PAT-A1008 Elevator 题目内容及题解

最新推荐文章于 2022-11-13 18:09:59 发布

Bourbon_Whiskey_

最新推荐文章于 2022-11-13 18:09:59 发布

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分类专栏： PAT甲级文章标签： PAT 甲级 OJ 编程学习

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这是一道来自PAT甲级编程考试的题目，要求模拟电梯根据给定的楼层请求运行并计算总耗时。电梯初始位于0楼，上楼每层用时6秒，下楼每层4秒，每停靠一层等待5秒。输入包含一个请求列表，输出电梯完成所有请求的总时间。给定示例输入为3 2 3 1，输出为41秒。

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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

题目大意

模拟电梯，题目给出一个请求序列，求出某电梯完成所有请求所需要的时间。

解题思路

按照给定条件模拟电梯运行即可。

代码

#include<stdio.h>

int main(){
    int N;
    int i;
    int current=0,goal,time=0;
    scanf("%d",&N);
    for(i=0;i<N;i++){
        scanf("%d",&goal);
        while(current>goal){
            current--;
            time+=4;
        }
        while(current<goal){
            current++;
            time+=6;
        }
        time+=5;//wait 
    }
    printf("%d\n",time);
    return 0;
}