HDU-1003 Max Sum（经典DP）

Problem Description
Given a sequence a[1],a[2],a[3]…a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output
Case 1:
14 1 4

Case 2:
7 1 6

定义一个数组sum，
sum[i]含义：所有以a[i]（i=3，表示以3结尾）为结尾的序列的序列和构成一个集合，此集合的最大值就是sum[i]。

如1,2,3.所有以a[3]=3为结尾的序列的序列和集合是{6,5,3},因而sum[3]=6.
sum的状态转移方程：sum[i] = max{sum[i-1]+a[i], a[i]}
ans必定是sum[0···(k-1)]之一。由于要记录起始位置和结束位置，引入s数组记录获得sum的序列的起始元素的位置，而由sum的定义，sum[i]的结束位置是i不用另外记录。

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