杭电1003 最大子串（第二次AC） 当作DP的训练吧

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 252815    Accepted Submission(s): 59950

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

 

题目意思很简单，就是在一个数组中找最大子串（这个是要连续的，之前那个是不连续的要求递增的）。

然后题目的精髓就在于: 一开始把first指针指向第一个位置，然后用一个temp变量也指向第一个位置，然后当有sum>maxsum的时候，就执行first=temp，保证了最大子串在后面依然可以被找到。计算子串的sum<0时就表示要重新找子串了，temp=j+1，sum=0，当找到下一个>0的数的时候，（因为有temp的实时更新）first指向该数，重新开始计算sum。

 

然后 就要注意下输出格式的问题。具体参考下第一次的AC代码