1007 Maximum Subsequence Sum（记忆化搜索，动态规划）

含泪补基础

1007 Maximum Subsequence Sum

1. 题目描述

Given a sequence of K integers { N​1​​ , N​2​​ , …, N​K​​ }. A continuous subsequence is defined to be { N​i​​ ,
​i+1​​ , …, N​j​​ } where 1≤i≤j≤K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

题目大意： 求最大连续子序列和，输出最大的和以及这个子序列的开始值和结束值。如果所有数都小于0，那么认为最大的和为0，并且输出首尾元素～

Input Specification:
Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (≤10000). The second line contains K numbers, separated by a space.

Output Specification:
For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

2. 解题思路

• 这题是动态规划的思想。我们需要求最大的连续子列和，那么我们从最简单的开始。
  假设输入的n个数存在nums[n] 中，以第n个数结尾的最大连续子列和存在dp[n]中，对应的连续子列的初始位置存在site[n]中。那么以第0个数结尾的最大连续子列和就是nums[0]本身，即dp[0]=nums[0]。计算以第1个数结尾的最大连续子列和，我们需要判断nums[1]大还是nums[1]+dp[0]。
• 这样，我们就可以推导出，逐一更新dp[n]的值时，只需要判断nums[n]和nums[n]+dp[n-1]。而更新每个子列的初始位置，只需要site[n]=site[n-1]。
• 在