UVA 11059 Maximum Product（枚举start+end）

题目大意：

　　就是说给你n个数字，然后求出连续序列的最大和。如果ans为负数就输出0，要么就输出这个ans。

解题思路：

　　其实我们只需要枚举起点和终点，然后考虑所有的情况就可以了，有点类似于求解最大连续和的问题。

唯一的不同就是每次都要初始化t = 1, t*=a[j]。注意long long。因为最多有18个数字，且每个数字的

最大值为10，那么他们的乘积是不会超过10^18的。

代码：

# include<cstdio>
# include<iostream>
# include<algorithm>
# include<functional>
# include<cstring>
# include<string>
# include<cstdlib>
# include<iomanip>
# include<numeric>
# include<cctype>
# include<cmath>
# include<ctime>
# include<queue>
# include<stack>
# include<list>
# include<set>
# include<map>

using namespace std;

const double PI=4.0*atan(1.0);

typedef long long LL;
typedef unsigned long long ULL;

# define inf 999999999
# define MAX 23

int a[MAX];

int n;

int main(void)
{
    int icase = 1;
    while ( cin>>n )
    {
        for ( int i = 1;i <= n;i++)
    {
        cin>>a[i];
    }
    LL ans = 0;
    for ( int i = 1;i <= n;i++ )
    {
        LL t = 1;
        //ans = max(ans,t);
        for ( int j = i;j <= n;j++ )
        {
            t*=a[j];
            ans = max(ans,t);
        }
    }
    printf("Case #%d: The maximum product is %lld.\n\n",icase++,ans);

    }


    return 0;
}