zoukankan      html  css  js  c++  java
  • HDU 1260 Tickets

     HDU 1260 Tickets

    kuangbin 专题十二:Problem H


    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 4219    Accepted Submission(s): 2128


    Problem Description
    Jesus, what a great movie! Thousands of people are rushing to the cinema. However,
    this is really a tuff time for Joe who sells the film tickets. He is wandering when could he go back home as early as possible.
    A good approach, reducing the total time of tickets selling, is let adjacent people buy tickets together.
     As the restriction of the Ticket Seller Machine, Joe can sell a single ticket or two adjacent tickets at a time.
    Since you are the great JESUS, you know exactly how much time needed for every person to buy a single ticket or two tickets for him/her.
     Could you so kind to tell poor Joe at what time could he go back home as early as possible?
     If so, I guess Joe would full of appreciation for your help.
     

    Input
    There are N(1<=N<=10) different scenarios, each scenario consists of 3 lines:
    1) An integer K(1<=K<=2000) representing the total number of people;
    2) K integer numbers(0s<=Si<=25s) representing the time consumed to buy a ticket for each person;
    3) (K-1) integer numbers(0s<=Di<=50s) representing the time needed for two adjacent people to buy two tickets together.
     

    Output
    For every scenario, please tell Joe at what time could he go back home as early as possible.
    Every day Joe started his work at 08:00:00 am. The format of time is HH:MM:SS am|pm.
     

    Sample Input
    2
    2
    20 25
    40
    1
    8
     

    Sample Output
    08:00:40 am
    08:00:08 am
    ///////////////////////////////


    d.K个人排队买票,单人买票花费Si时间,相邻两人一起买票花费Di时间,求售票所需最少时间。

    s.dp[i]表示前i个人买票所需最少时间

    dp[i]=min(dp[i-1]+S[i],dp[i-2]+D[i]);

    #include <cstdio>
    #include <iostream>
    #include <cstring>
    #include <algorithm>
    
    using namespace std ; 
    
    #define maxn 3000
    int main(){
        int t ; 
        scanf("%d" , &t) ; 
        while(t--){
            int n , num[maxn] , dp[maxn] , d[maxn];
            memset(num , 0 , sizeof(num)) ; 
            memset(dp , 0 , sizeof(dp)) ; 
            memset(d , 0 , sizeof(d)) ; 
            int hour = 8 , mintinue = 0 , sec=0 ; 
            scanf("%d" , &n) ; 
            for(int i=1 ; i<= n ; i++){
                scanf("%d" , &num[i]) ; 
            }
            for(int i=2 ; i<=n ; i++){
                scanf("%d" , &d[i]) ; 
            }
            dp[0] = 0 ; 
            dp[1] = num[1] ; 
            for(int i=2 ; i<=n ; i++){
                dp[i] = min(dp[i-1] + num[i] , dp[i-2] + d[i] ) ; 
            }
            
            sec = dp[n] %60 ; 
            mintinue = (dp[n]/60)%60 ; 
            hour = hour+(dp[n]/3600) ; 
            while(hour>=24){
                hour-=24 ;  
            } 
            printf("%02d:%02d:%02d " , hour , mintinue , sec ) ;
            if(hour<=12) printf("am
    ") ; 
            else printf("pm
    ") ;  
        }
        return 0 ; 
    } 



  • 相关阅读:
    win10 + uefi
    curl post请求方式
    Unable to process request: General SSLEngine problem.Unable to connect to neo4j at `localhost:7687`, because the certificate the server uses has changed.
    SpringMVC Ueditor1.4.3 未找到上传数据
    Unsupported major.minor version 52.0
    Spring官网改版后下载
    Linux防火墙(Iptables)的开启与关闭
    MySQLWorkbench里的稀奇事之timestamp的非空默认值
    秒杀系统
    java . 请在小于99999的正整数中找符合下列条件的数,它既是完全平方数,又有两位数字相同,如:144,676。
  • 原文地址:https://www.cnblogs.com/yi-ye-zhi-qiu/p/7667555.html
Copyright © 2011-2022 走看看