UVa 11059 Maximum Product

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本文介绍了一个编程问题——寻找给定整数序列中最大的连续正数乘积，并提供了一种解决方案。通过枚举所有可能的连续子序列并计算其乘积来找出最大值。需要注意的是，由于乘积可能很大，使用long long类型来存储结果。

Problem D - Maximum Product

Time Limit: 1 second

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.

Problem setter: Sérgio Queiroz de Medeiros

#include <cstdio>

int S[20];

int num_count = 0;

int main()
{
	long long max_product = 0;
	int g_count = 1;
	while(scanf("%d", &num_count) == 1)
	{
		max_product = 0;
		for(int i = 0; i < num_count; i++)
			scanf("%d", &S[i]);

		// 计算所有连续整数的乘积
		for(int i = 0; i < num_count; i++)
		{
			long long sub_product = S[i];
			if(sub_product > max_product)
				max_product = sub_product;
			for(int j = i+1; j < num_count; j++)
			{
				sub_product *= S[j];
				if(sub_product > max_product)
					max_product = sub_product;
			}
		}
		
		printf("Case #%d: The maximum product is %lld.\n\n", g_count, max_product);
		g_count++;
									
	}	
	return 0;	
}

这题枚举所有可能性就可以了，一开始老是通不过，后来查了答案才注意到，整个乘积可以达到10^18，超过了int范围，改用long long即可。

这里需要记住：int大概表示10位整数，最大为21亿左右，long long大概表示19位整数。