[PAT甲级]1008. Elevator (20)(求电梯运行时间)

最新推荐文章于 2022-06-30 16:34:57 发布

原创最新推荐文章于 2022-06-30 16:34:57 发布 · 482 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#PAT甲级 #1008 #Elevator #电梯运行时间 #C++

PAT甲级专栏收录该内容

17 篇文章

订阅专栏

本文介绍了一个关于电梯路径计算的问题，电梯初始位于0层，向上移动每层需6秒，向下移动每层需4秒，每次停靠需停留5秒。文章提供了一段C++代码，用于计算给定楼层序列所需的总时间。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1008. Elevator (20)

原题链接

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:

题目大意：

一个电梯，最初在0层，上一层需要6秒，下一层需要4秒，每次到达一层停留5秒，每次到达一个地方不用返回0层
给定一个数列，表示电梯停留的地方，求电梯走完这些地方需要的时间

思路：

分情况累加
注意开始是0层，每次完成都需要+5s，包括最后一次任务

代码：

#include <iostream>

using namespace std;

int main()
{
    int n;
    cin >> n;
    int sum=0;
    int after = 0;//表示上次停留的层数
    for(int i=0; i<n; i++){
        int temp;
        cin >> temp;
        if(temp > after)
            sum += (temp-after)*6;
        else
            sum += (after-temp)*4;
        sum += 5;
        after = temp;
    }
    cout << sum << endl;
    return 0;
}