UVA 11059

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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，就输出0（表示无解）。 1<=n<=18   -10<=S<=10

输出格式  每输出一组案例就空一行（注意）

题目分析：连续子序列有两个要素：起点和重点。所以只要枚举起点和终点就好。由于每个元素与的最大值不会超过10,且不超过18个元素，最大乘积不会超过10的18次方。所以可以用long long储存（试了一下，用int的话，输出18个10的结果不对，int存不下）

代码如下：（妈的，刚开始想到3重循环去了，还傻逼的循环了一个len表示子序列的长度，尽管试了很多案例都对了，就是不能过。重想了一下，抱着试一试的心态，写了，然后TM就过了，也是RLGL.....）

 1 #include <stdio.h>
 2 int a[20];
 3 int main()
 4 {
 5     int n,N=0,c2=0;
 6     while(scanf("%d",&n)==1)
 7     {
 8         long long c,c2=0;
 9         ++N;
10         for(int i=0; i<n; i++)
11             scanf("%d",&a[i]);
12         for(int q=0;q<n;q++)
13         {
14             c=1;
15             for(int z=q;z<n;z++)
16             {
17                 c*=a[z];
18                 if(c>c2)
19                     c2=c;
20             }
21         }
22         if(c2<=0)
23             printf("Case #%d: The maximum product is 0.\n\n",N);
24         else
25             printf("Case #%d: The maximum product is %lld.\n\n",N,c2);
26     }
27     return 0;
28 }

UVA 11059

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原文地址：http://www.cnblogs.com/huangguodong/p/4684279.html