A12438. Recursive Queries
编程题
普及/提高-
知识点
题目描述
You are given a permutation $p_1, p_2, \dots, p_n$ . You should answer $q$ queries. Each query is a pair $(l_i, r_i)$ , and you should calculate $f(l_i, r_i)$ .
Let's denote $m_{l, r}$ as the position of the maximum in subsegment $p_l, p_{l+1}, \dots, p_r$ .
Then $f(l, r) = (r - l + 1) + f(l, m_{l,r} - 1) + f(m_{l,r} + 1, r)$ if $l \le r$ or $0$ otherwise.
Let's denote $m_{l, r}$ as the position of the maximum in subsegment $p_l, p_{l+1}, \dots, p_r$ .
Then $f(l, r) = (r - l + 1) + f(l, m_{l,r} - 1) + f(m_{l,r} + 1, r)$ if $l \le r$ or $0$ otherwise.
输入格式
The first line contains two integers $n$ and $q$ ( $1 \le n \le 10^6$ , $1 \le q \le 10^6$ ) — the size of the permutation $p$ and the number of queries.
The second line contains $n$ pairwise distinct integers $p_1, p_2, \dots, p_n$ ( $1 \le p_i \le n$ , $p_i \neq p_j$ for $i \neq j$ ) — permutation $p$ .
The third line contains $q$ integers $l_1, l_2, \dots, l_q$ — the first parts of the queries.
The fourth line contains $q$ integers $r_1, r_2, \dots, r_q$ — the second parts of the queries.
It's guaranteed that $1 \le l_i \le r_i \le n$ for all queries.
The second line contains $n$ pairwise distinct integers $p_1, p_2, \dots, p_n$ ( $1 \le p_i \le n$ , $p_i \neq p_j$ for $i \neq j$ ) — permutation $p$ .
The third line contains $q$ integers $l_1, l_2, \dots, l_q$ — the first parts of the queries.
The fourth line contains $q$ integers $r_1, r_2, \dots, r_q$ — the second parts of the queries.
It's guaranteed that $1 \le l_i \le r_i \le n$ for all queries.
输出格式
Print $q$ integers — the values $f(l_i, r_i)$ for the corresponding queries.
输入输出样例
输入 #1
4 5 3 1 4 2 2 1 1 2 1 2 3 4 4 1
输出 #1
1 6 8 5 1
说明/提示
Description of the queries:
1. $f(2, 2) = (2 - 2 + 1) + f(2, 1) + f(3, 2) = 1 + 0 + 0 = 1$ ;
2. $f(1, 3) = (3 - 1 + 1) + f(1, 2) + f(4, 3) = 3 + (2 - 1 + 1) + f(1, 0) + f(2, 2) = 3 + 2 + (2 - 2 + 1) = 6$ ;
3. $f(1, 4) = (4 - 1 + 1) + f(1, 2) + f(4, 4) = 4 + 3 + 1 = 8$ ;
4. $f(2, 4) = (4 - 2 + 1) + f(2, 2) + f(4, 4) = 3 + 1 + 1 = 5$ ;
5. $f(1, 1) = (1 - 1 + 1) + 0 + 0 = 1$ .
1. $f(2, 2) = (2 - 2 + 1) + f(2, 1) + f(3, 2) = 1 + 0 + 0 = 1$ ;
2. $f(1, 3) = (3 - 1 + 1) + f(1, 2) + f(4, 3) = 3 + (2 - 1 + 1) + f(1, 0) + f(2, 2) = 3 + 2 + (2 - 2 + 1) = 6$ ;
3. $f(1, 4) = (4 - 1 + 1) + f(1, 2) + f(4, 4) = 4 + 3 + 1 = 8$ ;
4. $f(2, 4) = (4 - 2 + 1) + f(2, 2) + f(4, 4) = 3 + 1 + 1 = 5$ ;
5. $f(1, 1) = (1 - 1 + 1) + 0 + 0 = 1$ .