思维体操: HDU1008 Elevator

最新推荐文章于 2020-05-11 22:06:23 发布

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最新推荐文章于 2020-05-11 22:06:23 发布

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文章标签： java

原文链接：http://www.cnblogs.com/Pretty9/p/7347703.html

本文介绍了一个关于电梯算法的问题，通过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。考虑到电梯上行和下行的时间成本不同，以及在每层停留的时间。

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Elevator

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 73602 Accepted Submission(s): 40513

Problem 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
  

Author

ZHENG, Jianqiang

Problem : 1008 ( Elevator ) Judge Status : Accepted
RunId : 21242476 Language : G++ Author : hnustwanghe
Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta

#include<iostream> #include<cstring> #include<cstdio> using namespace std; int main(){ int n,ans,last,now,i; while(scanf("%d",&n)==1 &&n){ for(ans = last = 0,i=0;i<n;i++){ scanf("%d",&now); if(now - last > 0){ ans += (now - last) * 6; last = now; } else if(now - last < 0){ ans += (last - now) * 4; last = now; } } ans = ans + n*5; printf("%d\n",ans); } }

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

using namespace std;

int main(){
    int n,ans,last,now,i;
    while(scanf("%d",&n)==1 &&n){
        for(ans = last = 0,i=0;i<n;i++){
            scanf("%d",&now);
            if(now - last > 0){
                ans += (now - last) * 6;
                last = now;
            }
            else if(now - last < 0){
                ans += (last - now) * 4;
                last = now;
            }
        }
        ans = ans + n*5;
        printf("%d\n",ans);
    }
}