uva 11059Maximum Product (暴力)

本文介绍了一种算法,用于找出整数序列中连续子序列的最大正乘积,并提供了具体的输入输出样例及代码实现。

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Given a sequence of integers S = {S1, S2, . . . , Sn}, you should determine what is the value of themaximum 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 Siisan integer such that −10 ≤ Si ≤ 10. Next line will have N integers, representing the value of eachelement in the sequence. There is a blank line after each test case. The input is terminated by end offile (EOF).

Output

For each test case you must print the message: ‘Case #M: The maximum product is P.’, whereM is the number of the test case, starting from 1, and P is the value of the maximum product. Aftereach 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<bits/stdc++.h>
using namespace std;

int main()
{
    int n;
    int a[20];
    int flag = 0;
    long long ma;
    long long sum;
    while(~scanf("%d", &n))
    {

        for(int i = 0; i < n; i++)
            scanf("%d", &a[i]);
             ma=0;
        for(int i = 0; i < n; i++)
        {


            for(int j = i; j < n; j++)
            {
                sum = 1;
                for(int k = i; k <= j; k++)
                sum*=a[k];
                ma = max(ma, sum);
                //printf("sum = ---%d\n",ma);
            }
        }
         //if(flag)
            //printf("\n");
        flag++;
            printf("Case #%d: The maximum product is %lld.\n\n",flag, ma);
    }
}


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