Max Sum （dp）

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.

InputThe 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).
OutputFor 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
题意：找出最大的连续字串和；

#include<iostream>

#include<map>

using namespace std;

int main()

{

    int t,mark=,cnt=;

    cin >> t;

    while (t--)

    {

        if (cnt) cout << endl;

        cnt = ;

        mark++;

        int temp = , frist, end;

        int n,ko,sum=,max=-;

        cin >> n;

        for (int i = ; i < n; i++)

        {

            cin >> ko;

            sum += ko;

            if (sum > max)

            {

                max = sum; frist = temp; end = i + ;

            }

            if (sum < )

            {

                sum = ; temp = i + ;

            }

        }

        cout << "Case " << mark << ":" <<endl<< max << " " << frist << " " << end << endl;

    }

    return ;

}

注意：因为要输出下标，必须灵活应用temp这个中间值，刚开始因为一直没仔细考虑下标的情况，导致负数情况下不能出正确答案；

通过temp的加入后，可以在大于max的条件满足下更改最后的下标，小于零的情况下可以更改temp，到累计大于max时，直接改frist；

（经典DP题）

巴特西

Max Sum （dp）

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