原题：
As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world ^v^). But here comes a problem, there is only one railway where all the trains stop. So all the trains come in from one side and get out from the other side. For this problem, if train A gets into the railway first, and then train B gets into the railway before train A leaves, train A can't leave until train B leaves. The pictures below figure out the problem. Now the problem for you is, there are at most 9 trains in the station, all the trains has an ID(numbered from 1 to n), the trains get into the railway in an order O1, your task is to determine whether the trains can get out in an order O2.
题意：
进栈问题，给出原排列顺序和要出栈的顺序，问能不能得到这个出栈顺序，如果可以得到就输出进出栈顺序。
题解：
就是模拟一下进栈出栈过程，因为给出的序列中没有相同的数，所以只要栈顶元素和要求的序列元素相同就直接出栈，不相同就接着进栈。最后判断一下栈是否为空就可以了。
代码：AC
#include<stack>
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
int step[100000];
int main()
{
stack<int> s;
int n;
char A[200];
char C[200];
while(cin>>n>>A>>C)
{
while(!s.empty())
s.pop();
int i=0,j=0,k=0;
for(i=0;i<n;i++)
{
s.push(A[i]);
step[j]=1;
j++;
while(!s.empty()&&s.top()==C[k])
{
k++;
step[j]=0;
s.pop();
j++;
}
}
if(!s.empty())
cout<<"No."<<endl;
else
{
cout<<"Yes."<<endl;
for(i=0;i<j;i++)
{
if(step[i])
cout<<"in"<<endl;
else
cout<<"out"<<endl;
}
}
cout<<"FINISH"<<endl;
}
return 0;
}