1008. Elevator (20)

最新推荐文章于 2021-08-18 12:42:21 发布

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本文探讨了一个只有单一电梯的高楼中如何高效处理楼层请求的问题。通过分析电梯上下行时间和停留时间，设计了一种算法来计算完成一系列楼层请求所需的总时间。

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:

#include <iostream>

using namespace std;

int main()
{
    const int up = 6;
    const int down = 4;
    const int stay = 5;
    int n, time, pre, current;
    while(cin >> n)
    {
        time = 0;
        pre = 0;
        current = 0;
        for(int i=0; i<n; i++)
        {
            cin >> current;
            time += stay;
            time += ((current-pre)>0) ? (current-pre)*up : (pre-current)*down;
            pre = current;
        }
        cout << time << endl;
    }
    return 0;
}