uva11059（最大乘积）

Posted 2020-09-17 zsychanpin

Problem D - Maximum Product

Time Limit: 1 second

Given a sequence of integers S = {S₁, S₂, ..., S_n}, 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 S_i is an integer such that -10 ≤ S_i ≤ 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：

输入n个元素组成的序列S，你须要找一个成绩最大的连续子序列，假设最大子序列为负数。输出0.

思路：

确定七点和终点，进行枚举，然后推断符不符合要求。

代码：

#include<cstdio>
#include<algorithm>
using namespace std;
int S[20];
int main()
{
    int n,casex=1;
    long long ans,a;
    while(scanf("%d",&n)!=EOF)
    {
       ans=0;
        for(int i=1;i<=n;i++)
            scanf("%d",&S[i]);
        for(int i=1;i<=n;i++)
        {
            for(int j=i;j<=n;j++)
            {
                    a=1;
                for( int k=i;k<=j;k++)
                {
                    a*=S[k];
                }
                ans=max(ans,a);

            }
        }
            printf("Case #%d: The maximum product is %lld.\n\n",casex,ans);
            casex++;
    }
    return 0;
}