HDU 1003:Max Sum

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 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;}

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