题意：
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.
思路分析：
这道题关键在于给出计算的起点和终点。然后一直循环计算，最后再在所有的乘积中找到最大的数就KO了！
注意：
1、输出格式，空行
2、数字比较大，用long long类型
源代码：
1 #include<iostream>
2 #include<stdio.h>
3 #include<string>
4 #include<algorithm>
5 using namespace std;
6 int main()
7 {
8 int n;
9 int count = 0;
10 while (scanf_s("%d",&n)!=EOF) //输入方式
11 {
12 count++;
13 int a[25];
14 for (int i = 0; i < n; i++)
15 cin >> a[i];
16
17 long long sum = 0;
18 long long m = 0;
19 for (int i = 0; i < n; i++) //起始点
20 {
21 for (int j = i; j < n; j++) //终点
22 {
23 sum = 1; //每次一回合要重新开始
24
25 for (int k = i; k <= j; k++)
26 {
27 sum *= a[k];
28 m = max(sum, m); //找最大的数
29 }
30 }
31 }
32
33 printf("Case #%d: The maximum product is %lld.\n\n", count, m);
34 }
35 return 0;
36 }
心得：
做完这个题目心得还真没有，总之把代码写得让别人看得懂就好。题目用了三个循环，前两个是给出起点跟终点，第三个是计算乘积，每次出来一个乘积，就用max函数保存最大的数。