HDU 5929 Basic Data Structure

Mr. Frog learned a basic data structure recently, which is called stack.There are some basic operations of stack： 

∙

∙ PUSH x： put x on the top of the stack, x must be 0 or 1. 

∙

∙ POP： throw the element which is on the top of the stack. 

Since it is too simple for Mr. Frog, a famous mathematician who can prove "Five points coexist with a circle" easily, he comes up with some exciting operations： 

∙

∙REVERSE： Just reverse the stack, the bottom element becomes the top element of the stack, and the element just above the bottom element becomes the element just below the top elements... and so on. 

∙

∙QUERY： Print the value which is obtained with such way： Take the element from top to bottom, then do 

NAND operation one by one from left to right, i.e. If  

atop,atop−1,⋯,a1

atop,atop−1,⋯,a1 is corresponding to the element of the Stack from top to the bottom, 

value=atop

value=atop nand 

atop−1

atop−1 nand ... nand 

a1

a1. Note that the Stack 

will not change after QUERY operation. Specially, if the Stack is empty now，you need to print ” 

Invalid.”(without quotes). 

By the way, 

NAND is a basic binary operation： 

∙

∙ 0 nand 0 = 1 

∙

∙ 0 nand 1 = 1 

∙

∙ 1 nand 0 = 1 

∙

∙ 1 nand 1 = 0 

Because Mr. Frog needs to do some tiny contributions now, you should help him finish this data structure： print the answer to each QUERY, or tell him that is invalid. 

Input

T≤20

T≤20), which indicates the number of test cases. 

For each test case, the first line contains only one integers N (

2≤N≤200000

2≤N≤200000), indicating the number of operations. 

In the following N lines, the i-th line contains one of these operations below： 

∙

∙ PUSH x (x 

must be 0 or 1) 

∙

∙ POP 

∙

∙ REVERSE 

∙

∙ QUERY 

It is guaranteed that the current stack will not be empty while doing POP operation.

Output

For each test case, first output one line "Case #x：w, where x is the case number (starting from 1). Then several lines follow,  i-th line contains an integer indicating the answer to the i-th QUERY operation. Specially, if the i-th QUERY is invalid, just print " 

Invalid."(without quotes). (Please see the sample for more details.) 

Sample Input

2
8
PUSH 1
QUERY
PUSH 0
REVERSE
QUERY
POP
POP
QUERY
3
PUSH 0
REVERSE
QUERY      

Sample Output

Case #1:
1
1
Invalid.
Case #2:
0      

#include<cstdio>
#include<queue>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 1e5 + 10;
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int M = 3e5;
const int low(int x) { return x&-x; }
int T, n, x;
char s[N];
int a[N<<3];

int main(){
  int cas = 1;
  for (scanf("%d",&T);T--;cas++) {
    scanf("%d",&n);
    int l = M, r= M - 1,now = 1;
    printf("Case #%d:\n",cas);
    deque<int> b;
    while (n--) {
      scanf("%s",s);
      if (!strcmp(s,"PUSH")) {
        scanf("%d",&x);
        if (now) {
          a[++r] = x;
          if (!x) b.push_back(r);
        }
        else {
          a[--l] = x;
          if (!x) b.push_front(l);
        }
      }
      else if (!strcmp(s,"REVERSE")) {
        now^=1;
      }
      else if (!strcmp(s,"POP")) {
        if (now) {
          if (!a[r]) {
            b.pop_back();
          }
          --r;
        }
        else {
          if (!a[l]) {
            b.pop_front();
          }
          ++l;
        }
      }
      else {
        if (l>r) {
          printf("Invalid.\n");
        }
        else if (b.empty()) {
          printf("%d\n",(r-l+1)%2);
        }
        else if (now == 0) {
          if (b.back() == l) printf("%d\n",(r-l)%2);
          else printf("%d\n",(r-b.back()+1)%2);
        }
        else {
          if (b.front() == r) printf("%d\n",(r-l)%2);
          else printf("%d\n",(b.front()-l+1)%2);
        }
      }
    }
  }
  return 0;
}