zoukankan      html  css  js  c++  java
  • H

    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. 

    InputThere 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. 
    OutputFor 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
    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <string>
    #include <cstdio>
    #include <cstring>
    #include <fstream> 
    #include <sstream>
    using namespace std;
    #define MAXN 2003
    /*
    求解买票的最短时间,可以单独买票可以两个人合买,给出所有人单独买票和两个人合伙所需时间,最后
    输出时间注意一下格式即可
    一开始想的是区间DP,发现不对。
    dp[n] = max(dp[n-1]+a[n],dp[n-2]+a[n-1 到1])
    */
    int dp[MAXN],a[MAXN],b[MAXN];//a是单独买,b是两人一起!
    void Print(int time)
    {
        int hour = time/3600,minute = time/60%60,second = time%60;
        bool am = (hour<4)?true:false;
        if(!am) hour = (8+hour)%12;
        else hour = 8+hour;
        if(hour<10)
            printf("0%d:",hour);
        else
            printf("%d:",hour);
        if(minute<10)
            printf("0%d:",minute);
        else
            printf("%d:",minute);
        if(second<10)
            printf("0%d ",second);
        else
            printf("%d ",second);
        if(am)
            printf("am
    ");
        else
            printf("pm
    ");
    }
    int main()
    {
        int n,k;
        scanf("%d",&n);
        while(n--)
        {
            scanf("%d",&k);
            for(int i=1;i<=k;i++)
                scanf("%d",&a[i]);
            for(int i=1;i<k;i++)
                scanf("%d",&b[i]);
            dp[1] = a[1];
            for(int i=2;i<=k;i++)
                dp[i] = min(dp[i-1]+a[i],dp[i-2]+b[i-1]);
            Print(dp[k]);
        }
    }
  • 相关阅读:
    hibernate反向工程 (eclipse和myeclipse)【转】
    让你明白response.sendRedirect()与request.getRequestDispatcher().forward()区别
    Struts Tags
    在Eclipse中配置tomcat
    The Apache Tomcat Native library which allows optimal performance in production environments was not found on the java.library.path:
    如何将maven项目导入myeclipse中
    Hibernate配置文件详解
    网站怎么布局能解决不同浏览器对CSS解析的差异,使用css reset
    SqlServer:此数据库处于单用户模式,导致数据库无法删除的处理
    Myeclipse最全快捷键
  • 原文地址:https://www.cnblogs.com/joeylee97/p/6640961.html
Copyright © 2011-2022 走看看