A - I

A - I

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

Submit

Status

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

Sample Output

1

2

2

1

1

2

None

2

3

题意：栈与队列的基本运用。

思路：用栈和队列解决。

失误：没有清空栈和队列。

代码如下：

#include<iostream>
#include<cstdio>
#include<queue>
#include<stack>
#include<cstring>
#include<functional>
using namespace std;

int main()
{
int t,i,n,elem;
char str[39],str1[89],str2[99];
queue<int> qu;
stack<int> sta;
cin>>t;

while(t--)
{
cin>>n>>str;//输入字符串时越过空格读空格后的字符直到读到另一空格 然后在str后置'\0'.
//cin.getline可以读取空格，和gets的用法相似无限读取
for(i=1;i<=n;++i)
{
if(strcmp(str,"FIFO")==0)
{
cin>>str1;

if(strcmp(str1,"IN")==0)
{
cin>>elem;
qu.push(elem);
}

else
{
if(qu.empty())
cout<<"None"<<endl;
else
{
cout<<qu.front()<<endl;
qu.pop();
}
}

}

else
{
cin>>str2;

if(strcmp(str2,"IN")==0)
{
cin>>elem;
sta.push(elem);
}
else
{
if(sta.empty())
cout<<"None"<<endl;
else
{
cout<<sta.top()<<endl;
sta.pop();
}
}

}
}

while(!qu.empty())  //清空栈和队列
qu.pop();

while(!sta.empty())
sta.pop();
}
return 0;
}