Max Sum HDU - 1003

   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.
InputThe 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).
   OutputFor 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

1.动规：dp[i]=max(dp[i-1]+a[i],a[i]），dp[i]表示以i为末尾的连续序列的最大和。
2.非动规：请看代码~~~参考了大神的代码，我还是太弱了，orz！！！

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;

int n,m;
int a[100005];

int main()
{   cin>>n;
    for(int t=1;t<=n;t++){
        scanf("%d",&m);
        for(int i=1;i<=m;i++) scanf("%d",&a[i]);
        
        int sum=0,maxsum=-10000,l=1,r=1,temp=1;
        for(int i=1;i<=m;i++){
            sum+=a[i];
            if(sum>maxsum){
                maxsum=sum;
                l=temp;
                r=i;
            }
            if(sum<0){
                sum=0;
                temp=i+1;
            }
        }
        printf("Case %d:\n%d %d %d\n",t,maxsum,l,r);
        if(t!=n) printf("\n");
    }
}