HDU 1003 Max Sum

Given a sequence a11,a22,a33……ann, 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

大体思路：

本来是很简单的题目，硬是让我想复杂了。状态转移方程非常简单：
dp[i]=Max{dp[i-1]+a[i],a[i]}
然后需要注意的是标记开始与结尾。还有要注意都为负数的情况（因为没有在意负数wa了很多次，看别人的代码才注意到）。

代码：

#include<iostream>#include<cstdio>#define MAXN 100000+10using namespace std;int a[MAXN];int main(){    int T;    cin>>T;    for(int l=1;l<=T;++l){        int n;        cin>>n;        for(int i=0;i<n;++i)            cin>>a[i];        int maxsum=-1000,sum=0,head=0,tail=0,temp=1;//sum代替dp数组（因为题目简单）        for(int i=0;i<n;++i){//maxsum一定要是一个比较小的负数，不然会WA。            sum+=a[i];            if(sum>maxsum){                maxsum=sum;                head=temp;//利用temp记录                tail=i+1;            }            if(sum<0){                sum=0;                temp=i+2;            }        }        cout<<"Case "<<l<<":\n";        cout<<maxsum<<" "<<head<<" "<<tail<<"\n";        if(l!=T)            cout<<"\n";    }    return 0;}