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[ACM_动态规划] hdu1003 Max Sum [最大连续子串和]

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

???->>>>设某序列[x,j]的和大于等于0,则后面再来一个数肯定要接在上面;如果其小于0则后面来的数不要接在其后面,要把这个序列的开头设为当前输入的位置。

[ACM_动态规划] hdu1003 Max Sum [最大连续子串和]
[ACM_动态规划] hdu1003 Max Sum [最大连续子串和]

本文转自beautifulzzzz博客园博客,原文链接:http://www.cnblogs.com/zjutlitao/p/3674088.html,如需转载请自行联系原作者

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