HDU 1003:Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 199419    Accepted Submission(s): 46633

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

 

Author

Ignatius.L

 

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#include <iostream>
using namespace std;
int main()
{
    int T,N,n,j =1;
    cin >> T;
    while (T--)
    {
        int max = -1001,sum = 0,temp = 1,first,second;
        cin >> N;
        int *p = new int(N);
        for (int i = 1; i <= N; i++)
        {
            cin >> n;
            sum += n;
            if (max < sum)
            {
                max = sum;
                first = temp;
                second = i;
            }
            if (sum < 0)
            {
                sum = 0;
                temp = i +1;
            }
        }
        cout << "Case " << j++ << ":" << endl;
        cout << max << " " << first << " " << second << endl;
        if(T)cout << endl;
        delete p;
    }
    return 0;
}