HDU1008

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 2
3 2 3 1
0
 


Sample Output
17

41

乘电梯，每升一层需要6秒，下降一层需要4秒，停留一层需要5秒，电梯从0层开始，输入一串数字，代表所需停留的楼层，
求电梯运行完的时间。

#include<iostream>using namespace std;int main(){int s[110], n;int total,t;while (cin >> n&&n!=0){total = 0;memset(s,0, sizeof(0));for (int i = 0; i < n; i++)cin >> s[i];total =total+ (s[0] * 6 + 5);for (int i = 1; i < n; i++){if (s[i] > s[i - 1]){t = s[i] - s[i - 1];total +=( t * 6 + 5);}else{t = s[i - 1] - s[i];total += (t * 4 + 5);}}cout << total << endl;}}