C++ Pat甲级1008 Elevator （20 分）_1008 elevator (20 分) c++-优快云博客

本文介绍了一个关于电梯运行时间计算的问题，电梯在城市最高建筑中仅有一部，根据给定的楼层请求列表，计算电梯完成所有请求所需的总时间，包括上行、下行及停靠时间。

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1008 Elevator （20 分）

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:

输入：第一个数 n, 接下来再输入 n个数表示每次电梯需到达的楼层

输出：电梯在此期间经历的时间

题目中给出上升一层6秒，下降一层 4 秒，每层停留5秒

#include <iostream>
#include <vector>
using namespace std;
int main() {
	int n,temp=0;
	scanf("%d", &n);
	vector<int> v(n);
	int stop_time = n * 5;//中间停留的时间
	int run_time;
	int sum_time = 0;
	scanf("%d", &v[0]);
	temp =v[0];
	run_time =v[0]*6;//电梯上升到第一个位置的时间
	for (int i = 1; i < n; i++) {
		scanf("%d", &v[i]);
		if (v[i] > temp) {
			run_time += (v[i]-temp) * 6;
		} else {
			run_time += (temp-v[i] )* 4;
		}
		temp = v[i];
	}
	sum_time = run_time+stop_time;
	printf("%d",sum_time);
	return 0;
}