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
题解：就是求连续序列的最大和，遍历时不断比较更新下标即可。
代码如下：
#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <ctime>
#define maxn 1007
#define N 100005
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define eps 0.000000001
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define Debug(x) cout<<x<<" "<<endl
using namespace std;
typedef long long ll;
int main()
{
int t,cas=1;
cin>>t;
while(t--)
{
int n,a;
cin>>n;
int sum=0;
int start,end,temp=1,max1=-INF;
for(int i=0;i<n;i++)
{
cin>>a;
sum+=a;
if(sum>max1)
{
max1=sum;
start=temp;
end=i+1;
}
if(sum<0)
{
sum=0;
temp=i+2;
}
}
cout<<"Case "<<cas++<<":"<<endl;
cout<<max1<<" "<<start<<" "<<end<<endl;
if(t!=0)
cout<<endl;
}
return 0;
}