Maximum Product

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：给定一个序列，找出这个序列中的一个子序列，使得这个序列的乘积为这个序列中所有子序列的最大值，且这个值为正数
方法：使用穷举法，遍历所有的子序列，用一个变量保存当前的最大值，与之后所求得的值一一进行比较，求得最大值；可以通过定义序列的头尾位置，来依次遍历所有的子序列

#include <iostream>#include <cstdio>using namespace std;long long getMax(int num[], int st, int ed){    long long maxM = 0;    long long s = 1;    for(int i = st; i <=ed; i++){        s *= num[i];        maxM = maxM > s ? maxM:s;    }    return maxM;}int main(){    int n;    int cnt = 0;    while(scanf("%d", &n) == 1 && n){        int num[20];        long long maxM = 0;        for(int i = 0; i < n; i++){            scanf("%d,", &num[i]);        }        for(int i = 0; i < n; i++){            for(int j = i; j < n; j++){                long long s = getMax(num, i, j);                maxM = maxM > s ? maxM:s;            }        }        printf("Case #%d: The maximum product is %lld.\n\n", ++cnt, maxM);    }    return 0;}