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     }
32}