HDOJ 1008 Elevator_hdoj1008 java-优快云博客

本文链接：https://blog.youkuaiyun.com/RaAlGhul/article/details/50651410

本文探讨了电梯算法的时间复杂度计算，包括电梯上下楼层的运动时间和停留时间的综合影响，通过输入楼层请求列表计算完成请求所需总时间。

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Elevator

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58979 Accepted Submission(s): 32379

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

Sample Output

  
   17
41

Author

ZHENG, Jianqiang

Source

ZJCPC2004

题意：给你一个N，表示要停靠的楼层数量，再给你N个要停靠的楼层，从第0层开始(注意，这里有违反常理的地方，是从第0层开始的而不是从第一层开始的)，每上一层楼花费6秒，每下一层楼花费4秒，每停靠一次花费5秒，问总共需要多少时间。

题意简单明确，直接进行模拟即可。

#include<stdio.h>
int main()
{
	int n;
	while(scanf("%d",&n)&&(n!=0))
	{
		int s=n*5,t=0;
		while(n--)
		{
			int a,b,c;
			scanf("%d",&a);
			if((a-t)>0)
				s=s+(a-t)*6;	
			if((a-t)<0)
				s=s+(t-a)*4;
			t=a;
		}
		printf("%d\n",s);
	}
}