PAT甲级(Advanced Level)练习题 Elevator (20)

最新推荐文章于 2022-05-05 19:26:29 发布

ChenHsing22

最新推荐文章于 2022-05-05 19:26:29 发布

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分类专栏： PAT甲级练习题

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本文介绍了一个简单的电梯调度问题，通过计算电梯在不同楼层间的移动时间和停留时间来确定完成一系列请求所需的总时间。输入包括一系列正整数，表示电梯需要停靠的楼层。电梯从0楼开始，不需返回地面。文章提供了C++代码实现。

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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.

输入描述:

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.

输出描述:

For each test case, print the total time on a single line.

输入例子:

3 2 3 1

输出例子:

题目分析：

水题……会输入输出和赋值语句的人都会做

代码实现（C++版本）

#include <iostream>

using namespace std;

int main()
{
    int n;
    cin >> n;
    int start = 0;
    int sum = 0;

    for(int i =0;i<n;i++){
        int a;
        cin >> a;
        if(a>=start){
            sum+= (a-start)*6;
            start = a;
        }else if(a< start){
            sum += (start-a)*4;
            start = a;
        }
    }
    sum+=5*n;
    cout<<sum<<endl;
    return 0;
}