题目：

Given a sequence of integers S = {S1, S2, . . . , Sn}, you should determine what is the value of the

maximum positive product involving consecutive terms of S. If you cannot find a positive sequence,

you should consider 0 as the value of the maximum product.

Input

Each test case starts with 1 ≤ N ≤ 18, the number of elements in a sequence. Each element Si

is

an integer such that −10 ≤ Si ≤ 10. Next line will have N integers, representing the value of each

element in the sequence. There is a blank line after each test case. The input is terminated by end of

file (EOF).

Output

For each test case you must print the message: ‘Case #M: The maximum product is P.’, where

M is the number of the test case, starting from 1, and P is the value of the maximum product. After

each test case you must print a blank line.

Sample Input

3

2 4 -3

5

2 5 -1 2 -1

Sample Output

Case #1: The maximum product is 8.

Case #2: The maximum product is 20.

题目大意：输入个长度为n的序列，要你找到乘积最大的连续序列的积。（如果乘积小于0，就相当于乘积为0）

题目思路：既然要找连续的序列，给一个循环枚举起点，一个循环枚举终点，一个循环它起点到终点的元素乘起来，给个MAX变量赋值为0（因为乘积小于0点都会被赋值为0）。

X每个乘积都与MAX比较，比MAX大的，就把它赋值给MAX。循环完毕，输出MAX的值就行了。（还要注意输出案例时要连着输出2个换行）

代码：

1 #include
     
       2 #include
      
        3 #include
       
         4 using namespace std; 5 const int maxn=18+5; 6 int main() 7 { 8     int n,i,a[maxn],j,k,p=0; 9     long int  sum,max; 10     while(cin>>n&&n)11     {        12         for(i=0;i
        
         >a[i];14         max=0;15         for(i=0;i
         
          max)27                     max=sum;28             }29         }30         printf("Case #%d: The maximum product is %lld.\n\n",++p,max);31         32     }33     return 0;34 }