題目描述：
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.

Input Specification:
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.

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

Sample Input:
3 2 3 1
Sample Output:
41

題目大意：
剛開始在0層，給你一個電梯需要到達層數的順序，上升一層用6秒，下降一層用4秒，一層要停5秒，最後計算總的消耗時間。
這個題感覺有問題，如果連續兩個是在同一層下的話，它停留時間按的10秒（5*2）算的。其中的三個測試點都是這個問題。

程式碼如下：

#include <iostream>
using namespace std;
 

int main()
{
    int k,sum=0,now=0;
    cin>>k;
    while(cin>>k){
        if(now<k) sum+=6*(k-now)+5;
        else  sum+=4*(now-k)+5;
        now=k;
    }
    cout<<sum<<endl;
    return 0;
}

原本寫的(如果連續好幾個人都是一層的話，只按5秒計算)：
其實題意是每有一個人就多5秒。

 if(now<k) sum+=6*(k-now)+5;
 else if(now>k) sum+=4*(now-k)+5;