首页 > 代码库 > 2016CCPC东北地区大学生程序设计竞赛1008/HDU 5929 模拟

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,atop1,?,a1 is corresponding to the element of the Stack from top to the bottom, value=atop nand atop1 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 (T20), which indicates the number of test cases.

For each test case, the first line contains only one integers N (2N200000), 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                     else106                     {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                                 else125                                     printf("1\n");126                             }127                             else128                             {129                                 if(l==r+1)130                                     printf("%d\n",mp[exm]);131                                 else132                                 {133                                     if(exm==(l-1))134                                         exm--;135                                     if((exm-r+1)%2)136                                         printf("1\n");137                                     else138                                         printf("0\n");139                                 }140                             }141                         }142                         else143                         {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                                 else166                                     printf("1\n");167                             }168                             else169                             {170                                 int hhh=1;171                                 if(l==r-1)172                                     printf("%d\n",mp[exm]);173                                 else174                                 {175                                     if(exm==(l+1))176                                         exm++;177                                     if((r-exm+1)%2)178                                         printf("1\n");179                                     else180                                         printf("0\n");181                                 }182                             }183                         }184                     }185                 }186             }187         }188     }189     return 0;190 }191 /*192 2193 6194 PUSH 1195 PUSH 1196 PUSH 1197 PUSH 1198 REVERSE199 QUERY200 5201 PUSH 1202 PUSH 1203 PUSH 1204 PUSH 1205 QUERY206 */

 

2016CCPC东北地区大学生程序设计竞赛1008/HDU 5929 模拟