PAT甲级真题 1008 Elevator (20分) C++实现(测试点2、3、6有坑)

PAT甲级试题1008:Elevator C++解决方案
该博客介绍了PAT甲级考试中的一道编程题——1008 Elevator,讨论了如何计算电梯在给定停靠楼层完成请求所需的时间。作者指出,电梯在每个楼层停留5秒,上一层楼花费6秒,下一层楼花费4秒,且即使楼层重复也要计算停留时间。博主提供了问题的思路和解决方案,并给出了一个测试案例及其输出结果。

题目

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

思路

C++源代码 注重类的交互 片段 #include using namespace std; #include "elevator.h" //Elevator class definition #include "person.h" //Person class definition #include "floor.h" //Floor class definition //constants that represent time required to travel //between floors and direction of the elevator const int Elevator::ELEVATOR_TRAVEL_TIME = 5; const int Elevator::UP = 0; const int Elevator::DOWN = 1; //constructor Elevator::Elevator( Floor &firstFloor, Floor &secondFloor) : elevatorButton( * this ), currentBuildingClockTime( 0 ), moving( false ), direction( UP ), currentFloor( Floor::FLOOR1 ), arrivalTime( 0 ), floor1NeedsService( false ), floor2NeedsService( false ), floor1Ref( firstFloor ), floor2Ref( secondFloor ), passengerPtr( 0 ) { cout << "elevator constrcuted" <<endl; }// end Elevator constructor //destructor Elevator::~Elevator() { delete passengerPtr; cout << "elevator destructed" << endl; }//end Elevator destructor //give time to elevator void Elevator::processTime( int time ) { currentBuildingClockTime = time; if ( moving ) //elevator is moving processPossibleArrival(); else processPossibleDeparture(); if ( !moving ) cout << "elevator at rest on floor " << currentFloor << endl; }// end function processTime // when elevator is moving, determine if it should stop void Elevator::processPossibleArrival() { //if elevator arrives at destination floor if ( currentBuildingClockTime == arrivalTime ) { currentFloor = ( currentFloor == Floor::FLOOR1 ? Floor::FLOOR2 : Floor::FLOOR1); //update current floor direction = ( currentFloor == Floor::FLOOR1 ? UP : DOWN ); //update direction cout << "elevator arrives on floor " << currentFloor <<endl; // process arrival at currentFloor arriveAtFloor( currentFloor == Floor::FLOOR1 ? floor1Ref : floor2Ref); return; }//end if //elevator still moving cout << "elevator moving " << ( direction == UP ? "UP" : "DOWN"
### PAT 甲级 1034 测试点 4 的解析 PAT 甲级 1034 题目名为 **"Elevator"**,主要涉及电梯运行过程中的时间计算以及路径规划问题。对于测试点 4 的注意事项和可能的陷阱如下: #### 输入数据范围 测试点 4 可能会包含大量的输入数据或者极端情况下的边界条件。例如,可能存在非常大的楼层差值或较多的人数请求。因此,在实现算法时需特别关注变量的数据类型选择,建议使用 `long` 或者更大的整型来存储中间结果以防止溢出[^1]。 #### 边界条件处理 该测试点可能会考察一些特殊的边界场景,比如所有人都在同一层楼等待的情况、目标楼层为负数或者是零等情况。程序应该能够正确识别并合理处理这些特殊情况而不应报错或返回错误的结果。 #### 时间复杂度优化 由于此题涉及到多次循环迭代来进行模拟运算,如果效率不够高,则容易超时。针对这一点,可以考虑采用更高效的数据结构如优先队列 (priority queue) 来管理乘客请求顺序,从而减少不必要的比较操作次数,提高整体性能表现。 以下是基于上述析的一个 Python 实现方案: ```python def elevator(): n = int(input()) floors = list(map(int, input().split())) total_time = abs(floors[0]) * 6 current_floor = floors[0] for i in range(1, len(floors)): diff_floors = abs(current_floor - floors[i]) if floors[i] > current_floor: total_time += diff_floors * 6 + 5 elif floors[i] < current_floor: total_time += diff_floors * 4 + 5 else: continue current_floor = floors[i] print(total_time) elevator() ``` 以上代码片段展示了如何通过遍历每一对相邻的目标楼层之间的距离,并根据不同方向调整所需的时间增量完成整个行程耗时统计的任务逻辑。
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