Description
Given a sequence of integers S = {S1, S2, . . . , Sn}, you should determine what is the value of themaximum 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.
题意：
输入n个元素组成的序列S，你需要找出一个乘积最大的连续子序列。如果这个最大的乘积不是正数，应输出0（表示无解）。1 ≤ N ≤ 18，−10 ≤ Si ≤ 10。
分析：
1.连续子序列有两个要素：起点和终点，只要枚举起点和终点即可。
2.每个元素的绝对值不超过10且不超过18个元素，最大可能的乘积不会超过10^18。
3.最大可能超过 int 的范围，可用 long long 进行存储。
4.每个案例用空格隔开。
代码：
1 #include<cstdio>
2 #include<iostream>
3 using namespace std;
4
5 int main()
6 {
7 int n,a[20];
8 int i,j;
9 int M=0;
10 while(scanf("%d",&n)!=EOF)
11 {
12 long long P=0;//最大乘积
13 for(i=0;i<n;i++)
14 scanf("%d",&a[i]);
15 for(i=0;i<n;i++)
16 {
17 long long mid_ans=1;//中间乘积
18 for(j=i;j<n;j++)
19 {
20 mid_ans=mid_ans*a[j];
21 if(mid_ans>P)
22 P=mid_ans;
23 }
24 }
25 M++;
26 if(P>0)
27 printf("Case #%d: The maximum product is %lld.\n\n",M,P);
28 else
29 printf("Case #%d: The maximum product is 0.\n\n",M);
30 }
31 return 0;
32 }
心得：
long long 类型在VC里不能编译成功，在VC里应用_int64 类型。后来用code：：blocks编译通过了。观察几个提交的程序发现其实有些不必要的语句可以省略，这样可以使程序更简化，更清晰易懂。