1008 Elevator(20)

本文介绍了一个关于电梯在特定请求列表中移动的时间计算问题。通过分析楼层间的移动方向和距离,结合电梯上下楼所需时间和停靠时间,计算出完成一系列楼层请求所需的总时间。

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1008. Elevator (20)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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

//
//  main.cpp
//  elevator
//
//  Created by Hui Du on 2017/10/1.
//  Copyright © 2017年 Hui Du. All rights reserved.
//

#include <iostream>
#include <string.h>
#include <cstdio>
using namespace std;
int main(int argc, const char * argv[]) {
    int n;
    cin >> n;

    int first, second;
    int time = 0;

    cin >> first;
    time += first*6 +5;

    for(int i = 1; i < n; i++){
        cin >> second;
        if(second > first){
            time += (second - first) * 6 + 5;
        }

        else {
            time += (first - second)*4 + 5;
        }

        first = second;
    }
    cout << time << endl;
    return 0;
}
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