杭电ACM OJ 1008 Elevator 其实就是简单的加减法 有点水

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 23 2 3 10

 

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