[Codeforces 863D]Yet Another Array Queries Problem

Description

You are given an array a of size n, and q queries to it. There are queries of two types:

• 1 l_i r_i — perform a cyclic shift of the segment [l_i, r_i] to the right. That is, for every x such that l_i ≤ x < r_i new value of a_x + 1 becomes equal to old value of a_x, and new value of a_li becomes equal to old value of a_ri;
• 2 l_i r_i — reverse the segment [l_i, r_i].

There are m important indices in the array b₁, b₂, ..., b_m. For each i such that 1 ≤ i ≤ m you have to output the number that will have index b_i in the array after all queries are performed.

Input

The first line contains three integer numbers n, q and m (1 ≤ n, q ≤ 2·10⁵, 1 ≤ m ≤ 100).

The second line contains n integer numbers a₁, a₂, ..., a_n (1 ≤ a_i ≤ 10⁹).

Then q lines follow. i-th of them contains three integer numbers t_i, l_i, r_i, where t_i is the type of i-th query, and [l_i, r_i] is the segment where this query is performed (1 ≤ t_i ≤ 2, 1 ≤ l_i ≤ r_i ≤ n).

The last line contains m integer numbers b₁, b₂, ..., b_m (1 ≤ b_i ≤ n) — important indices of the array.

Output

Print m numbers, i-th of which is equal to the number at index b_i after all queries are done.

Sample Input

6 3 5
1 2 3 4 5 6
2 1 3
2 3 6
1 1 6
2 2 1 5 3

Sample Output

3 3 1 5 2 

题解

官方给出的标签是数据结构...确实可以打个链表，但你发现$m<=100$，我们甚至暴力模拟都不会超时。

我们离线所有的操作，我们每读入一个询问，倒序操作模拟一遍就好了。

//It is made by Awson on 2017.9.30
#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Max(a, b) ((a) > (b) ? (a) : (b))
using namespace std;
const int N = 2e5;
void read(int &x) {
  char ch; bool flag = 0;
  for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
  for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
  x *= 1-2*flag;
}

struct Query {
  int id, l, r;
}query[N+5];
int n, q, m, p;
int a[N+5];

void work() {
  read(n), read(q), read(m);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 1; i <= q; i++)
    read(query[i].id), read(query[i].l), read(query[i].r);
  while (m--) {
    scanf("%d", &p);
    for (int i = q; i >= 1; i--) {
      if (query[i].l <= p && query[i].r >= p) {
    if (query[i].id == 1) {
      if (query[i].l == p) p = query[i].r;
      else p--;
    }
    else {
      p = query[i].r-(p-query[i].l);
    }
      }
    }
    printf("%d ", a[p]);
  }
  printf("
");
}
int main() {
  work();
  return 0;
}