SDUT1009——Elevator

最新推荐文章于 2024-04-27 21:33:58 发布

原创最新推荐文章于 2024-04-27 21:33:58 发布 · 310 阅读

1 ·

CC 4.0 BY-SA版权

文章标签：

#SDUT

SDUT 专栏收录该内容

17 篇文章

订阅专栏

博客围绕电梯运行时间计算问题展开，给出问题描述，即电梯初始在0层，上升一层6秒、下降一层4秒、每层停留5秒。还说明了输入输出格式，有多个测试用例，以N = 0结束输入，最后给出示例输入输出及题意解释。

Elevator
Time Limit: 1000 ms Memory Limit: 32768 KiB

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
Print the total time on a single line for each test case.

Sample Input
1 2
3 2 3 1
0

Sample Output
17
41

**题意：**电梯厨师位置在0层位置，每次上升一层需要6s，每次下降一层需要4s，在一层停止5s。每一次输入，第一个数n，后面跟着n个数。输出该过程需要的总时间

AC代码：


#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,x,s,t,i;
    while(cin>>n&&n>0)
    {
        s=0,t=0;
        for(i=0; i<n; i++)
        {
            cin>>x;
            if(t<x)
            {
                s+=(x-t)*6+5;
                t=x;
            }
            else
            {
                s+=(t-x)*4+5;
                t=x;
            }
        }
        cout<<s<<endl;
    }
    return 0;
}

余生还请多多指教！