Problem A

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 206947    Accepted Submission(s):
48397

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

简单题意：

　　求最大子段和，，并记录位置。。。

简单思路：

　　第一个DP，写的比较艰难，，，，用从第一个开始求最大子段和，再判断下一个数的最大子段和

#include <iostream>
#include <fstream>
#include <cstdio>
using namespace std;

int main()
{
    //fstream cin("aaa.txt");
    int t, jishu = 1;
    cin >> t;
    while (t--)
    {

        int n, dp = 0, max = -1000001, r = 1, l = 1, temp = 1;
        cin >> n;
        for(int i = 1; i <= n; i++)
        {
            int tep;
            cin >> tep;
            dp += tep;
            if (dp > max)
            {
                max = dp;
                r = temp;
                l = i;
            }
            if (dp < 0)
            {
                dp = 0;
                temp = i + 1;
            }
        }
        printf("Case %d:\n%d %d %d\n", jishu++, max, r, l);
        if (t != 0)
            cout << endl;
    }
    return 0;
}