queue的使用-Hdu 1702的更多相关文章

  1. hdu 1702 ACboy needs your help again!

    题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=1702 ACboy needs your help again! Description ACboy w ...

  2. HDU - 1702 ACboy needs your help again!(栈和队列)

    Description ACboy was kidnapped!! he miss his mother very much and is very scare now.You can't image ...

  3. HDU 1702 http://acm.hdu.edu.cn/showproblem.php?pid=1702

    #include<stdio.h> #include<string.h> #include<queue> #include<stack> #define ...

  4. HDU 1702 队列与栈的简单运用http://acm.hdu.edu.cn/showproblem.php?pid=1702

    #include<stdio.h> #include<string.h> #include<queue> #include<stack> #define ...

  5. hdu 1702 栈和队列的简单应用

    #include<stdio.h> #include<string.h> #include<queue> #include<stack> using n ...

  6. HDU题解索引

    HDU 1000 A + B Problem  I/O HDU 1001 Sum Problem  数学 HDU 1002 A + B Problem II  高精度加法 HDU 1003 Maxsu ...

  7. 第一周训练 | STL和基本数据结构

    A - 圆桌问题: HDU - 4841 #include<iostream> #include<vector> #include<stdio.h> #includ ...

  8. hdu 栈题1022&1702

    http://acm.hdu.edu.cn/showproblem.php?pid=1022 http://blog.csdn.net/swm8023/article/details/6902426此 ...

  9. hdu 1387(Team Queue) STL

    Team Queue Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total ...

随机推荐

  1. jave的安装

    1.此电脑-属性-高级系统设置-环境变量2.点下面那个 新建-  JAVA_HOME3. 双击PATH变量,新建一个参数 4.新建CLASSPATH环境变量

  2. rest_framework:版本控制

    一.作用: 用于版本的控制,区分不同的功能,对于一些新增功能的添加更新就会在新的版本中体现,有点类似于,新版本就是软件发布的新版本,和老版本相比新增或者修改了一些功能 二.内置的版本控制类: from ...

  3. windows redis启动

    1.下载redis 2.启动redis 3.启动redis客户端并设置protected-mode为false

  4. 汉语诗词 LaTeX 排版样式

    清世何须忧庙廊——汉语诗词 LaTeX 排版样式 作者想一些中国古典诗歌,发现大多数早期的例子都是为了英文诗而创作的环境. 下面是作者给出唐诗选集的布局实例. 它不是一般解决方案,而只是一个特定的例子 ...

  5. AST抽象语法树——最基础的javascript重点知识,99%的人根本不了解

    AST抽象语法树——最基础的javascript重点知识,99%的人根本不了解 javascriptvue-clicommonjswebpackast  阅读约 27 分钟 抽象语法树(AST),是一 ...

  6. android 获取webview内容真实高度(webview上下可滚动距离)

    正常获取: mainWebView.getContentHeight()//获取html高度 mainWebView.getScale()//手机上网页缩放比例 mainWebView.getHeig ...

  7. Java-POJ1003-Hangover

    题目大意: 给出一个浮点数a,求出使得 不等式 1/2 + 1/3 + ... + 1/(n+1) ≥ a 成立的最小值 大水题,由于数据范围小,给出了确认上界5.20,满足二分答案 但是我懒啊,直接 ...

  8. apache http server 和tomcat的区别 以及nginx

    Apache HTTP Server(简称Apache)是Apache软件基金会的一个开放源码的网页服务器. 要明白他们之间的区别,我们首先需要明白HTTP协议.HTML页面.JSP.Servlet之 ...

  9. java.lang.NoClassDefFoundError: org/apache/commons/lang/exception/NestableRuntimeException

    运行项目时出现java.lang.NoClassDefFoundError: org/apache/commons/lang/exception/NestableRuntimeException时,只 ...

  10. Ehab and a Special Coloring Problem

    You're given an integer nn. For every integer ii from 22 to nn, assign a positive integer aiai such ...