hdu 1008 Elevator

最新推荐文章于 2020-02-09 09:15:44 发布

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本文探讨了在特定建筑中，电梯从一个楼层移动到另一个楼层的最优调度策略，包括上行、下行以及停靠时间的计算。通过输入一系列楼层请求，算法计算完成所有请求所需的总时间。

http://acm.hdu.edu.cn/showproblem.php?pid=1008

1、题目描述：

Elevator

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status Practice ZOJ 2108

Description

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

There are multiple test cases. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100. A test case with N = 0 denotes the end of input. This test case is not to be processed.

Output

Print the total time on a single line for each test case.

Sample Input

1 2
3 2 3 1
0

Sample Output

17
41

2、代码：

#include<stdio.h>
int a[105];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
        break;
        int sum=0;
        for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
        {
            if(a[i]-a[i-1]>0)
            sum+=(a[i]-a[i-1])*6;
            else
            sum+=(a[i-1]-a[i])*4;
        }
        sum+=5*n;
        printf("%d\n",sum);
    }
    return 0;
}