PAT:1008 Elevator (20分)

最新推荐文章于 2024-12-17 09:09:57 发布

原创最新推荐文章于 2024-12-17 09:09:57 发布 · 121 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#c++

PAT 专栏收录该内容

19 篇文章

订阅专栏

本文探讨了一个城市中最高建筑内仅有一部电梯的调度问题。通过分析电梯上下楼层所需时间和停靠时间，提供了一种计算完成特定楼层请求列表总时间的方法。输入包含一系列正数，代表电梯需要停靠的楼层，输出则是完成所有请求所需的总时间。

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:

这道题是比较简单的，只要每次记录下在第几次就OK了

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,b=0,a,sum=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&a);
        if(a>b){
            sum+=(a-b)*6;
        }
        else if(a<b){
            sum+=(b-a)*4;
        }
        b=a;
    }
    sum+=n*5;
    cout<<sum;
}