//错的莫名其妙的O w O

第二个的格式也是莫名其妙的

 

Input

测试输入包含若干测试用例，每个测试用例占2行，第1行给出正整数K( < 10000 )，第2行给出K个整数，中间用空格分隔。当K为0时，输入结束，该用例不被处理。

Output

对每个测试用例，在1行里输出最大和、最大连续子序列的第一个和最后一个元

素，中间用空格分隔。如果最大连续子序列不唯一，则输出序号i和j最小的那个（如输入样例的第2、3组）。若所有K个元素都是负数，则定义其最大和为0，输出整个序列的首尾元素。

Sample Input

6

-2 11 -4 13 -5 -2

10

-10 1 2 3 4 -5 -23 3 7 -21

6

5 -8 3 2 5 0

1

10

3

-1 -5 -2

3

-1 0 -2

0

Sample Output

20 11 13

10 1 4

10 3 5

10 10 10

0 -1 -2

0 0 0

 

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 #include 
         
           3 int main() 4 { 5       int first,last,temp,n,i,j,flag,thissum; 6       int a[22222]; 7       while(scanf("%d",&n)&&n) 8       { 9           flag=0;10           thissum=0;11           memset(a,0,sizeof(a));12           for(i=1;i<=n;i++)13           {14                 scanf("%d",&a[i]);15                 if(a[i]>=0)16                     flag=1;17           }18                  first=last=temp=1;19           if(!flag)20           {21               printf("0 %d %d\n",a[1],a[n]);22               continue;23           }24           int max=-33333;25           for(i=1;i<=n;i++)26           {27               thissum+=a[i];28                 if(thissum>max)29                 {30                     max=thissum;31                     first=temp;32                     last=i;33                 }34                 if(thissum<0)35                 {36                     thissum=0;37                     temp=i+1;38                 }39           }40           printf("%d %d %d\n",max,a[first],a[last]);41       }42       return 0;43 }