problem F

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

1 2
3 2 3 1
0

 

Sample Output

17
41
简单题意：
  在一栋大楼里面只有一台电梯，电梯每爬升一层需要6秒的时间，下降一层需要4秒的时间，并且在每一层停留的时间是5秒钟。给出一串要求记录，编写一个算法求所需的时间。
解题思路形成过程：
  这同样也是一个特别简单的算法。给出的数据，如果后一个的层数比前一个大，a[i+1]-a[i]乘以6秒即可，同理，小的话，乘以4.最后在每层停留的时间都是5秒，乘以n个要求记录即可。
感想：
  算法一直没错，在输入那块有点问题，所以导致一直WA，最后while（cin>>n&&n!=0）还有total+=x全部改为total=total+x,AC了。网上评判系统，也是够了。
AC代码：
#include <iostream>
using namespace std;
int main()
{
    int n,a[100];
    while(cin>>n&&n!=0)
    {
        for(int i=0;i<n;i++)
            cin>>a[i];
        int total=a[0]*6;
        for(int i=1;i<n;i++)
        {
            if(a[i]>a[i-1])
                {
                total=total+(a[i]-a[i-1])*6;
            }
            else total=total+(a[i-1]-a[i])*4;
        }
        total=total+n*5;//
        cout<<total<<endl;
    }
    return 0;
}