HDU 1702 FIFO FILO

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


ACboy needs your help again!

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5806    Accepted Submission(s): 3055


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


数据结构栈和队列水题


//hdu 1702 FIFO,FILO#include <iostream>#include <cstring>#include <string>#include <queue>#include <stack>using namespace std;int main(){int n;cin >> n;while (n--){int t;string data_struct;cin >> t;cin >> data_struct;if (data_struct == "FIFO"){queue<int> q;string str;int num;while (t--){cin >> str;if (str == "IN"){cin >> num;q.push(num);}else{if (q.empty()){puts("None");}else{printf("%d\n", q.front());q.pop();}}}}else if (data_struct == "FILO"){stack<int> s;string str;int num;while (t--){cin >> str;if (str == "IN"){cin >> num;s.push(num);}else{if (s.empty()){puts("None");}else{printf("%d\n", s.top());s.pop();}}}}}system("pause");return 0;}

//发现加不加system("pause")  对结果的影响不大



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