HDU 5929 Basic Data Structure 模拟

Basic Data Structure

Time Limit: 7000/3500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2200    Accepted Submission(s): 477

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 notchange 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.

 

Source

2016CCPC东北地区大学生程序设计竞赛 - 重现赛

 

 

双端队列加数组模拟

查询的时候找到最后一个零的位置可以直接推出答案

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN     freopen("input.txt","r",stdin);
#define FOUT    freopen("output.txt","w",stdout);
#define INF     0x3f3f3f3f
#define INFLL   0x3f3f3f3f3f3f3f
#define lson    l,m,rt<<1
#define rson    m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int, int> PII;
using namespace std;

int p[600000];

int lef = 300000, righ = 300001;

int main() {
    //FIN
    int T;
    int cnt = 1;
    scanf("%d", &T);
    while(T--) {
        int n;
        deque<int> deq;
        lef = 300000;
        righ = 300001;
        char op[10];
        printf("Case #%d:
", cnt++);
        scanf("%d", &n);
        int flag = 1;
        while(n--) {
            scanf("%s", op);
            int num;
            if(op[0] == 'P' && op[1] == 'U') {
                scanf("%d", &num);
                if(flag) {
                    if(num == 0) deq.push_front(lef);
                    p[lef] = num;
                    lef--;
                }
                else {
                    if(num == 0) deq.push_back(righ);
                    p[righ] = num;
                    righ++;
                }
            } else if(op[0] == 'P' && op[1] == 'O') {
                if(flag) {
                    lef++;
                    if(p[lef] == 0) deq.pop_front();
                }
                else {
                    righ--;
                    if(p[righ] == 0) deq.pop_back();
                }
            } else if(op[0] == 'R') {
                if(flag) flag = 0;
                else flag = 1;
            } else if(op[0] == 'Q') {
                if(righ - 1 == lef) {
                    printf("Invalid.
");
                }
                else if(deq.empty()) {
                    int ans = (righ - lef - 1) % 2;
                    printf("%d
", ans);
                }
                else {
                    if(flag) {
                        int tmp = deq.back();
                        //cout << tmp <<"  flag "<<righ - tmp - 1<<endl;
                        tmp = righ - tmp - 1 + (deq.back() != lef + 1);
                        printf("%d
", tmp % 2);
                    }
                    else {
                        int tmp = deq.front();
                        //cout << tmp <<" !flag "<< lef <<endl;
                        tmp = tmp - lef - 1 + (deq.front() != righ - 1);
                        printf("%d
", tmp % 2);
                    }


                }


            }
        }


    }
    return 0;
}