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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
∙ 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.
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.
Invalid.“(without quotes). (Please see the sample for more details.)
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.
#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:\n", 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.\n");
}
else if(deq.empty()) {
int ans = (righ - lef - 1) % 2;
printf("%d\n", 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\n", tmp % 2);
}
else {
int tmp = deq.front();
//cout << tmp <<" !flag "<< lef <<endl;
tmp = tmp - lef - 1 + (deq.front() != righ - 1);
printf("%d\n", tmp % 2);
}
}
}
}
}
return 0;
}
转载于:https://www.cnblogs.com/Hyouka/p/7351251.html
发布者：全栈程序员-站长，转载请注明出处：https://javaforall.net/108242.html原文链接：https://javaforall.net
