PAT甲级 1008 Elevator (20 分)

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于 2021-06-23 21:15:49 发布

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分类专栏： PAT甲级

本文链接：https://blog.youkuaiyun.com/DoMoreSpeakLess/article/details/118162702

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这篇博客探讨了一座城市最高建筑中只有一个电梯的情况。文章通过输入规格和输出规格详细介绍了电梯运行的规则，包括上楼、下楼的时间成本以及每站停留时间。给出一个样例输入和输出，展示如何计算完成请求列表所需总时间。源代码实现展示了如何计算这一时间，突出了算法和效率问题。

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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:

#include <iostream>


using namespace std;


int main() {
    int sum = 0;
    int N, floor;
    int recent = 0;
    cin >> N;
    for (int i = 0; i < N; ++i) {
        cin >> floor;
        sum += 5;
        if (floor > recent)
            sum += (floor - recent) * 6;
        else
            sum += (recent - floor) * 4;
        recent = floor;
    }
    cout << sum;

}