1008. Elevator (20)

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本文介绍了一个简单的算法问题：计算电梯按指定楼层顺序运行所需的总时间。电梯从0楼出发，向上移动每层花费6秒，向下移动每层花费4秒，并在每个停靠层停留5秒。输入为一序列正整数，代表电梯需要停靠的楼层。

1008. Elevator (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

水题一道。。。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;

int main(){
	
	int N;
	scanf("%d",&N);
	int sum = 0;
	int floor,cur = 0;
	for(int i=1; i<=N; i++){
		scanf("%d",&floor);
		if(cur<floor)
			sum += 6*(floor-cur);
		else
			sum += 4*(cur-floor);
		sum += 5;
		cur = floor;
	}
	printf("%d\n",sum);
	return 0;
}