hdu1022（stack）

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.

Input

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.

Output

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.

Sample Input

3 123 3213 123 312

Sample Output

Yes.inininoutoutoutFINISHNo.FINISH          For the first Sample Input, we let train 1 get in, then train 2 and train 3.So now train 3 is at the top of the railway, so train 3 can leave first, then train 2 and train 1.In the second Sample input, we should let train 3 leave first, so we have to let train 1 get in, then train 2 and train 3.Now we can let train 3 leave.But after that we can't let train 1 leave before train 2, because train 2 is at the top of the railway at the moment.So we output "No.".

/*题意：给你两个数字串，每个串的长度已知，然后让你判断能否按照栈的方式将第一个串按照第二个串的方式输出

思路：用栈将第一个串读入，每读入一个字符做一次判断判断当前栈顶元素是否为第二个串的n--个元素且保证栈不为空，通过计数法若将第二个串的整个长度遍历说明yes，否则no，至于in，out的输出用一个标记数组即可

*/

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <stack>using namespace std;char a[50];char b[50];int c[50];int main(){    int n;    while(scanf("%d", &n) != EOF){        getchar();        scanf("%s", a);        getchar();        scanf("%s", b);        memset(c, -1, sizeof(c));        stack<char>q;        int k = 0;        int j = 0;        for(int i = 0; i < n; i++){            q.push(a[i]);            c[k++] = 1;            while(!q.empty() && q.top() == b[j]){                c[k++] = 0;                q.pop();                j++;            }        }        if(j != n){            printf("No.\nFINISH\n");        }        else{            printf("Yes.\n");            for(int i = 0; i < 2*n; i++){                if(c[i] == 1) printf("in\n");                if(c[i] == 0) printf("out\n");            }            printf("FINISH\n");        }    }    return 0;}