1008 Elevator (20 point(s))

最新推荐文章于 2021-12-14 14:38:09 发布

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本文详细介绍了一个基于电梯调度的算法问题，电梯在城市最高建筑中运行，仅有一部电梯需要完成一系列楼层请求。算法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，以计算完成所有请求所需的总时间。

1008 Elevator (20 point(s))

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(void){
	int n;
	int arr[107]={0};
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&arr[i]);
	int res = 0;
	for(int i=1;i<=n;i++){
		res+=5;
		if(arr[i]>arr[i-1]) res=res+6*(arr[i]-arr[i-1]);
		else res=res+4*(arr[i-1]-arr[i]);
	}
	printf("%d\n",res);
	return 0;
}