1008 Elevator (20)

最新推荐文章于 2021-08-18 12:42:21 发布

原创最新推荐文章于 2021-08-18 12:42:21 发布 · 139 阅读

0 ·

CC 4.0 BY-SA版权

PAT甲级专栏收录该内容

54 篇文章

订阅专栏

本文介绍了一个电梯调度算法的具体实现过程，该算法通过计算不同楼层请求之间的移动时间来确定总时间消耗。文章提供了一段C++代码示例，展示了如何根据输入的楼层请求序列计算电梯完成所有请求所需的总时间。

1008 Elevator (20)
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
解题思路：
按照题目要求对输入序列进行处理即可。
用before记录前一层，初始为0，now记录要去的层，做差就知道要往哪个方向多少层，累加计算ans即可。

#include<cstdio> 
int main(){
    int n,before=0,now,ans=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&now);
        if(now>before){
            ans=ans+(now-before)*6+5;
        }else{
            ans=ans+(before-now)*4+5;
        }
        before=now;
    }
    printf("%d",ans);
    return 0;
}