ACM-C - ACboy needs your help again!

C - ACboy needs your help again!
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
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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 

#include<stdio.h>#include<string.h>#include<algorithm>#include<stack>#include<queue>using namespace std;int main(){int n;char e[5]="FIFO",w[5]="FILO";char in[3]="IN",out[4]="OUT";scanf("%d",&n);while(n--){    stack<int> s;    queue<int> q;    int z;char a[10];char tem[5];    int dl,zhan,i;    scanf("%d",&z); scanf("%s",&a);    if(strcmp(a,e)==0)    { // printf("%s",a);        for(i=0;i<z;i++){        scanf("%s",&tem);        if(strcmp(tem,in)==0)        {// printf("in");            scanf("%d",&dl);q.push(dl);        }        if(strcmp(tem,out)==0)        {  //printf("out\n");          if(q.empty()) printf("sorry\n");           else {printf("%d\n",q.front());q.pop();}        }        }    }    if(strcmp(a,w)==0)    {//printf("%s\n",a);    for(i=0;i<z;i++){        scanf("%s",&tem);     if(strcmp(tem,in)==0)        { //printf("in");            scanf("%d\n",&dl);s.push(dl);        }        if(strcmp(tem,out)==0)        {  //printf("out\n");          if(s.empty()) printf("None\n");           else {printf("%d\n",s.top());s.pop();}        }    }    }}return 0;}

比较麻烦的地方就是判断是进栈还是出栈，麻烦但是不难- -、

出现了两个小漏洞- -、empty的时候我又用了pop，出现re,非法读取内存，

然后把题目看错了，None我看成sorry了。