#P1519D. Maximum Sum of Products
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ID: 5898
Type: RemoteJudge
2000ms
256MiB
Tried: 0
Accepted: 0
Difficulty: (None)
Uploaded By:
Hydro
Tags> Maximum Sum of Products
No submission language available for this problem.
Description
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.
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 $b_1, b_2, \dots, b_n$ ($1 \le b_i \le 10^7$).
Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of $a$.
Input
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 $b_1, b_2, \dots, b_n$ ($1 \le b_i \le 10^7$).
Output
Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of $a$.
Samples
5
2 3 2 1 3
1 3 2 4 2
29
2
13 37
2 4
174
6
1 8 7 6 3 6
5 9 6 8 8 6
235
Note
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 $1 \cdot 5 + 8 \cdot 9 + 3 \cdot 6 + 6 \cdot 8 + 7 \cdot 8 + 6 \cdot 6 = 235$.