B - Max Sum

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



本题要注意的一个是最大值的查找，为防止超时不能用两个for暴力搜索，可以设置一个当前最大值maxv，左标l，右标r，和一个预设左标k，还有当前的求和值sum，当sum>maxv时将maxv、l、r值更新，当sum<0时说明以当前的左标无论如何也找不到更大的值了，所以预设左标k要更新为i+1.
本体坑点，输出时不按格式容易PE。
#include<iostream>
#include<cstdio>
using namespace std;

typedef long long ll;

const long maxn=100005;

int main()
{
    int t,no=1;
    int a[maxn];
    cin>>t;
    while(t--)
    {
        long n;
        cin>>n;
        for(long i=0;i<n;i++)
            cin>>a[i];
        ll sum=0,maxv=-100000000;
        long l=0,r=0,k=0;
        for(long i=0;i<n;i++)
        {
            sum+=a[i];
            if(sum>maxv){l=k;r=i;maxv=sum;}
            if(sum<0){sum=0;k=i+1;}
        }
        printf("Case %d:\n",no++);
        cout<<maxv<<" "<<l+1<<" "<<r+1<<endl;
        //if(t!=0)cout<<endl;
    }
    return 0;
}