PAT 甲级 1008 Elevator （20 分）

最新推荐文章于 2024-12-17 09:09:57 发布

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最新推荐文章于 2024-12-17 09:09:57 发布

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本文探讨了一个简单的电梯调度问题，通过给定楼层请求列表，计算电梯完成所有请求所需的总时间。考虑到电梯上下楼的时间成本和在每一层停留的时间，提供了一段C++代码实现，旨在优化电梯运行效率。

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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 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:

这题比较简单，代码如下：

#include<iostream>
using namespace std;
int main(){
  int n, p=0, pnxt,step, time=0;
  cin>>n;
  for(int i=0;i<n;i++){
    cin>>pnxt;
    step=pnxt-p;
    p=pnxt;
    if(step>0){
      time=time+step*6;
    }else if(step<0){
      time=time-step*4;
    }
  }
  time=time+n*5;
  cout<<time;
  return 0;
  
}