Train Problem I
时间限制：3000 ms | 内存限制：65535 KB
难度：2
描述
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.
输入
The input contains several test cases. Each test case consists of an integer, the number of trains, and two strings, the order of the trains come in:O1, and the order of the trains leave:O2. The input is terminated by the end of file. More details in the Sample Input.
输出
The output contains a string "No." if you can't exchange O2 to O1, or you should output a line contains "Yes.", and then output your way in exchanging the order(you should output "in" for a train getting into the railway, and "out" for a train getting out of the railway). Print a line contains "FINISH" after each test case. More details in the Sample Output.
样例输入
3 123 321
3 123 312
样例输出
Yes.
in
in
in
out
out
out
FINISH
No.
FINISH
题意：给出两个字符串，问第一个字符串能否经过栈转化为第二个字符串。 如果可以，输出进栈和出栈的顺序。
这是一个关于栈的问题。栈是允许在同一端进行插入和删除操作的特殊线性表。允许进行插入和删除操作的一端称为栈顶(top)，另一端为栈底(bottom)；栈底固定，而栈顶浮动；栈中元素个数为零时称为空栈。插入一般称为进栈（PUSH），删除则称为退栈（POP）。栈也称为后进先出表。
[cpp #include
#include
#include
using namespace std; /*千万不要忘记*/
int main()
{
int n,i,j,k,flag[20];
char s1[10],s2[10];
stack
while(~scanf("%d%s %s",&n,s1,s2))
{
memset(flag,-1,sizeof(flag));
j=k=0;
for(i=0;i
s.push(s1[i]);
flag[k++]=1; /*1为进栈，in*/
while(!s.empty()&&s.top()==s2[j]) /*栈非空且栈顶元素与s2中的元素相等*/
{
flag[k++]=0; /*0为出栈，out*/
s.pop(); /*清除栈顶元素*/
j++; /*与s2中的下一个元素比较*/
}
}
if(j==n) /*可以转换*/
{
printf("Yes.\n");
for(i=0;i
if(flag[i]) /*flag[i]==1*/
printf("in\n");
else
printf("out\n");
}
printf("FINISH\n");
}
else /*不能转换*/
printf("No.\nFINISH\n");
}
return 0;
}
#include
#include
#include
using namespace std; /*千万不要忘记*/
int main()
{
int n,i,j,k,flag[20];
char s1[10],s2[10];
stack
while(~scanf("%d%s %s",&n,s1,s2))
{
memset(flag,-1,sizeof(flag));
j=k=0;
for(i=0;i
s.push(s1[i]);
flag[k++]=1; /*1为进栈，in*/
while(!s.empty()&&s.top()==s2[j]) /*栈非空且栈顶元素与s2中的元素相等*/
{
flag[k++]=0; /*0为出栈，out*/
s.pop(); /*清除栈顶元素*/
j++; /*与s2中的下一个元素比较*/
}
}
if(j==n) /*可以转换*/
{
printf("Yes.\n");
for(i=0;i
if(flag[i]) /*flag[i]==1*/
printf("in\n");
else
printf("out\n");
}
printf("FINISH\n");
}
else /*不能转换*/
printf("No.\nFINISH\n");
}
return 0;
}