PTA-1008——Elevator

本文分析了一种基于电梯调度的算法,通过计算电梯在不同楼层间的移动时间和停留时间,实现了对请求列表的有效处理。输入包含一系列正数,代表电梯需要停靠的楼层,输出为完成所有请求所需总时间。

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题目:

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:

41

题目分析:

水题

代码:

 1 #include<iostream>
 2 using namespace std;
 3 int sum;
 4 int n;
 5 int now,nex;
 6 int main(){
 7     cin>>n;
 8     sum=0;
 9     if(n!=0){
10         cin>>now;
11         sum+=now*6+5;
12     }
13     for(int i=1;i<n;i++){
14         cin>>nex;
15         if(nex>now){
16             sum+=(nex-now)*6;
17         }else{
18             sum+=(now-nex)*4;
19         }
20         sum+=5;
21         now=nex;
22     }
23     cout<<sum;
24     return 0;
25 }

 

 

转载于:https://www.cnblogs.com/orangecyh/p/10286241.html

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