[ACM]Elevator

最新推荐文章于 2020-03-16 15:14:31 发布

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最新推荐文章于 2020-03-16 15:14:31 发布

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原文链接：http://www.cnblogs.com/sr1993/p/3697838.html

本文介绍了一个关于计算电梯在特定路径上所需时间的问题。电梯在不同楼层间的移动时间取决于上下行方向，同时每层楼还需要额外的停靠时间。通过算法实现，可以计算出完成一系列楼层请求所需的总时间。

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Problem Description

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

There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.

Output

Sample Input

1 2
3 2 3 1
0

Sample Output

17
41

Author

ZHENG, Jianqiang

Source

ZJCPC2004

解题思路：

开一个数组用来保存输入的楼层数，变量time为总共的时间，初始化为0,从数组的第个数开始，依次比较当前数与前一个数的大小，分情况time+=(6或4）*(大数-小数）,最后time再加上数组中的一个数（楼层数）所用的时间，再加上所有楼层停留的时间。

代码：

#include <iostream>
using namespace std;
int fl[102];
int main()
{
    int n;
    while(cin>>n&&n)
    {
        int i;
        for(i=0;i<n;i++)
            cin>>fl[i];
        int time=0;
        for(i=1;i<n;i++)
        {
            if(fl[i]>fl[i-1])//比较
                time=time+6*(fl[i]-fl[i-1]);
            if(fl[i]<fl[i-1])
                time=time+4*(fl[i-1]-fl[i]);
        }
        cout<<time+6*fl[0]+5*n<<endl;
    }
    return 0;
}

运行截图：