HDU--Elevator（水题）

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本文介绍了一个简单的编程问题：计算电梯按指定楼层顺序运行所需的总时间。电梯从0楼出发，根据输入的楼层列表向上或向下移动，每经过一层楼都有相应的停留时间和移动时间。

Elevator

Time Limit: 1000ms Memory limit: 32768K 有疑问？点这里^_^

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

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.

输出

Print the total time on a single line for each test case.

示例输入

1 2
3 2 3 1
0

示例输出

17
41

水题，but坑了我一顿，本渣以为这电梯上到顶就会下来不在上去了呢。。。所以一开始开数组做的WA了

#include <iostream>
#include <cstring>
#include <algorithm>
#include <stdlib.h>
#include <vector>
using namespace std;
int main()
{
  int n,i,start,x,sum;
  while(cin>>n)
  {
  	if(!n)break;
  	sum=0;start=0;
  	for(i=1;i<=n;i++)
	{
		cin>>x;
		if(x>start) sum+=6*(x-start);
		else sum+=4*(start-x);
		sum+=5;
		start=x;
	}
	cout<<sum<<endl;
  }
  return 0;
}