Maximum Product UVA - 11059

最新推荐文章于 2022-01-23 22:37:10 发布

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最新推荐文章于 2022-01-23 22:37:10 发布

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原文链接：http://www.cnblogs.com/l609929321/p/6900885.html

本文介绍了一种求解整数序列中最大连续子序列正乘积的方法，并提供了一个具体的C++实现方案。该方案通过枚举每个可能的子序列来计算其乘积，最终找出最大的乘积作为答案。

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.

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
using namespace std;
int main()
{
    int n,t=1,a[20],i,j;
    long long maxn,sum;
    while(cin >> n)
    {
        memset(a,0,sizeof(a));
        for(i=0;i<n;i++)
            cin >> a[i];
        for(i=0,maxn=0;i<n;i++)
        {
            for(j=i,sum=1;j<n;j++)
            {
                sum *= a[j];//直接枚举起点到每个可能的终点的乘积
                if(sum>maxn)
                    maxn = sum;
            }
        }
        cout << "Case #" << t++ << ": The maximum product is " << maxn << "." << endl;
        cout << endl;
    }
    return 0;
}