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
用变量记录开头和结尾，然后从第一个不断累加到最后一个
在途中，不断更新和维护【最大连续子序和】和他的开头和结尾的下表，
如果临时的连续子序和小于零 就赋0 再累加下去。。。
#include<iostream>
#include<algorithm>
using namespace std;
class M{
public:
int a,b,sum;
};
int x[];
int main(){
int t;cin>>t;
for(int k=;k<=t;k++){
if(k!=)cout<<endl;
int n;cin>>n;
M t; t.a=; t.b=; t.sum=-;
int ta=; int sum=;
for(int i=;i<=n;i++){
cin>>x[i];
sum+=x[i];
if(sum>t.sum){
t.a=ta;
t.b=i;
t.sum=sum;
}
if(sum<){
sum=;
ta=i+;
}
}
cout<<"Case "<<k<<":"<<endl;
cout<<t.sum<<" "<<t.a<<" "<<t.b<<endl;
}
return ;
}