原题：
Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 217385 Accepted Submission(s): 51290
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
14 1 4
Case 2:
7 1 6
7 1 6
Author
Ignatius.L
每加入新的队列数前，判断前一加和数是否小于０，如果小于０，显然要重新计算加和，装入数组中，否则继续加，并将新的加和存入数组对应下标的空间。加和数组下标对应该加和子队列的最后一位。在加和数组中找到最大值，记录此时下标作为目标末位，接着寻找起始位即可。