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Problem

portal: 1008 Elevator (20 分)

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

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

For each test case, print the total time on a single line.

Sample

Sample Input

3 2 3 1

Sample Output

41

Solution

Analysis

?电梯向上走一层需要6s，向下走一层需要4s，每一次停止时停留5s，最开始电梯是在第0层，结束后电梯无需返回底层。所以在给出了列表之后，只需要判断电梯是向上还是向下，然后用对应的楼层差值乘所需要的时间，便是电梯移动需要的时间花费，最后加上停留需要的时间，将列表中的每一个楼层都处理完，便是整体所需要的时间。

Code

#include <bits/stdc++.h>
using namespace std;

int main(void) {
    int n, last, now; // last是上一次停留的楼层, now是这次要去的楼层
    int result = 0;

    cin >> n; 
    last = 0;
    for (int i = 0; i < n; i++) {
        cin >> now;
        if (now > last) { // 此时电梯向上运动
            result += 6 * (now - last) + 5;
        } else { // 此时电梯向下运动
            result += 4 * (last - now) + 5;
        }
        last = now; // 当前停留楼层更新
    }
    cout << result << endl;
}

[PTA] PAT(A) 1008 Elevator (20 分)

标签：last   second   test   solution   class   note   problems   line   order   

原文地址：https://www.cnblogs.com/by-sknight/p/11444492.html