1008 Elevator （20 分）（模拟）

最新推荐文章于 2020-03-23 14:18:24 发布

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本文介绍了一个基于电梯调度的算法模拟问题，通过输入一系列楼层请求，计算电梯完成所有请求所需的总时间。考虑到电梯上下楼的时间成本及停靠时间，提供了一段C++代码实现，适用于城市中最高建筑仅有一部电梯的情况。

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:

题解：看懂题意模拟过程即可，水题

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>

using namespace std;

int main() {
	long long int sum=0;
	int n;

	while(scanf("%d",&n)!=EOF) {
		int a[10005]= {0};
		sum=5*n;
		scanf("%d",&a[0]);
		sum+=6*a[0];
		for(int t=1; t<n; t++) {
			scanf("%d",&a[t]);
			if(a[t]-a[t-1]>0) {
				sum+=(a[t]-a[t-1])*6;
			} else {
				sum+=abs(a[t]-a[t-1])*4;
			}
		}
		cout<<sum<<endl;
	}
	return 0;
}