题意:n,m,k,表示有一个长度为 n 的序列,有 m 个操作,操作有 2 种,第一种是 ADD 在前面添加一个串,第二种是把前 k 个进行翻转,问你最后的序列是什么样的。
析:很明显,如果直接模拟,肯定会超时,由于 k 是固定的,我们就可以前 k 个串,如果没有翻转,那么就把添加的串方法直接放到双端队列前面,然后把双端队列后面那个串再拿出去,如果翻转了,就把添加的串方法直接放到双端队列后面,然后把双端队列前面那个串再拿出去。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <sstream> #include <list> #include <assert.h> #include <bitset> #include <numeric> #define debug() puts("++++") #define gcd(a, b) __gcd(a, b) #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define fi first #define se second #define pb push_back #define sqr(x) ((x)*(x)) #define ms(a,b) memset(a, b, sizeof a) #define sz size() #define be begin() #define ed end() #define pu push_up #define pd push_down #define cl clear() #define lowbit(x) -x&x //#define all 1,n,1 #define FOR(i,n,x) for(int i = (x); i < (n); ++i) #define freopenr freopen("in.in", "r", stdin) #define freopenw freopen("out.out", "w", stdout) using namespace std; typedef long long LL; typedef unsigned long long ULL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const LL LNF = 1e17; const double inf = 1e20; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 10; const int maxm = 1e6 + 10; const LL mod = 1000000007; const int dr[] = {-1, 1, 0, 0, 1, 1, -1, -1}; const int dc[] = {0, 0, 1, -1, 1, -1, 1, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c) { return r >= 0 && r < n && c >= 0 && c < m; } inline int readInt(){ int x; scanf("%d", &x); return x; } vector<string> v; stack<string> st; deque<string> q; int main(){ int k; scanf("%d %d %d", &n, &m, &k); char s[30]; int cnt = 0; for(int i = 0; i < n; ++i){ scanf("%s", s); if(q.sz < k) q.push_back(s); else v.pb(s); } char t[30]; while(m--){ scanf("%s", t); if(t[0] == 'R') cnt ^= 1; else { t[strlen(t)-1] = 0; char *s = t + 4; if(cnt == 0){ q.push_front(s); if(q.sz > k){ st.push(q.back()); q.pop_back(); } } else{ q.push_back(s); if(q.sz > k){ st.push(q.front()); q.pop_front(); } } } } if(cnt) while(!q.empty()){ printf("%s\n", q.back().c_str()); q.pop_back(); } else while(!q.empty()){ printf("%s\n", q.front().c_str()); q.pop_front(); } while(!st.empty()){ printf("%s\n", st.top().c_str()); st.pop(); } for(int i = 0; i < v.sz; ++i) printf("%s\n", v[i].c_str()); return 0; }