11059 - Maximum Product(暴力)

最新推荐文章于 2024-02-06 15:19:22 发布

原创最新推荐文章于 2024-02-06 15:19:22 发布 · 640 阅读

0 ·

CC 4.0 BY-SA版权

|--UVaOJ 专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一道在线评测平台上的算法题目——求解序列中连续子序列的最大乘积。通过提供完整的C语言实现代码，展示了如何遍历序列并计算所有可能的连续子序列乘积，最终找到最大的乘积值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=22&problem=2000&mosmsg=Submission+received+with+ID+14210883

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.

#include<stdio.h>
#include<string.h>
int main()
{
    int N,t=1,a[20],i,j;
	long long max,sum;
    while(scanf("%d",&N)!=EOF)
    {
        memset(a,0,sizeof(a));
        for(i=0;i<N;i++)
        scanf("%d",&a[i]);
        for(i=0,max=0;i<N;i++)
        {
            for(j=i,sum=1;j<N;j++)
            {
                sum*=a[j];
                if(sum>max)
                max=sum;
            }
        }
        printf("Case #%d: The maximum product is %lld.\n",t++,max);
        printf("\n");
    }

    return 0;
}