1008. Elevator (20)——PAT (Advanced Level) Practise

最新推荐文章于 2019-01-03 19:57:39 发布

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本文介绍了一个关于电梯在特定楼层请求列表下移动的时间计算问题。电梯从0楼开始，依次访问指定楼层，上行每层耗时6秒，下行每层4秒，并在每个停靠层停留5秒。文章提供了计算总耗时的C++实现代码。

题目信息：

1008. Elevator (20)

时间限制

400 ms

内存限制

32000 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.

For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.

Input Specification:

Each input file contains one test case. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100.

Output Specification:

For each test case, print the total time on a single line.

Sample Input:

3 2 3 1

Sample Output:

代码如下：

#include <iostream>
using namespace std;

int main()
{
	int n;
	cin >> n;
	int t = 0, f;
	int sum = 0;
	for (int i = 0; i < n; i++)
	{
		f = t;
		cin >> t;
		if (t - f > 0)
			sum += 6 * (t - f);
		else if (t - f < 0)
			sum += 4 * (f - t);
		sum += 5;
	}
	cout << sum << endl;
	return 0;
}