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:
    async await 了解
    vi 命令
    mysql 相关操作
    mac下配置python的虚拟环境virtualenv和虚拟环境管理包virtualenvwrapper
    ip 域名 和端口号
    脱离 flask 上下文,使用 jinja2 来动态渲染模板
    使用 vue-cli 3.0 创建项目
    p 标签和 span 标签
    el-table 更改表格行高和列髋
    使用 axios 传参问题
  • 原文地址:https://www.cnblogs.com/yi-ye-zhi-qiu/p/7667555.html
Copyright © 2011-2022 走看看