Maximum Product UVA - 11059

最新推荐文章于 2020-08-06 14:33:42 发布

原创最新推荐文章于 2020-08-06 14:33:42 发布 · 424 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#UVA oj

oj题专栏收录该内容

74 篇文章

订阅专栏

本文介绍了一个算法问题，即给定一个整数序列，如何找出一个乘积最大的连续子序列，并提供了一段C++代码实现。该问题适用于编程竞赛和技术面试等场景。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.

题目大意：

输入n个元素组成的序列S，找出一个乘积最大的连续子序列。如果这个最大的乘积不是正数，应输出0。

1<=n<=18，-10<=Si<=10。

解析：枚举起点和终点，求最大乘积。

#include <cstdio>
#include <algorithm>
#include <bits/stdc++.h>
using namespace std;

int main()
{
    int n;
    int i,j,k,ct=0,a[20];
    long long int sum,_max;
    while(cin>>n)
    {
        ct++;_max=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
        }
        for(i=0;i<n;i++)
        {
            for(j=i;j<n;j++)
            {
                sum=1;
                for( k=i;k<=j;k++)
                {
                    sum*=a[k];
                }
                if(sum>_max)
                    _max=sum;
            }
        }
        cout<<"Case #"<<ct<<": The maximum product is "<<_max<<'.'<<endl<<endl;

    }
}