1008. Elevator (20)

最新推荐文章于 2021-08-18 12:42:21 发布

orange_melon

最新推荐文章于 2021-08-18 12:42:21 发布

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分类专栏： pat 文章标签： PAT甲级 1008. Elevator

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本文针对一个只有单一电梯的高楼场景，解析了一种电梯调度算法。该算法通过计算不同楼层请求间的移动时间及停留时间，来确定完成所有请求所需的总时间。输入为一系列正整数的楼层请求列表，输出则是完成这些请求所需的总秒数。

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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:

#include<cstdio>
using namespace std;

int main()
{
	int n,now,i,x;
	int sum=0;
	scanf("%d",&n);
	now=0;
	while (n--)
	{
		scanf("%d",&x);
		//if (x==now) continue;
		if (x>now){
			sum+=(x-now)*6+5;
		}
		else {
			sum+=(now-x)*4+5;
		}
		now=x;
	}
	printf("%d\n",sum);
}

不清楚连续在同一楼层会怎么样。如果当前楼层等于上一楼层直接continue的话，只能拿14分，去掉这个判断便AC