通过枚举每个点作为最小值，再通过动态规划求出以每个点作为最小值的左右区间。

代码如下：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #define MAXN 100005using namespace std;typedef long long int Int64;Int64 seq[MAXN], sum[MAXN], ret;int L[MAXN], R[MAXN], N;int main(){    while (scanf("%d", &N) == 1) {        ret = -(1LL << 60);        for (int i = 1; i <= N; ++i) {            scanf("%I64d", &seq[i]);                sum[i] = sum[i-1] + seq[i];        }        for (int i = 1; i <= N; ++i) {            int len = 0, j = i - 1;             while (j >= 1) {                if (seq[j] < seq[i]) break;                len += L[j] + 1;                j = i - len - 1;            }            L[i] = len;        }        for (int i = N; i >= 1; --i) {            int len = 0, j = i + 1;            while (j <= N) {                if (seq[j] < seq[i]) break;                len += R[j] + 1;                j = i + len + 1;            }                R[i] = len;        }        for (int i = 1; i <= N; ++i) {            ret = max(ret, (sum[i+R[i]]-sum[i-L[i]-1]) * seq[i]);        }        printf("%I64d\n", ret);    }    return 0;    }