栈与队列的pop和push

本文介绍了一个编程挑战,玩家需要通过解决一系列使用FIFO(先进先出)和FILO(先进后出)数据结构的问题来营救虚构角色ACboy。文章提供了一段C++代码示例,展示了如何使用队列和栈实现这些操作。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

E - ACboy needs your help again!
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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
 
 1 #include<cstdio>
 2 #include<string>
 3 #include<stack>
 4 #include<queue>
 5 #include<iostream>
 6 using namespace std;
 7 
 8 int main ()
 9 {
10     int t,n,a;
11     string s, s1;
12     scanf("%d",&t);
13     while(t--)
14     {
15         cin >> n >> s;
16         if(s == "FIFO")
17         {
18             queue<int>que;
19             while(n--)
20             {
21                 cin >> s1;
22                 if(s1 == "IN")
23                 {
24                     scanf("%d",&a);
25                     que.push(a);
26                 }
27                 else if(s1 == "OUT")
28                 {
29                     if(!que.empty())
30                     {
31                         printf("%d\n",que.front());
32                          que.pop();
33                     }
34                     else printf("None\n");
35                 }
36             }
37         }
38         else if(s == "FILO")
39         {
40             stack<int>st;
41             while(n--)
42             {
43                 cin >> s1;
44                 if(s1 == "IN")
45                 {
46                     scanf("%d",&a);
47                     st.push(a);
48                 }
49                 else if(s1 == "OUT")
50                 {
51                     if(!st.empty())
52                     {
53                         printf("%d\n",st.top());
54                         st.pop();
55                     }
56                     else printf("None\n");
57                 }
58             }
59         }
60     }
61     return 0;
62 }

 

转载于:https://www.cnblogs.com/LK1994/archive/2013/03/22/2976335.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值