A14180. Maximum Sum of Products
编程题
普及/提高-
知识点
题目描述
You are given two integer arrays $a$ and $b$ of length $n$ .
You can reverse at most one subarray (continuous subsegment) of the array $a$ .
Your task is to reverse such a subarray that the sum $\sum\limits_{i=1}^n a_i \cdot b_i$ is maximized.
You can reverse at most one subarray (continuous subsegment) of the array $a$ .
Your task is to reverse such a subarray that the sum $\sum\limits_{i=1}^n a_i \cdot b_i$ is maximized.
输入格式
The first line contains one integer $n$ ( $1 \le n \le 5000$ ).
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ( $1 \le a_i \le 10^7$ ).
The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ( $1 \le b_i \le 10^7$ ).
The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ( $1 \le a_i \le 10^7$ ).
The third line contains $n$ integers $b_1, b_2, \dots, b_n$ ( $1 \le b_i \le 10^7$ ).
输出格式
Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of $a$ .
输入输出样例
输入 #1
5 2 3 2 1 3 1 3 2 4 2
输出 #1
29
输入 #2
2 13 37 2 4
输出 #2
174
输入 #3
6 1 8 7 6 3 6 5 9 6 8 8 6
输出 #3
235
说明/提示
In the first example, you can reverse the subarray $[4, 5]$ . Then $a = [2, 3, 2, 3, 1]$ and $2 \cdot 1 + 3 \cdot 3 + 2 \cdot 2 + 3 \cdot 4 + 1 \cdot 2 = 29$ .
In the second example, you don't need to use the reverse operation. $13 \cdot 2 + 37 \cdot 4 = 174$ .
In the third example, you can reverse the subarray $[3, 5]$ . Then $a = [1, 8, 3, 6, 7, 6]$ and $1 \cdot 5 + 8 \cdot 9 + 3 \cdot 6 + 6 \cdot 8 + 7 \cdot 8 + 6 \cdot 6 = 235$ .
In the second example, you don't need to use the reverse operation. $13 \cdot 2 + 37 \cdot 4 = 174$ .
In the third example, you can reverse the subarray $[3, 5]$ . Then $a = [1, 8, 3, 6, 7, 6]$ and $1 \cdot 5 + 8 \cdot 9 + 3 \cdot 6 + 6 \cdot 8 + 7 \cdot 8 + 6 \cdot 6 = 235$ .