Max Sum（最大子序和）

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 #include <stdio.h>
 3 using namespace std;
 4 int main()
 5 {
 6     int T,c=1;
 7     scanf("%d",&T);
 8     while(T--)
 9     {
10         int N,i,sum=0,max=-1001,t=1,start,end,n;
11         cin>>N;
12         for(i=1;i<=N;i++)
13         {
14             scanf("%d",&n);
15             sum=sum+n;
16             if(sum>max)
17             {
18                 max=sum;
19                 start=t;
20                 end=i;
21             }
22             if(sum<0)
23             {
24                 sum=0;t=i+1;
25             }
26         }
27         cout<<"Case "<<c++<<":"<<endl;
28         cout<<max<<" "<<start<<" "<<end<<endl;
29         if(T!=0)
30             cout<<endl;
31     }