1008 Elevator

最新推荐文章于 2022-11-25 09:22:02 发布

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本文探讨了一个具体的电梯调度问题，即在一栋高楼中，只有一部电梯如何高效地响应多个楼层的请求。通过给出的示例代码，我们计算了完成所有楼层请求所需的总时间，考虑了电梯上行和下行的时间成本，以及在每层停留的时间。

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:

题解：

娱乐题，注意有时楼层不发生移动时，依旧需要停留5s；

#include <iostream>
using namespace std;

int main()
{
    int stay = 0, next;
    int sum = 0, temp;
    int k;
    cin>>k;
    for(int i=0;i<k;i++){
        cin>>next;
        temp = next - stay;
        if(temp<0)
            sum += (-temp)*4 + 5;
        else if(temp>0)
            sum += temp*6 + 5;
        else
            sum += 5;
        stay = next;
    }
    cout<<sum<<endl;
}