PAT 甲级 1008 Elevator (20)（20 分）

最新推荐文章于 2024-12-17 09:09:57 发布

UKnowNothing_

最新推荐文章于 2024-12-17 09:09:57 发布

阅读量163

点赞数

CC 4.0 BY-SA版权

分类专栏： PAT 甲级

本文链接：https://blog.youkuaiyun.com/Hello_Mr_X/article/details/81772298

PAT 甲级专栏收录该内容

41 篇文章

订阅专栏

本文介绍了一个基于特定电梯路径请求的算法实现，该算法通过计算电梯完成一系列楼层请求所需的总时间来优化其运行效率。考虑到电梯上行和下行的时间差异及在各楼层停留的时间，文章提供了一段C++代码示例，用于解决实际问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1008 Elevator (20)（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>
# include <cstdio>
using namespace std;

int main(){
	int N, f, bf = 0;
	long long sum = 0;
	scanf("%d", &N);
	while(N--){
		scanf("%d", &f);
		if(f > bf){
			sum += (f - bf) * 6 + 5;
		}
		else if(f < bf){
			sum += (bf - f) * 4 + 5;
		}
		else 
			sum += 5;
		bf = f;
	} 
	printf("%ld", sum);
	return 0;
}