题目：
ACboy needs your help again!
**Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6094 Accepted Submission(s): 3197
**
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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
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
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Sample Output
1
2
2
1
1
2
None
2
3
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Source
代码：
#include <stdio.h>
#include <stack>
#include <cstring>
#include <queue>
using namespace std;
int main()
{
int t;
scanf("%d",&t);//数据的组数
while(t--)
{
int n;
queue<int> q;//先声明一个队列和一个栈
stack<int> s;
char str[10];
scanf("%d %s",&n,str);
if(strcmp(str,"FIFO")==0)//判断输入要求的是队列
{
while(n--)//指令的组数
{
char zl[10];
int x;
scanf("%s",zl);
if(strcmp(zl,"IN")==0)
{
scanf("%d",&x);
q.push(x);//如果是输入的指令，就把输入的值加入队列
}
else if(strcmp(zl,"OUT")==0)//如果是出的指令
{
if(!q.empty())//当队列非空时
{
printf("%d\n",q.front());//打印队首，并抛弃队首元素
q.pop();
}
else
{
printf("None\n");
}
}
}
}
else if(strcmp(str,"FILO")==0)//判断输入要求的是栈
{
while(n--)//命令的组数
{
char zl[10];
int x;
scanf("%s",zl);
if(strcmp(zl,"IN")==0)
{
scanf("%d",&x);
s.push(x);//如果是输入的指令，就把输入的值加入栈
}
else if(strcmp(zl,"OUT")==0)
{
if(!s.empty())//当栈非空时
{
printf("%d\n",s.top());//打印栈顶元素，并出栈
s.pop();
}
else
{
printf("None\n");
}
}
}
}
}
return 0;
}
PS:队列和栈的一道很简单的练习题，帮助理解概念
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