1008 Elevator （20 分）

最新推荐文章于 2021-08-18 12:42:21 发布

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#PAT #甲级 #1008

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本文探讨了一个具体的电梯调度问题，通过给定的楼层请求列表，计算电梯完成所有请求所需的总时间。考虑到电梯上行和下行的不同速度，以及在每一楼层停留的时间，文章提供了一段C++代码实现，用于高效计算总耗时。

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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:

#include <iostream>
using namespace std;
int main(int argc, char** argv) 
{
	int n,a1=0,a2,sum=0;
	cin>>n;
	cin>>a1;
	sum=a1*6+5;	
	for(int i=1;i<n;i++)
	{
		cin>>a2;
		if(a2>a1)
		{
			sum+=(a2-a1)*6+5; 
		}
		else
		{
			sum+=(a1-a2)*4+5;
		}
		a1=a2;
	} 
	cout<<sum;
	return 0;
}