1008 Elevator (20分)_or a given request list, you are to compute the to-优快云博客

本文介绍了一个关于电梯在多个楼层间移动的时间计算问题。电梯从0层出发，按给定的楼层列表顺序移动，每上升一层需要6秒，下降一层需要4秒，每层停留5秒。文章提供了一个C++代码实现，用于计算电梯完成所有楼层请求所需的总时间。

测试地址：☞

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:

题意：一个电梯从 0 层开始，给出该电梯依次按顺序停的楼层数，已知每上升一层需要 6s，每下降一层需要 4s，每一层停留 5s，问电梯最后走完所有楼层数的总时间。

思路：上升层数*6，下降层数*4，停留层数*5，进行累加求出总花费时间。

c++代码：

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