LeetCode Maximum Product Subarray

Description

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，你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）。每个案例要用空格隔开。

题解：

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。

只要给最大值赋初值0，大于0才给最大值重新赋值，则任何负数最后输出都会为0，不需分开考虑。

注意：最大值要用long long类。

#include<iostream>
#include<cstdio>
using namespace std;
int a[30];
int main()
{
    int k=0,t;
    while(scanf("%d",&t)==1)
    {
        long long maxn=0;
        for(int i=0; i<t; i++)
            cin>>a[i];
        for(int i=0; i<t; i++)
        {
            long long q=1;
            for(int j=i; j<t; j++)
            {
                q=q*a[j];
                if(q>maxn)
                    maxn=q;
            }
        }
        k++;
            printf("Case #%d: The maximum product is %lld.\n",k,maxn);
            printf("\n");                // 注意格式啊
    }
    return 0;
}