Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 85953 Accepted Submission(s): 19912
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
1 #include <iostream>
2 using namespace std;
3
4
5 int main()
6 {
7 int numbers;
8 cin>>numbers;
9 for(int k =0;k<numbers;k++)
10 {
11 int num;
12 cin>>num;
13
14
15 int * number = new int [num];
16 int * start = new int [num];
17 int * b = new int [num];
18 for(int i=0;i<num;i++)
19 cin>>number[i];
20
21 b[0]= number[0];
22 start[0]=0;
23 for(int i=1;i<num;i++)
24 {
25 if(b[i-1]+number[i]>=number[i])
26 {
27 b[i]=b[i-1]+number[i];
28 start[i]= start[i-1];
29 }
30 else
31 {
32 b[i]= number[i];
33 start[i]= i;
34 }
35 }
36 int max = b[0];
37 int pos=0;
38 for(int i=0;i<num;i++)
39 {
40 if(b[i]>max)
41 {
42 max=b[i];
43 pos = i;
44 }
45 }
46 cout<<"Case "<<k+1<<":"<<endl<< max<<" "<<start[pos]+1<<" "<<pos+1<<endl;
47 if(k!=numbers-1)
48 cout<<endl;
49 delete number;
50 delete start;
51 delete b;
52
53
54 }
55
56
57 return 0;
58
59
60
61 }