LeetCode Maximum Product Subarray

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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。。。。。

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 const int Max = 20;
 8 const int INF = 1000000000;
 9 int a[Max], n;
10 
11 int main()
12 {
13     int temp = 0;
14     while(~scanf("%d",&n))
15     {
16         memset(a,0,sizeof(a));
17         for(int i = 0; i < n; i++)
18             scanf("%d",&a[i]);
19         long long sum, ans = 0;
20         for(int i = 0; i < n; i++)
21         {
22             for(int j = i; j < n; j++)
23             {
24                 sum = 1;
25                 for(int k = i; k <= j; k++)
26                     sum *= a[k];
27                 ans = max(ans,sum);
28             }
29         }
30         printf("Case #%d: The maximum product is %lld.\n\n",++temp,ans);
31     }
32     return 0;
33 }

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LeetCode Maximum Product Subarray

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原文地址：http://www.cnblogs.com/demodemo/p/4690573.html