HDU 1108 Elevator

最新推荐文章于 2020-02-09 09:15:44 发布

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最新推荐文章于 2020-02-09 09:15:44 发布

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本文介绍了一个关于电梯路径的时间计算问题，电梯在不同楼层间的移动时间（上行6秒/层，下行4秒/层），以及在每层停留5秒的时间。通过模拟电梯运行过程，计算并输出完成一系列楼层请求所需的总时间。

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Elevator

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 58340 Accepted Submission(s): 31975

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

题意：电梯上行一层花6s，下行一层花4s，停留花5s，计算总时间。

模拟一下，在输入的同时计算就可以。

#include <stdio.h>
int main()
{
	int T;
	int sum;
	int i,a[2];
	
	while(scanf("%d",&T),T)
	{
		a[0]=0;
		sum=5*T;
		for(i=1;i<=T;i++)
		{
			scanf("%d",&a[1]);
			if(a[1]>a[0])
				sum+=(a[1]-a[0])*6;
			else if(a[0]>a[1])
				sum+=(a[0]-a[1])*4;
			a[0]=a[1];
		}
		printf("%d\n",sum);
	}
	return 0;
}