HDU Basic Data Structure 2016CCPC东北地区大学生程序设计竞赛 - 重现赛

Basic Data Structure
Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 28    Accepted Submission(s): 0

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 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.

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 (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
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.

序列最后一个值为0时

ans=0        //序列长度为1
ans=1        //other

所以 用一个双向队列记录序列里0的下标

query时

当序列是正向时 找到序列中最右边的一个0的下标 此时序列该下标右边全是1

例如

1011 0 111->1 111

0 1111 -> 1 111

此时的值会不断地0,1,0,1,0循环 判断1的个数即可

反向也的同理

#include<iostream>
#include<stdlib.h>
#include<stdio.h>
#include<string>
#include<vector>
#include<deque>
#in
4000
clude<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<time.h>
#include<math.h>
#include<list>
#include<cstring>
#include<fstream>
//#include<memory.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define pii pair<int,int>
#define INF 1000000007
#define pll pair<ll,ll>
#define pid pair<int,double>
//#define CHECK_TIME

const int N=200000+3;

struct S{
int elem[2*N];
bool reverse;
int st,end;
deque<int>index_of_0;
S(){
reverse=false;
st=end=N;
}
void push(int x){
if(reverse){
elem[--st]=x;
if(x==0)
index_of_0.push_front(st);
}
else{
elem[end]=x;
if(x==0)
index_of_0.push_back(end);
++end;
}
}
void pop(){
if(reverse){
if(index_of_0.front()==st)
index_of_0.pop_front();
++st;
}
else{
if(index_of_0.back()==end-1)
index_of_0.pop_back();
--end;
}
}

void reverseS(){
reverse=false==reverse;
}
bool empty(){
return end==st;
}
int size(){
return end-st;
}
int query(){
if(empty())
return -1;
if(size()==1)
return elem[st];
if(index_of_0.empty())
return size()&1;
if(!reverse){//注意方向问题 在此WA 4次....
if(index_of_0.front()!=end-1)
return (index_of_0.front()-st+1)&1;
else
return (index_of_0.front()-st)&1;
}
else{
if(index_of_0.back()!=st)
return (end-index_of_0.back())&1;
else
return ((end-index_of_0.back())&1)==0;
}
}
void clear(){
st=end=N;
index_of_0.clear();
reverse=false;
}
};

int main()
{
//freopen("/home/lu/文档/r.txt","r",stdin);
//freopen("/home/lu/文档/w.txt","w",stdout);
int T,n,x;
char in[6];
S mystack;
scanf("%d",&T);
for(int t=1;t<=T;++t){
scanf("%d",&n);
printf("Case #%d:\n",t);
mystack.clear();
for(int i=0;i<n;++i){
scanf("%s",in);
if(in[2]=='S'){
scanf("%d",&x);
mystack.push(x);
}
else
if(in[2]=='P')
mystack.pop();
else
if(in[2]=='V')
mystack.reverseS();
else
if(in[2]=='E'){
x=mystack.query();
if(x<0)
printf("Invalid.\n");
else
printf("%d\n",x);
}
}
}
return 0;
}