PAT-1008 Elevator

1008 Elevator （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:

41

算法說明：

不要存儲好樓層再計算，很差控制究竟是上升仍是降低，會得不全分的。

// 1008 Elevator.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
using namespace std;

int main(int argc, char* argv[])
{
    int floor,stay=0;
    int N;
    int i,time=0;
    
    cin >>N;
    for(i=1;i<=N;i++){
        cin >>floor;
        if(stay==floor){
            time+=5;
            continue;
        }else if(floor>stay){
            time+=(floor-stay)*6+5;
        }else{
            time+=(stay-floor)*4+5;
        }
        stay=floor;
    }
    cout << time <<endl;
    return 0;
}

2020考研打卡第十天，你可知道星辰之變，驕陽豈是終點？美好生活等着咱們呢zyy&cdl 加油！！！
努力不必定有收穫，可是不努力絕對沒有收穫，天上也沒有掉餡餅那麼美好的事情，出來混老是要還的。