POJ 2796 Feel Good

Feel Good

Time Limit: 3000MS		Memory Limit: 65536K
Total Submissions: 10949		Accepted: 2990
Case Time Limit: 1000MS		Special Judge

Description

Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated to studying how good or bad days influent people's memories about some period of life. 

A new idea Bill has recently developed assigns a non-negative integer value to each day of human life. 

Bill calls this value the emotional value of the day. The greater the emotional value is, the better the daywas. Bill suggests that the value of some period of human life is proportional to the sum of the emotional values of the days in the given period, multiplied by the smallest emotional value of the day in it. This schema reflects that good on average period can be greatly spoiled by one very bad day. 

Now Bill is planning to investigate his own life and find the period of his life that had the greatest value. Help him to do so.

Input

The first line of the input contains n - the number of days of Bill's life he is planning to investigate(1 <= n <= 100 000). The rest of the file contains n integer numbers a1, a2, ... an ranging from 0 to 10 ⁶ - the emotional values of the days. Numbers are separated by spaces and/or line breaks.

Output

Print the greatest value of some period of Bill's life in the first line. And on the second line print two numbers l and r such that the period from l-th to r-th day of Bill's life(inclusive) has the greatest possible value. If there are multiple periods with the greatest possible value,then print any one of them.

Sample Input

6
3 1 6 4 5 2

Sample Output

60
3 5

Source

Northeastern Europe 2005

 

题目大意：

给你一段序列，要你求出所有子序列中（在这段区间内的最小值*这段区间所有元素之和）最大的一个子序列，并要求输出结果。

 

大致思路：

这道也算是很明显的单调栈题目了。何谓单调栈？可以这样理解：它就是以某一个值为最小（或最大）值，将区间向这个值的两侧不断延伸，直到遇到小于（或大于）该值的值时才停止扩展。

这样就可以在O(n)的复杂度下算出所有可能的区间范围。

之后就是利用前缀和来算出区间元素的和了(一开始卡在这里良久，后来脑袋一拍，这不是前缀和就可以了么QAQ)，而算这个的复杂度也是O(n）。

 

代码如下：

 

//Feel Good.cpp -- POJ 2796
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