Elevator http://acm.hdu.edu.cn/showproblem.php?pid=1008-优快云博客

本文介绍了一个简单的算法问题：计算电梯完成一系列楼层请求所需的总时间。电梯向上移动一层需要6秒，向下移动一层需要4秒，并且在每个停留的楼层会等待5秒。文章通过示例解释了如何计算总时间。

Elevator

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 34433 Accepted Submission(s): 18770

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

Author

ZHENG, Jianqiang

Source

ZJCPC2004

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#include<stdio.h>
int main()
{
	int n;
	while(scanf("%d",&n),n!=0)
	{
		int i,m,start=0,sum=0;
		for(i=0;i<n;i++)
		{
			scanf("%d",&m);
			if(m>start)
				sum+=6*(m-start)+5;
			else
			{
				sum+=4*(start-m)+5;
			}
			start=m;
		}
		printf("%d\n",sum);
	}
	return 0;
}

题目比较简单，意思就是坐电梯，上去一层用6秒，如果中间停了，时间间隔就是5秒，下去一层用4秒。第一个数据n为输入的个数，后面是n个数。第一个层数从0层开始算。

转载于:https://www.cnblogs.com/wangyouxuan/p/3265745.html