UVa 11059(枚举)

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.

//思路：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。

AC源码：

#include <iostream>#include <cmath>#include <cstdio>using namespace std;typedef long long LL;#define LOCALconst int MAXN=25;int A[MAXN],n;LL solve(){LL ans=-(1e15);for(int i=0;i<n;++i){for(int j=i;j<n;++j){LL sum=1;for(int k=i;k<=j;++k)sum*=A[k];ans=max(max(sum,ans),(LL)0);}}return ans;}int main(){#ifdef LOCALfreopen("data.in","r",stdin);freopen("data.out","w",stdout);#endifint kase=1;while(~scanf("%d",&n)){int tmp;for(int i=0;i<n;++i){scanf("%d",&tmp);A[i]=tmp;}printf("Case #%d: The maximum product is %lld.\n\n",kase++,solve());}return 0;}