牛客红包OI赛 B 小可爱序列

Description

连接：https://ac.nowcoder.com/acm/contest/224/B
来源：牛客网

”我愿意舍弃一切，以想念你，终此一辈子。“
”到后来，只能将记忆拼凑。“ ——QAQ
小可爱刚刚把KR的序列切开了，可是她尚未玩够，因而就又双叒叕打乱了佳佳刚刚买回来的序列。
可是还好，佳佳经过监控记录下来了小可爱的打乱方式，因而把小可爱送回家以后，如今佳佳要还原这个序列。
佳佳须要维护一个长度为n的序列，小可爱只用了如下两种操做：
a.将最后一个数挪到第一位
b.将序列第3位挪到第一位
你须要给出最后的序列

Solution

某人跟我说是链表
而后就没读题无脑作
而后TLE以后发现不能无脑作

而后有脑一下

Code

#include <stdlib.h>
#include <stdio.h>
#include <string.h>

struct Node {
    int v; Node *nxt, *pre;
    Node(int _v, Node *_n = nullptr, Node *_p = nullptr) :
        v(_v), nxt(_n), pre(_p) { }
} *head, *tail;

void Delete(const Node* b, const Node* e = nullptr) {
    if (not e) e = b;
    b->pre->nxt = e->nxt;
    e->nxt->pre = b->pre;
}
void InsertNxt(Node* p, Node* b, Node* e = nullptr) {
    if (not e) e = b;
    Node* Pre = p;
    Node* Nxt = p->nxt;
    Pre->nxt = b;
    Nxt->pre = e;
    b->pre = Pre;
    e->nxt = Nxt;
}
void InsertPre(Node* p, Node* b, Node* e = nullptr) {
    if (not e) e = b;
    Node* Pre = p->pre;
    Node* Nxt = p;
    Pre->nxt = b;
    Nxt->pre = e;
    b->pre = Pre;
    e->nxt = Nxt;
}

void Sol1(int n) {
    Node* Beg = tail, *End = tail->pre;
    while (n--) { Beg = Beg->pre; }
    Delete(Beg, End);
    InsertNxt(head, Beg, End);
}
void sol2() {
    Node *now = head->nxt->nxt->nxt;
    Delete(now);
    InsertNxt(head, now);
}
void Show() {
    Node* now = head->nxt;
    while (now != tail) {
        printf("%d ", now->v);
        now = now->nxt;
    }
    puts("");
}

int main () {
    int n, m;
    scanf("%d%d", &n, &m);
    int u;
    tail = new Node(0); head = new Node(0);
    tail->pre = head, head->nxt = tail;
    for (int i = 1; i <= n; i += 1) {
        scanf("%d", &u);
        InsertPre(tail, new Node(u));
    }
    for (int i = 1; i <= m; i += 1) {
        char ch; int u;
        scanf("%d%c", &u, &ch);
        if (ch == 'a') {
            u %= n;
            if (u) Sol1(u);
        }
        else {
            u %= 3;
            while(u--) sol2();
        }
    }
    Show();
    return 0;
}