Maximum Sum of Products

题意翻译

给定两个长为 $n$ 的序列 $a$ 和 $b$。你可以对 $a$ 的一段区间翻转，也可以不翻转，要求翻转后 $a$ 与 $b$ 对应位置之积的和最大。即求下式的值最大： $$\sum_{i=1}^na_i\times b_i$$ 其中 $n\le 5000$，$a_i,b_i\le10^7$。 translated by @zc_Ethan

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.

输入输出格式

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 $ ).

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

输入样例 #1

5
2 3 2 1 3
1 3 2 4 2

输出样例 #1

29

输入样例 #2

2
13 37
2 4

输出样例 #2

174

输入样例 #3

6
1 8 7 6 3 6
5 9 6 8 8 6

输出样例 #3

235

说明

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 $ .