PAT -- 甲级1008（1008 Elevator）

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本文介绍了一个电梯调度问题的解决方法，通过模拟算法计算电梯完成一系列楼层请求所需的总时间。考虑到电梯上行、下行及停留时间，文章提供了一段C++代码实现，用于计算基于特定楼层请求序列的总耗时。

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:

给一个数n后面跟着n个数，表示电梯的层数，刚开始从0层开始，上升1层用6秒，下降一层用4秒，停一层用5秒，问按照这个顺序执行需要用几秒，电梯最后不必返回0层

思路：模拟

Code：

#include<bits/stdc++.h>

using namespace std;

int val[105],n;

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> val[i];
	}
	int res = val[0] * 6, tmp;
	for (int i = 1; i < n; i++) {
		tmp = val[i] - val[i - 1];
		if (tmp > 0) {
			res += tmp * 6;
		} else {
			res -= tmp * 4;
		}
	}
	cout << res + n * 5 << endl;
	return 0;
}