(hdu step 8.1.1)ACboy needs your help again!(STL中棧和隊列的基本使用)，hduacboy

(hdu step 8.1.1)ACboy needs your help again!(STL中棧和隊列的基本使用)，hduacboy

題目：

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 DescriptionACboy 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!

InputThe 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

Source2007省賽集訓隊練習賽（1）

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題目分析：

              棧和隊列的基本使用，簡單題。其實出題人的意思可能是讓我們自己手寫一個棧和隊列。但是，作為一個早就知道STL的渣渣來說，是沒有耐心再去寫stack和queue了。。。哎哎。。

代碼如下：

/* * a.cpp * 棧和隊列的類比 * *  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;}