Maximum Product(简单枚举)

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本文介绍了一个关于寻找整数序列中最大连续子序列乘积的问题，并提供了一种枚举起点和终点的方法来求解该问题。使用long long类型变量以确保能够存储最大可能的乘积。

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Maximum Product

Time Limit: 1 second

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.

///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的18次方，可以用long long来存；
#include<stdio.h>
int main()
{
    int n,i,j,a[20],f=1;
    long long int ans,max1;
    while(~scanf("%d",&n))
    {
        max1=0;
        for(i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(i=0;i<n;i++)///枚举起点；
        {
            ans=1;
            for(j=i;j<n;j++)///枚举终点；
            {
                ans*=a[j];
                if(ans>max1)///找到较大的乘积就赋值给max1；
                    max1=ans;
            }
        }
        printf("Case #%d: The maximum product is ",f++);
        if(max1<0)printf("0.\n");
        else
            printf("%lld.\n\n",max1);
    }
}