杭电ACM OJ 1008 Elevator 其实就是简单的加减法 有点水

本文介绍了一个基于电梯调度的算法问题,电梯从0层出发,向上移动每层需6秒,向下移动每层需4秒,每层停留5秒。通过计算不同楼层请求列表,得出完成所有请求所需的总时间。

Elevator

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 76354    Accepted Submission(s): 41998


Problem Description
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
There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.
 

Output
Print the total time on a single line for each test case. 
 

Sample Input
1 2 3 2 3 1 0
 

Sample Output
17
41
主要翻译下题
public class Elevator {

    public static void main(String[] args) throws Exception {

        initData();

        int totalSecond = caculate(list);

        System.out.println(totalSecond + "");

    }

    private static List<Integer> list;

    private static void initData() {
        list = new ArrayList<>();

        list.add(2);
    }

    private static int caculate(List<Integer> list) {//3   2 3 1 (from 0)
        int current = 0;
        int total = 0;
        for (int i : list) {
            if (i > current) {
                total+= (i - current) * 6 + 5;
                current = i;
            } else if (i < current) {
                total+= (current - i) * 4 + 5;
            }
        }
        return total;
    }
}
目:从0层开始,每上升一层,6s,停留在一层,5s,下降一层,4s。例子中的第一个数字是有几个停留点

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值