HUAS Summer Trainning #3 B

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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<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 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<n;i++)
13             cin>>a[i];
14         max=0;
15         for(i=0;i<n;i++)
16         {
17             for(j=i;j<n;j++)
18             {
19                 sum=1;
20                 for(k=i;k<=j;k++)
21                 {
22                     sum*=a[k];
23                 }
24                 if(sum<0)
25                     sum=0;
26                 if(sum>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 }

HUAS Summer Trainning #3 B

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原文地址：http://www.cnblogs.com/huaxiangdehenji/p/4694124.html