uva11059(最大乘积) 解题心得

原题：

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.

 

 分析：直接进行n^2暴力循环就可以比较出最大值了

　　注意结果要用long long 并且所有与结果“同级”的也要用long long 

 

我的代码

// UVa11059 Maximum Product
// Rujia Liu
#include<iostream>
#include<cstdio>
using namespace std;

int main() {
    int S[20], kase = 0, n;
    while (cin >> n && n) 
    {
        kase++;
        for (int i = 0; i < n; i++)        cin >> S[i];
        long long ans = 0;
        for (int i = 0; i < n; i++) 
        {
            long long v = 1;
            for (int j = i; j < n; j++) 
            {
                v *= S[j];
                if (v > ans) ans = v;
            }
        }
        printf("Case #%d: The maximum product is %lld.\n\n", kase, ans);
    }
    return 0;
}