2016CCPC东北地区大学生程序设计竞赛1008/HDU 5929 模拟

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
感谢评论区 hack数据
1
10
PUSH 0
REVERSE
QUERY
PUSH 1
PUSH 1
REVERSE
POP
QUERY
POP
QUERY

  代码中已经标记更改

  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:
",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.
");
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(exm>=l)/*评论区hack点 更正 */
121                                 gg=0;
122                             if(gg==0)
123                             {
124                                 if((l-r)%2==0)
125                                     printf("0
");
126                                 else
127                                     printf("1
");
128                             }
129                             else
130                             {
131                                 if(l==r+1)
132                                     printf("%d
",mp[exm]);
133                                 else
134                                 {
135                                     if(exm==(l-1))
136                                         exm--;
137                                     if((exm-r+1)%2)
138                                         printf("1
");
139                                     else
140                                         printf("0
");
141                                 }
142                             }
143                         }
144                         else
145                         {
146                             int exm;
147                             int gg=0;
148                             if(se.size()!=0)
149                             {
150                                 it=--se.end();
151                                 for(;; it--)
152                                 {
153                                     if(*it<=r)
154                                     {
155                                         exm=*it;
156                                         gg=1;
157                                         break;
158                                     }
159                                     if(it==se.begin())
160                                         break;
161                                 }
162                             }
163                             if(exm<=l)/*评论区hack点 更正*/
164                                 gg=0;
165                             if(gg==0)
166                             {
167                                 if((r-l)%2==0)
168                                     printf("0
");
169                                 else
170                                     printf("1
");
171                             }
172                             else
173                             {
174                                 int hhh=1;
175                                 if(l==r-1)
176                                     printf("%d
",mp[exm]);
177                                 else
178                                 {
179                                     if(exm==(l+1))
180                                         exm++;
181                                     if((r-exm+1)%2)
182                                         printf("1
");
183                                     else
184                                         printf("0
");
185                                 }
186                             }
187                         }
188                     }
189                 }
190             }
191         }
192     }
193     return 0;
194 }
195 /*
196 2
197 6
198 PUSH 1
199 PUSH 1
200 PUSH 1
201 PUSH 1
202 REVERSE
203 QUERY
204 5
205 PUSH 1
206 PUSH 1
207 PUSH 1
208 PUSH 1
209 QUERY
210 */
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原文地址：https://www.cnblogs.com/hsd-/p/5934206.html