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.

题目是求最大连续子串乘积；

枚举起点和终点即可。。。。。

#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

const int Max = 20;
const int INF = 1000000000;
int a[Max], n;

int main()
{
    int temp = 0;
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof(a));
        for(int i = 0; i < n; i++)
            scanf("%d",&a[i]);
        long long sum, ans = 0;
        for(int i = 0; i < n; i++)
        {
            for(int j = i; j < n; j++)
            {
                sum = 1;
                for(int k = i; k <= j; k++)
                    sum *= a[k];
                ans = max(ans,sum);
            }
        }
        printf("Case #%d: The maximum product is %lld.

",++temp,ans);
    }
    return 0;
}