暴力求解最大乘积

题意：

      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.

思路分析：

           这道题关键在于给出计算的起点和终点。然后一直循环计算，最后再在所有的乘积中找到最大的数就KO了！

           注意：

           1、输出格式，空行

           2、数字比较大，用long long类型

源代码：

        

#include<iostream>
#include<stdio.h>
#include<string>
#include<algorithm>
using namespace std;
int main()
{
    int n;
    int count = 0;
    while (scanf_s("%d",&n)!=EOF)            //输入方式
    {
        count++;
        int a[25];
        for (int i = 0; i < n; i++)
            cin >> a[i];

        long long sum = 0;
        long long m = 0;
        for (int i = 0; i < n; i++)        //起始点
        {
            for (int j = i; j < n; j++)        //终点
            {
                sum = 1;                           //每次一回合要重新开始

                for (int k = i; k <= j; k++)
                {
                    sum *= a[k];
                    m = max(sum, m);                //找最大的数
                }
            }
        }
    
    printf("Case #%d: The maximum product is %lld.

", count, m);
    }
    return 0;
}

 

心得：

       做完这个题目心得还真没有，总之把代码写得让别人看得懂就好。题目用了三个循环，前两个是给出起点跟终点，第三个是计算乘积，每次出来一个乘积，就用max函数保存最大的数。