​​传送门​​ 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 ∑i=1nai⋅bi is maximized.

Input

The first line contains one integer n (1≤n≤5000).

The second line contains n integers a1,a2,…,an (1≤ai≤107).

The third line contains n integers b1,b2,…,bn (1≤bi≤107).

Output

Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of a.

思路：

dp(x,y)：表示翻转x，y区间所获得的额外价值，则区间dp过程中记录翻转所得到的最大价值，最后再加上原有的价值。

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
ll a[5010];
ll b[5010];
ll dp[5010][5010];

int main()
{
  int n;
  cin>>n;
  ll ans = 0;
  ll sum = 0;
  for(int i = 1; i <= n; i++)cin>>a[i];
  for(int i = 1; i <= n; i++)cin>>b[i];
  for(int i = 1; i <= n; i++)sum += a[i]*b[i];
  for(int i = n-1; i >= 1; i--)
  {
    for(int j = i+1; j <= n; j++)
    {
      dp[i][j] = dp[i+1][j-1] + a[i]*b[j] + a[j]*b[i] - a[i]*b[i] - a[j]*b[j]; 
      ans = max(ans,dp[i][j]);
    }
  }
  printf("%lld\n",ans+sum);
}