Basic Data Structure
Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 78 Accepted Submission(s): 12
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.
题意：维护一个栈，支持往栈里塞 0/1 ，弹栈顶，翻转栈，询问从栈顶到栈底按顺序 NAND 的值。
题解：只要知道最后的 0后面 1的个数的奇偶性就行。
我这里是用set 存储0的位置 比较麻烦 也可以和存储0/1一样 维护一个 0的位置 的栈
、
1 /******************************
2 code by drizzle
3 blog: www.cnblogs.com/hsd-/
4 ^ ^ ^ ^
5 O O
6 ******************************/
7 #include<bits/stdc++.h>
8 #include<map>
9 #include<set>
10 #include<cmath>
11 #include<queue>
12 #include<bitset>
13 #include<math.h>
14 #include<vector>
15 #include<string>
16 #include<stdio.h>
17 #include<cstring>
18 #include<iostream>
19 #include<algorithm>
20 #pragma comment(linker, "/STACK:102400000,102400000")
21 using namespace std;
22 #define A first
23 #define B second
24 const int mod=1000000007;
25 const int MOD1=1000000007;
26 const int MOD2=1000000009;
27 const double EPS=0.00000001;
28 //typedef long long ll;
29 typedef __int64 ll;
30 const ll MOD=1000000007;
31 const int INF=1000000010;
32 const ll MAX=1ll<<55;
33 const double eps=1e-5;
34 const double inf=~0u>>1;
35 const double pi=acos(-1.0);
36 typedef double db;
37 typedef unsigned int uint;
38 typedef unsigned long long ull;
39 int t;
40 char str[10];
41 map<int,int> mp;
42 int l,r;
43 int l0,r0;
44 int exm;
45 int flag=0;
46 int n;
47 set<int>se;
48 set<int>::iterator it;
49 int biao=0;
50 int main()
51 {
52 scanf("%d",&t);
53 {
54 for(int k=1; k<=t; k++)
55 {
56 se.clear();
57 mp.clear();
58 biao=0;
59 scanf("%d",&n);
60 l=r=0;
61 printf("Case #%d:\n",k);
62 for(int i=1; i<=n; i++)
63 {
64 scanf("%s",str);
65 if(strcmp(str,"PUSH")==0)
66 {
67 scanf("%d",&exm);
68 if(exm==0)
69 se.insert(l);
70
71 mp[l]=exm;
72 if(biao==0)
73 l++;
74 else
75 l--;
76 }
77 if(strcmp(str,"POP")==0)
78 {
79 if(biao==0)
80 l--;
81 else
82 l++;
83 }
84 if(strcmp(str,"REVERSE")==0)
85 {
86 if(biao==0)
87 {
88 l--;
89 r--;
90 swap(l,r);
91 biao=1;
92 }
93 else
94 {
95 l++;
96 r++;
97 swap(l,r);
98 biao=0;
99 }
100 }
101 if(strcmp(str,"QUERY")==0)
102 {
103 if(l==r)
104 printf("Invalid.\n");
105 else
106 {
107 if(biao==0)
108 {
109 int exm;
110 int gg=0;
111 for(it=se.begin(); it!=se.end(); it++)
112 {
113 if(*it>=r)
114 {
115 exm=*it;
116 gg=1;
117 break;
118 }
119 }
120 if(gg==0)
121 {
122 if((l-r)%2==0)
123 printf("0\n");
124 else
125 printf("1\n");
126 }
127 else
128 {
129 if(l==r+1)
130 printf("%d\n",mp[exm]);
131 else
132 {
133 if(exm==(l-1))
134 exm--;
135 if((exm-r+1)%2)
136 printf("1\n");
137 else
138 printf("0\n");
139 }
140 }
141 }
142 else
143 {
144 int exm;
145 int gg=0;
146 if(se.size()!=0)
147 {
148 it=--se.end();
149 for(;; it--)
150 {
151 if(*it<=r)
152 {
153 exm=*it;
154 gg=1;
155 break;
156 }
157 if(it==se.begin())
158 break;
159 }
160 }
161 if(gg==0)
162 {
163 if((r-l)%2==0)
164 printf("0\n");
165 else
166 printf("1\n");
167 }
168 else
169 {
170 int hhh=1;
171 if(l==r-1)
172 printf("%d\n",mp[exm]);
173 else
174 {
175 if(exm==(l+1))
176 exm++;
177 if((r-exm+1)%2)
178 printf("1\n");
179 else
180 printf("0\n");
181 }
182 }
183 }
184 }
185 }
186 }
187 }
188 }
189 return 0;
190 }
191 /*
192 2
193 6
194 PUSH 1
195 PUSH 1
196 PUSH 1
197 PUSH 1
198 REVERSE
199 QUERY
200 5
201 PUSH 1
202 PUSH 1
203 PUSH 1
204 PUSH 1
205 QUERY
206 */