Elevator (20)

最新推荐文章于 2020-07-18 18:54:18 发布

原创最新推荐文章于 2020-07-18 18:54:18 发布 · 255 阅读

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本文深入探讨了一种电梯调度算法的实现细节，通过分析电梯在不同楼层间的移动成本，计算出完成一系列请求所需的总时间。文章提供了一个具体的代码示例，展示了如何根据电梯上行和下行的耗时差异来优化电梯调度，确保高效地处理所有楼层请求。

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Elevator (20)

时间限制 1000 ms 内存限制 32768 KB 代码长度限制 100 KB 判断程序 Standard (来自小小)

题目描述

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.

输入描述:

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.

输出描述:

For each test case, print the total time on a single line.

输入例子:

3 2 3 1

输出例子:

#include<stdio.h>
int main()
{
	int n,i,sum;
	int a[105]; 
	scanf("%d",&n);
	for(i=0;i<n;i++)
		scanf("%d",&a[i]);
	sum=6*a[0]+5;
	for(i=1;i<n;i++)
	{
		if(a[i]>a[i-1])
		   sum+=(a[i]-a[i-1])*6+5;
		else
		 sum+=(a[i-1]-a[i])*4+5;
	}
	printf("%d\n",sum);
	return 0;
}

太水了。。。。还是记录一下吧