杭电第1008题-----Elevator

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 23 2 3 10

 

Sample Output

1741
/*这题目一开始我没看清题目意思，以为1 2 两个数据，其实1 是说1后面只有一个数字2；3后面有三个数字2 3 1； 然后怎么折腾都弄不出来，后来上网查了代码，就这代码看的比较舒服，这算是我的杭电的第一题吧，虽然没做出来                                                                ____________2015/3/1  */
#include<iostream>
<span style="font-size:18px;"><strong>using namespace std;int main(){int i,n,sum,num[101];<span style="color:#6600cc;">/*利用数组存输入的数据，还便于比较前后数据的大小关系</span>while(cin>>n&&n)<span style="color:#3366ff;">/*表示不能以0为开头，但可以输入多个数据的一个条件</span>{sum=0;for(i=0;i<n;i++){cin>>num[i];if(i==0)sum=sum+num[i]*6+5;<span style="color:#6633ff;">/*一开始看不懂，其实就是从0层升到num[i]层所用的时间</span>else if(num[i]>num[i-1])sum=sum+(num[i]-num[i-1])*6+5;else sum=sum+(num[i-1]-num[i])*4+5;}cout<<sum<<endl;}  return 0;}</strong></span>