PAT 1008 Elevator [模拟]

最新推荐文章于 2025-07-25 15:37:39 发布

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本文探讨了一个城市中最高建筑内仅有的一部电梯的调度问题。通过分析电梯在不同楼层间的移动成本，包括上行和下行的时间消耗，以及在指定楼层停留的时间，设计了一种算法来计算完成一系列楼层请求所需的总时间。输入包含一系列正整数，代表电梯需要停靠的楼层，输出则是完成所有请求所需的时间。

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:

------------------------------------这是题目和解题的分割线------------------------------------

#include<cstdio>

int main()
{
	int n,i,a[1000] = {};
	scanf("%d",&n);
	for(i=1;i<=n;i++)
		scanf("%d",&a[i]);
	//不是每一层楼都要停留，在输入的层数停留就够了 
	int time = n*5;
	for(i=1;i<=n;i++)
	{
		if(a[i]-a[i-1]>0) time += (a[i]-a[i-1])*6; //上楼
		else time += (a[i-1]-a[i])*4; //下楼
	}
	printf("%d",time);
}