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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
   
    #include
    
     #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
     
      >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
      
       maxsum){                maxsum=sum;                head=temp;//利用temp记录                tail=i+1;            }            if(sum<0){                sum=0;                temp=i+2;            }        }        cout<<"Case "<
       
        <<":\n";        cout<
        
         <<" "<<<" "<
         
          <<"\n"; if(l!=T) cout<<"\n"; } return 0;}
         
        
       
      
     
    
   

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