1008 Elevator (20)（20 分）

1008 Elevator (20)（20 分）

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:

在一座高楼上只有一个电梯，而后有 n 个电梯将要中止的楼层 电梯上升一层 用时 6 秒 ， 降低一层用时 4 秒， 在每层中止时 停下5秒
初始位置为 0 层， 当电梯中止在最后一个位置后再也不返回 0 层
问 电梯须要多长时间完成 n 个任务请求

41

#include <iostream>
#include <algorithm>

using namespace std; 

#define maxn 200

int n , floors[maxn] ; 

int main(){

    while(cin >> n ){
        for(int i=1 ; i<=n ; i++){
            cin >> floors[i] ; 
        }
        int sum = 0 ; 
        floors[0] = 0 ; // 第 0 层

        for(int i=1 ; i<=n ; i++){
            if(floors[i] - floors[i-1] >= 0 ){ // 上升
                sum += (floors[i] - floors[i-1]) * 6 ; 
            }else { // 降低
                sum += abs(floors[i] - floors[i-1]) * 4 ; 
            }
        }
        sum += n * 5 ; // 每层楼中止时 停下 5 秒
        
        cout << sum << endl ;
    }
    return 0 ; 
}