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 |
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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! |
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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. |
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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.
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Sample Input
4 |
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Sample Output
1 |
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Source
2007省赛集训队练习赛（1）
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Recommend
lcy
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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;
}