Maximum Product UVA - 11059

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本文介绍了一种算法，用于找出整数序列中具有最大正乘积的连续子序列。该算法通过遍历序列并计算所有可能连续子序列的乘积来确定最大乘积。

注意：是序列，有顺序的，有顺序的，有顺序的。

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.

#include <stdio.h>

int main()
{
    int num[100000];
    int h = 1, n, i, j;
    while(scanf("%d", &n) != EOF)
    {
        for(i = 0; i < n; i++)
        {
            scanf("%d", &num[i]);
        }
        long long maxn = 0;
        for(i = 0; i < n; i++)
        {
            long long product = 1;
            for(j = i; j < n; j++)
            {
                product *= num[j];
                if(product > maxn)
                {
                    maxn = product;
                }
            }
        }
        printf("Case #%d: The maximum product is %lld.\n", h++, maxn);
        printf("\n");
    }
    return 0;
}