PAT甲级1008

最新推荐文章于 2021-07-03 10:17:23 发布

倾鸢染

最新推荐文章于 2021-07-03 10:17:23 发布

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分类专栏：刷题题解文章标签： PAT 甲级 1008

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本文解析了PAT甲级1008题电梯问题，详细介绍了如何计算电梯在给定楼层请求列表中运行的总时间，包括电梯上下楼时间和在各楼层停留时间的计算方法。

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PAT甲级1008

题目
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:
41

题目意思是有个电梯，每上升一楼需要6秒，下降一楼需要4秒，在每个目的地需要停留5秒，问总共需要多少时间

思路
看输入，首先给个n，告诉你有多少个需求，并且题目强调，最开始电梯在第0楼，最后也不需要返回到第一楼，所以直接算就好了。不过有个坑，就是下一个需求还是该楼是还是需要停留5秒。

代码如下

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

int main()
{
    int n;
    while(~scanf("%d ",&n))
    {
        int k=0,sum=0;
    int hx[103];
        for(int i=0; i<n; i++)
        {
            scanf("%d",&hx[i]);
            if(hx[i]-k>0)
                sum=sum+(hx[i]-k)*6+5;
            else if(hx[i]-k<0)
                sum=sum+(k-hx[i])*4+5;
            else
                sum=sum+5;
            k=hx[i];
        }
        printf("%d\n",sum);
    }
    return 0;
}