PAT甲级1008 Elevator

最新推荐文章于 2025-07-25 15:37:39 发布

原创最新推荐文章于 2025-07-25 15:37:39 发布 · 236 阅读

0 ·

CC 4.0 BY-SA版权

PAT甲级专栏收录该内容

13 篇文章

订阅专栏

本文介绍了一个关于电梯在特定路径上运行的时间计算问题。电梯从0层开始，根据给定的楼层请求列表移动，每次上升或下降一层分别需要6秒和4秒，且在每一层停留5秒。文章提供了一段C++代码实现，用于计算完成所有楼层请求所需的总时间。

题目描述：
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

题目大意：
刚开始在0层，给你一个电梯需要到达层数的顺序，上升一层用6秒，下降一层用4秒，一层要停5秒，最后计算总的消耗时间。
这个题感觉有问题，如果连续两个是在同一层下的话，它停留时间按的10秒（5*2）算的。其中的三个测试点都是这个问题。

代码如下：

#include <iostream>
using namespace std;
int main()
{
    int k,sum=0,now=0;
    cin>>k;
    while(cin>>k){
        if(now<k) sum+=6*(k-now)+5;
        else  sum+=4*(now-k)+5;
        now=k;
    }
    cout<<sum<<endl;
    return 0;
}

原本写的(如果连续好几个人都是一层的话，只按5秒计算)：
其实题意是每有一个人就多5秒。

 if(now<k) sum+=6*(k-now)+5;
 else if(now>k) sum+=4*(now-k)+5;