1008 Elevator (20 分)-优快云博客

这篇博客探讨了一个城市最高建筑的电梯调度问题。给定一个包含N个正整数的请求列表，表示电梯需要停靠的楼层，算法需要计算完成所有请求所需的时间。电梯上行一层耗时6秒，下行一层耗时4秒，每停靠一层会停留5秒。输入案例中，电梯初始位于0楼，无需返回地面。输出为总时间。示例输入为3个请求2、3、1，输出为41秒，表示电梯完成请求的总时间。

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1008 Elevator (20 分)

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:

英文依旧看不懂，感谢web的英文翻译，大体看懂了

#include <bits/stdc++.h>
using namespace std;
int num[105];
int n,a,ans=0;
int main()
{
    cin>>n;
    a = 0;
    for(int i = 0;i<n;i++)
    {
        cin>>num[i];
        if(num[i]>a)
        {
            ans += (num[i]-a)*6;
        }
        else if(num[i]<a)
        {
            ans += (a-num[i])*4;
        }
        a=num[i];
    }
    cout<<ans+n*5<<endl;
    return 0;
}

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