zoukankan      html  css  js  c++  java
  • 思维体操: HDU1008 Elevator

    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
    ",ans);
        }
    }
    

  • 相关阅读:
    Linux 常用命令--来自B站Up主codesheep
    如何区别调用python2和python3
    fastp 使用
    使用bash shell删除目录中的特定文件的3种方法
    python 正则表达式 finditer
    vcf format
    vcf文件(call variants得来的)怎么看变异是纯合还是杂合
    js Object.preventExtensions()
    js 对象的属性特征
    shell基础 以及 sed、awk
  • 原文地址:https://www.cnblogs.com/Pretty9/p/7347703.html
Copyright © 2011-2022 走看看