/*Max Sum
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 37083 Accepted Submission(s): 7979
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
*/
#include<iostream>
using namespace std;
int main(){
int N;cin>>N;
for(int k=1;k<=N;k++){
if(k-1)cout<<endl;
int n;cin>>n;
int i,j,beg,end,max=-11111111,sum,fig;
for(i=j=1,sum=0;j<=n;j++){
cin>>fig;
sum+=fig;
if(sum>max){max=sum;beg=i;end=j;}//a只在此处改变beg,和end 的大小
if(sum<0){i=j+1;sum=0;}//如果sum小于零了，那么说明刚加进来的fig特别的小（小到这种“特殊”的程度）
//试探性的以次为起点，如果后来的sum 比max 大，则前面的一连串就毫无意义（因为加上一个负值肯定会变小的）
}
cout<<"Case "<<k<<":\n";
cout<<max<<" "<<beg<<" "<<end<<endl;
}
return 0;
}
说白了是一种贪心算法：
Int MaxSubsequenceSum(const int A[], int N)
{
int ThisSum, MaxSum, j;
ThisSum = MaxSum = 0;
For(j=0; j < N; j++)
{
ThisSum += A[j];
If (ThisSum > MaxSum)
MaxSum = ThisSum;
Else if(ThisSum < 0)
ThisSum = 0;
}
return MaxSum;
}