hdu1506(dp减少重复计算）

可以算出以第i个值为高度的矩形可以向左延伸left[i]，向右延伸right[i]的长度

那么答案便是 (left[i] + right[i] + 1) * a[i] 的最大值

关键left[i] 和right[i]的计算

如果a[i] > a[i-1]  ,  那么left[i] = 0

如果a[i] <=a[i-1],  那么left[i] = left[i-1] + 1,   但是位置i-1-left[i-1]-1及其往左的元素没有比较过，所以需要继续去比较

同理right[i]的计算也是一样的

 #pragma warning(disable:4996)
 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include <stdio.h>
 #include <string.h>
 #include <time.h>
 #include <math.h>
 #include <map>
 #include <set>
 #include <queue>
 #include <stack>
 #include <vector>
 #include <bitset>
 #include <algorithm>
 #include <iostream>
 #include <string>
 #include <functional>
 #include <unordered_map>
 typedef __int64 LL;
 const int INF = ;

 const int N =  + ;
 int a[N];
 LL left[N], right[N];
 int main()
 {
     int n;
     while (scanf("%d", &n), n)
     {
         for (int i = ;i <= n;++i)
         {
             scanf("%d", &a[i]);
         }
         left[] = right[n] = ;
         int l,r;
         for (int i = ;i <= n;++i)
         {
             l = i - ;
             left[i] = ;
             while (l>= && a[i] <= a[l])
             {
                 left[i] += left[l] + ;
                 l = l - left[l] - ;
             }
         }
         for (int i = n - ;i >= ;--i)
         {
             right[i] = ;
             r = i + ;
             while (r <= n && a[i] <= a[r])
             {
                 right[i] += right[r] + ;
                 r = r + right[r] + ;
             }
         }
         LL ans = ;
         for (int i = ;i <= n;++i)
         {
             ans = std::max(ans, (left[i] + right[i] + )*a[i]);
         }
         printf("%I64d\n", ans);
     }
     return ;
 }