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 44 FIFOIN 1IN 2OUTOUT4 FILOIN 1IN 2OUTOUT5 FIFOIN 1IN 2OUTOUTOUT5 FILOIN 1IN 2OUTIN 3OUT |
| Sample Output 122112None23 |
| 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#include #include #include using namespace std;int main(){ int t; scanf("%d",&t); while(t--){ int n; string type; cin >> n >> type; if(type == "FIFO"){ queue 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 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;}