CF1519D.Maximum Sum of Products

普及/提高-

通过率：0%

AC君温馨提醒

该题目为【codeforces】题库的题目，您提交的代码将被提交至codeforces进行远程评测，并由ACGO抓取测评结果后进行展示。由于远程测评的测评机由其他平台提供，我们无法保证该服务的稳定性，若提交后无反应，请等待一段时间后再进行重试。

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#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 $1 \cdot 5 + 8 \cdot 9 + 3 \cdot 6 + 6 \cdot 8 + 7 \cdot 8 + 6 \cdot 6 = 235$ .

首页