D - Maximum Product

UVA - 11059

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.

解法：對于序列S，找到相鄰的子序列，使子序列的乘積最大。范圍很小，用枚舉法，枚舉起點 i，終點 j，對之間的數求積。

忘記在每次循環前初始化ans，所以 WA 很多次

代碼：

#include<iostream>
using namespace std;

int main()
{
    long long a[25];
    int n,num=0;
    while(scanf("%d",&n)!=EOF){
        long long ans=0,mul;
        num++;
        for(int i=0;i<n;i++)
            cin>>a[i];
        for(int i=0;i<n;i++){
            for(int j=i;j<n;j++){
                mul=1;
                for(int k=i;k<=j;k++){
                    mul*=a[k];
                }
                if(mul>ans)
                    ans=mul;
            }
        }
        cout<<"Case #"<<num<<": The maximum product is "<<ans<<"."<<endl<<endl;
    }
    return 0;
}