Maximum Product (UVA 11059)解题

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，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。

注意：每行输出后有一行空白行

思路：连续子序列有两个要素：起点和终点，因此只需枚举起点和终点即可。由于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，可以用long long存储。

#include<stdio.h>
int main()
{
    int n,a[20],q=0;
    long long max,k;
    while(scanf("%d",&n)!=EOF){
        q++;
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
        }
        max=0;
        for(int i=0;i<n;i++){        //枚举起点
            for(int j=i;j<n;j++){     //枚举终点
                k=1;
                for(int t=i;t<=j;t++){
                    k=k*a[t];
                    if(max<k){
                        max=k;
                    }
                }
            }
        }
        printf("Case #%d: The maximum product is %lld.\n",q,max);
        printf("\n");
    }    
}