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

雖然是道水題…..但剛看題還以為用線段樹做….想了想線段樹…..後來發現什麼鬼啊，直接從頭遍歷一遍，如果遇到這個時候的總和小於0就將此節點更新為起點，如果這個時候的總和大於遍歷過的值的最大和，就將max更新為sum,然後把這個點設為結束點，注意的是剛開始不能把max設為0，因為更新的最大值可能為負值
AC程式碼：

#include<bits/stdc++.h>
using namespace
 
 std;
const int INF=0x3f3f3f3f;
int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int T,temp,n;
    cin>>T;
    for(int k=1;k<=T;k++) {
        cin>>n;
        int max=-INF,sum=0;
        int start=1,end=1,start_t=1;
        for(int i=1;i<=n;i++) {
            cin>>temp;
            sum+=temp;
            if(sum>max) {
                max=sum;
                end=i;
                start=start_t;
            }
            if(sum<0) {
                sum=0;
                start_t=i+1;
            }
        }
        if(k!=1) cout<<endl;
        cout<<"Case "<<k<<":"<<endl;
        cout<<max<<" "<<start<<" "<<end<<endl;
    }
    return 0;
}

四級考完….該好好準備期末和刷題了