Basic Data Structure
Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
∙ 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 is corresponding to the element of the Stack from top to the bottom, value=atop nand atop−1 nand ... nand 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.
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 (x must be 0 or 1)
∙ POP
∙ REVERSE
∙ QUERY
It is guaranteed that the current stack will not be empty while doing POP operation.
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.)
8
PUSH 1
QUERY
PUSH 0
REVERSE
QUERY
POP
POP
QUERY
3
PUSH 0
REVERSE
QUERY
1
1
Invalid.
Case #2:
0
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<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=4e5+,M=4e6+,inf=1e9+,mod=1e9+;
const ll INF=1e18+;
char ch[];
int main()
{
int T,cas=;
scanf("%d",&T);
while(T--)
{
printf("Case #%d:\n",cas++);
int st=,en=,f=;
int n;
scanf("%d",&n);
set<int>s;
set<int>::iterator it;
for(int i=;i<=n;i++)
{
scanf("%s",ch);
if(ch[]=='P')
{
if(ch[]=='U')
{
int x;
scanf("%d",&x);
if(x==)
s.insert(en);
if(f)
en--;
else
en++;
}
else
{
if(f)
{
if(!s.empty())
{
it=s.begin();
if(*it<=en+)
s.erase(it);
}
en++;
}
else
{
if(!s.empty())
{
it=s.end();
it--;
if(*it>=en-)
s.erase(it);
}
en--;
}
}
}
else if(ch[]=='Q')
{
if(st==en)
{
printf("Invalid.\n");
}
else if(s.empty())
{
printf("%d\n",abs((en-st)%));
}
else if(f)
{
int p;
it=s.end();
it--;
p=*it;
int ans=st-p;
if(p>en+)ans++;
printf("%d\n",ans%);
}
else
{
int p;
it=s.begin();
p=*it;
int ans=p-st;
if(p<en-)ans++;
printf("%d\n",ans%);
}
}
else
{
if(f)
{
int temp=st;
st=en+;
en=temp+;
f=;
}
else
{
int temp=st;
st=en-;
en=temp-;
f=;
}
}
}
}
return ;
}