PAT.A1008. Elevator

最新推荐文章于 2022-02-07 22:19:51 发布

克拉拉123

最新推荐文章于 2022-02-07 22:19:51 发布

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本文链接：https://blog.youkuaiyun.com/mokena1/article/details/80013219

本文探讨了一个只有单一电梯的高楼中如何高效处理楼层请求的问题。通过分析电梯上下行时间和停靠时间，提出了一种计算总时间消耗的方法，并给出了具体的实现代码。

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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:

#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <iostream>
using namespace std;

int main(){
	int n;
	int sum=0,fl,t=0;
	scanf("%d", &n);
for (int i = 0; i < n; i++) {
		scanf("%d", &fl);
		if (fl > t) sum +=(fl-t)* 6 + 5;
		else sum += (t-fl) * 4 + 5;
		t = fl;
	}
printf("%d", sum);

	return 0;
	
}