PAT甲级 1008.Elevator(20) 题目翻译与答案

题目来源自PAT网站  https://www.patest.cn/

题目描述：

1008. Elevator (20)

The highestbuilding in our city has only one elevator. A request list is made up with Npositive numbers. The numbers denote at which floors the elevator will stop, inspecified order. It costs 6 seconds to move the elevator up one floor, and 4seconds to move down one floor. The elevator will stay for 5 seconds at eachstop.

For a given requestlist, you are to compute the total time spent to fulfill the requests on thelist. The elevator is on the 0th floor at the beginning and does not have toreturn to the ground floor when the requests are fulfilled.

InputSpecification:

Each input filecontains one test case. Each case contains a positive integer N, followed by Npositive numbers. All the numbers in the input are less than 100.

OutputSpecification:

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

SampleInput:

3 2 3 1

SampleOutput:

41

题目翻译：

1008.电梯

在我们的城市里，最高的建筑物里只有一部电梯。有一份由N个正数组成的请求列表。这些数表示电梯将会以规定的顺序在哪些楼层停下。电梯升高一层需要6秒，下降一层需要4秒。每次停下电梯将花费5秒。

给你一个请求列表，你需要计算出完成列表里的请求总共花费的时间。一开始电梯在0层。当请求全部完成时，电梯不需要回到底层。

输入说明：

每个输入文件包含一个测试实例。每个实例包含一个正整数N，后面跟着N个数字。所有输入的数字小于100。

输出说明：

对于每个测试实例，在一行中输出总时间。

样例输入：

3 2 3 1

样例输出：

41

答案代码：

#include<cstdio>
int main()
{
	int stay=0,t;
	int N,i,total=0;
	scanf("%d",&N);
	for(i=0;i<N;++i)
	{
		scanf("%d",&t);
		if(t==stay)
		{
			total+=5;
			continue;
			
		}
		if(t>stay)
			total+=(t-stay)*6+5;
		else
			total+=(stay-t)*4+5;
		stay=t;
	}
	printf("%d",total);
	return 0;
}