1008 Elevator （20 分）

最新推荐文章于 2022-02-10 21:00:30 发布

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本文深入探讨了一种电梯调度算法的实现细节，通过分析电梯在不同楼层间的移动成本，包括上升和下降的时间消耗，以及在每层停留的时间，来计算完成一系列楼层请求所需的总时间。文章提供了一个具体的代码示例，展示了如何根据输入的楼层请求列表，计算出电梯完成所有任务所需的时间。

1008 Elevator （20 分）

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 Npositive 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:

小小提示：在给出的点中，一定要停5分钟

#include <bits/stdc++.h>
using namespace std;
int a[101];
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0; i<n; i++)
    {
        scanf("%d",&a[i]);
    }
    int ans=0;
    int t=0;
    for(int i=0; i<n; i++)
    {
        int s=a[i];
        if(s==t)
            ans+=5;
        else if(s>t)
        {
            while(s!=t)
            {
                ans+=6;
                t++;
            }
            ans+=5;
        }
        else if(s<t)
        {
            while(s!=t)
            {
                ans+=4;
                t--;
            }
            ans+=5;
        }
    }
    cout<<ans<<endl;
    return 0;
}