Uva - 11059 - Maximum Product

Given a sequence of integers S = {S₁, S₂, ..., S_n}, 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 S_i is an integer such that -10 ≤ S_i ≤ 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.

二重循环遍历过去就搞定了，因为每个数绝对值不超过10，不超过18个数，最大乘积不会超过10的18次方，这样就可以把乘积用long long存放。

不知道为什么用printf打印结果不管怎么弄都是WA，都快疯了，最后用了cout却AC了，真是各种无语。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <string>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <stack>
#include <queue>
#include <bitset> 
#include <cassert> 

using namespace std;

int s[20];

int main()
{
	ios::sync_with_stdio(false);
	int n;
	int kase = 0;
	while (cin >> n && n) {
		for (int i = 0; i < n; i++) {
			cin >> s[i];
		}
		long long pro = 0;
		for (int star = 0; star < n; star++) {
			long long proTem = 1;
			for (int end = star; end < n; end++) {
				proTem = proTem * (long long)s[end];
				if (proTem > pro) {
					pro = proTem;
				}
			}
		}
		//printf("Case #%d: The maximum product is %d.\n\n", ++kase, pro);
		cout << "Case #" << ++kase << ": The maximum product is " << pro << ".\n\n";
	}

	return 0;
}