PAT甲级真题 1008 Elevator (20分) C++实现（测试点2、3、6有坑）

zhang35

于 2020-01-25 22:46:33 发布

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分类专栏： PAT 文章标签：算法 leetcode

本文链接：https://blog.youkuaiyun.com/zhang35/article/details/104085136

该博客介绍了PAT甲级考试中的一道编程题——1008 Elevator，讨论了如何计算电梯在给定停靠楼层完成请求所需的时间。作者指出，电梯在每个楼层停留5秒，上一层楼花费6秒，下一层楼花费4秒，且即使楼层重复也要计算停留时间。博主提供了问题的思路和解决方案，并给出了一个测试案例及其输出结果。

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题目

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:
41