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

 

1 #include 
       
         2  3 int main(){ 4     int T; 5     int n; 6     int number[100001]; 7     int i; 8     int sum; 9     int max;10     int start;11     int end;12     int time;13     int temp;14 15     16     scanf("%d",&T);17     time=1;18 19     while(T--){20         sum=0;21         max=-1000;22         start=0;23         end=0;24         temp=0;25 26         scanf("%d",&n);27 28         for(i=0;i
        
         max){35                 max=sum;36                 start=temp;37                 end=i;38             }39 40             if(sum<0){   //关键是这里，当求和小于0时，便把下一个数值作为开头再找最大值41                 sum=0;42                 temp=i+1;43             }44         }45 46         printf("Case %d:\n",time);47         time++;48         printf("%d %d %d\n",max,start+1,end+1);49         if(T!=0)50             printf("\n");51     }52 53     return 0;54 }