HDU 5929 Basic Data Structure 【模拟】 （2016CCPC东北地区大学生程序设计竞赛）

Basic Data Structure

Time Limit: 7000/3500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 872 Accepted Submission(s): 236


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.)

[align=left]Sample Input[/align]

2
8
PUSH 1
QUERY
PUSH 0
REVERSE
QUERY
POP
POP
QUERY
3
PUSH 0
REVERSE
QUERY

[align=left]Sample Output[/align]

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.

[align=left]Source[/align]
2016CCPC东北地区大学生程序设计竞赛 - 重现赛

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1 //
2 //by coolxxx
3 //#include<bits/stdc++.h>
4 #include<iostream>
5 #include<algorithm>
6 #include<string>
7 #include<iomanip>
8 #include<map>
9 #include<stack>
10 #include<queue>
11 #include<set>
12 #include<bitset>
13 #include<memory.h>
14 #include<time.h>
15 #include<stdio.h>
16 #include<stdlib.h>
17 #include<string.h>
18 //#include<stdbool.h>
19 #include<math.h>
20 #pragma comment(linker,"/STACK:1024000000,1024000000")
21 #define min(a,b) ((a)<(b)?(a):(b))
22 #define max(a,b) ((a)>(b)?(a):(b))
23 #define abs(a) ((a)>0?(a):(-(a)))
24 #define lowbit(a) (a&(-a))
25 #define sqr(a) ((a)*(a))
26 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
27 #define mem(a,b) memset(a,b,sizeof(a))
28 #define eps (1e-10)
29 #define J 10000
30 #define mod 1000000007
31 #define MAX 0x7f7f7f7f
32 #define PI 3.14159265358979323
33 #define N 100004
34 using namespace std;
35 typedef long long LL;
36 double anss;
37 LL aans;
38 int cas,cass;
39 int n,m,lll,ans;
40 int b,t;
41 int a[N*3][2];
42 char ch[10];
43 void PUSH()
44 {
45     int i,j,x;
46     scanf("%d",&x);
47     n++;
48     if(n==1)
49         a[++t][0]=x,a[t][1]=1;
50     else if(a[t][0]==x)
51         a[t][1]++;
52     else if(b>t)
53         a[--t][0]=x,a[t][1]=1;
54     else
55         a[++t][0]=x,a[t][1]=1;
56 }
57 void POP()
58 {
59     if(!n)return;
60     n--;
61     if(--a[t][1]==0)
62     {
63         if(b>t)t++;
64         else t--;
65     }
66 }
67 void REVERSE()
68 {
69     if(!n)return;
70     swap(b,t);
71 }
72 void QUERY()
73 {
74     if(!n)puts("Invalid.");
75     else if(n==1)
76         printf("%d\n",a[b][0]);
77     else
78     {
79         if(a[b][0])
80         {
81             if(n<=a[b][1]+1)printf("%d\n",a[b][1]&1);
82             else printf("%d\n",a[b][1]&1^1);
83         }
84         else puts("1");
85     }
86 }
87 int main()
88 {
89     #ifndef ONLINE_JUDGEW
90 //    freopen("1.txt","r",stdin);
91 //    freopen("2.txt","w",stdout);
92     #endif
93     int i,j,k;
94     int x,y,z;
95 //    init();
96 //    for(scanf("%d",&cass);cass;cass--)
97     for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
98 //    while(~scanf("%s",s))
99 //    while(~scanf("%d%d",&n,&m))
100     {
101         printf("Case #%d:\n",cass);
102         scanf("%d",&m);
103         n=0;
104         b=N;t=N-1;
105         for(k=1;k<=m;k++)
106         {
107             scanf("%s",ch);
108             if(ch[2]=='S')PUSH();
109             else if(ch[2]=='P')POP();
110             else if(ch[2]=='V')REVERSE();
111             else if(ch[2]=='E')QUERY();
112         }
113     }
114     return 0;
115 }
116 /*
117 //
118
119 //
120 */