HDU5929-Basic Data Structure
2017-07-03 09:29:34         来源：wang_128的博客

# Basic Data Structure

Time Limit: 7000/3500 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2057Accepted Submission(s): 451

Problem Description 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 doNANDoperation one by one from left to right, i.e. Ifatop,atop?1,?,a1is corresponding to the element of the Stack from top to the bottom,value=atopnandatop?1nand ... nanda1. Note that the Stackwill notchange after QUERY operation. Specially, if the Stack is empty now，you need to print ”Invalid.”(without quotes).

By the way,NANDis 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 The first line contains only one integer T (T≤20), which indicates the number of test cases.

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

Hint

In the first sample： during the first query, the stack contains only one element 1, so the answer is 1. then in the second query, the stack contains 0, l (from bottom to top), so the answer to the second is also 1. In the third query, there is no element in the stack, so you should output Invalid.

```#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include

using namespace std;
const int INF = 0x3f3f3f3f3f;
#define LL long long

int main()
{
int t, cas = 0;
scanf("%d", &t);
while (t--)
{
int n;
scanf("%d", &n);
dequeq;
char ch;
int x, flag = 0;
printf("Case #%d:\n", ++cas);
for (int i = 1;i <= n;i++)
{
scanf("%s", ch);
if (!strcmp(ch, "PUSH"))
{
scanf("%d", &x);
if (!flag)
{
if (x == 0) q.push_back(x);
else
{
if (!q.empty() && q.back()!= 0)
{
int k = q.back() + 1;
q.pop_back();
q.push_back(k);
}
else q.push_back(x);
}
}
else
{
if (x == 0) q.push_front(x);
else
{
if (!q.empty() && q.front() != 0)
{
int k = q.front() + 1;
q.pop_front();
q.push_front(k);
}
else q.push_front(x);
}
}
}
else if (!strcmp(ch, "POP"))
{
if (!flag)
{
if (q.back() == 0) q.pop_back();
else
{
int k = q.back()-1;
q.pop_back();
if(k>0) q.push_back(k);
}
}
else
{
if (q.front() == 0) q.pop_front();
else
{
int k = q.front() - 1;
q.pop_front();
if (k>0) q.push_front(k);
}
}
}
else if (!strcmp(ch, "REVERSE")) flag ^= 1;
else
{
if (q.empty()) printf("Invalid.\n");
else
{
if (flag)
{
if (q.back() == 0 && q.size() > 1) printf("1\n");
else if (q.back() == 0) printf("0\n");
else if (q.size() <= 2)
{
if (q.back() % 2 == 1) printf("1\n");
else printf("0\n");
}
else
{
if (q.back() % 2 == 1) printf("0\n");
else printf("1\n");
}
}
else
{
if (q.front() == 0 && q.size() > 1)
printf("1\n");
else if (q.front() == 0)
printf("0\n");
else if (q.size() <= 2)
{
if (q.front() % 2 == 1) printf("1\n");
else printf("0\n");
}
else
{
if (q.front() % 2 == 1) printf("0\n");
else printf("1\n");
}
}
}
}
}
}
return 0;
}```