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.
題意：輸入n個元素組成的序列S，你需要找出一個乘積最大的連續子序列 。如果這個最大的乘積不是正數，就輸出0（表示無解）。 1<=n<=18 -10<=S<=10
輸出格式 每輸出一組案例就空一行（注意）
題目分析：連續子序列有兩個要素：起點和重點。所以只要枚舉起點和終點就好。由於每個元素與的最大值不會超過10,且不超過18個元素，最大乘積不會超過10的18次方。所以可以用long long儲存（試了一下，用int的話，輸出18個10的結果不對，int存不下）
代碼如下：（媽的，剛開始想到3重循環去了，還傻逼的循環了一個len表示子序列的長度，盡管試了很多案例都對了，就是不能過。重想了一下，抱著試一試的心態，寫了，然後TM就過了，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 }
