n个酒桶，每个都有价值Vi，每隔一天它们的价值都会上升1，每天都需要从首部或者尾部取出一桶酒，求得到的最大的价值是多少。C++，动态规划

每一次求得的最大值是a【0，n】=max（a[1,n]+a[0]*count ,a[0,n-1]+a[n]*count）;

所以得到如下：

import java.io.*;
import java.util.*;
public class Main
{
  static int [][] maxValue;
  private static int getMax(int []value,int left, int right,int count)
  {
      if (left <= right)
    {
      if (maxValue[left][right] != -1)
      {
          return maxValue[left][right];
      }
        int a = getMax(value, left + 1, right, count + 1) + count*value[left];
      int b = getMax(value, left, right - 1, count + 1) + count*value[right];
      if (a > b) {
        maxValue[left][right] = a;
      return a;
      } else {
        maxValue[left][right] = b;
      return b;
      }
    }
    return 0;
  }
    public static void main(String args[])
    {
        Scanner cin = new Scanner(System.in);
     int n = cin.nextInt();   
      int []value = new int[n];
        for (int i = 0;i < n;i++)
        {
            value[i] = cin.nextInt();
        }
      int left = 0;
      int right = n - 1;
      maxValue = new int[n][n];
      for (int i = 0 ;i < n; i++)
      {
          for (int j = 0; j < n; j++)
        {
                maxValue[i][j] = -1;        
        }
      }
      int sum = 0;
      sum = getMax(value, 0, n - 1, 1);
      System.out.println(sum);
    }
}