标签:invalid 位置 math include 操作 XML cst type structure
题意:给定一种二进制操作nand,为
0 nand 0 = 1
0 nand 1 = 1
1 nand 0 = 1
1 nand 1 = 0
现在要你模拟一个队列,实现PUSH x 往队头塞入x,POP队尾退出,REVERSE翻转,QUERY询问队头到队尾的nand和。
思路:其他都可以模拟,但是n为2e5,如果直接求nand和必超时,找规律可得,任何串和0nand都为1,那么我们维护一个双端队列保存0出现的位置,然后每次就计算离队尾最近的0和队尾间有几个1。
代码:
#include<set> #include<map> #include<stack> #include<cmath> #include<queue> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> typedef long long ll; using namespace std; const int maxn = 200000 + 100; const int seed = 131; const ll MOD = 109; const int INF = 0x3f3f3f3f; deque<int> q; //0 int main(){ int t, x, l, r, ca = 1, cnt; char order[10]; scanf("%d", &t); while(t--){ printf("Case #%d:\n", ca++); int n; int turn = 1; cnt = 0; l = 0, r = -1; while(!q.empty()) q.pop_back(); scanf("%d", &n); while(n--){ scanf("%s", order); if(order[2] == ‘S‘){ //push scanf("%d", &x); if(turn){ cnt++; r++; if(x == 0) q.push_back(r); } else{ cnt++; l--; if(x == 0) q.push_front(l); } } else if(order[2] == ‘P‘){ //pop if(turn){ if(!q.empty() && q.back() == r) q.pop_back(); r--; } else{ if(!q.empty() && q.front() == l) q.pop_front(); l++; } cnt--; } else if(order[2] == ‘V‘){ //reverse turn ^= 1; } else{ //query int ans; if(cnt == 0) printf("Invalid.\n"); else{ if(q.empty()) ans = (r - l + 1) % 2; else{ if(turn) ans = (q.front() - l + 1 - (q.front() == r)) % 2; else ans = (r - q.back() + 1 - (q.back() == l)) % 2; } printf("%d\n", ans); } } } } return 0; }
HDU 5929 Basic Data Structure(模拟 + 乱搞)题解
标签:invalid 位置 math include 操作 XML cst type structure
原文地址:https://www.cnblogs.com/KirinSB/p/10051292.html