Tickets-白红宇

Tickets

阅读量：192 次

发布时间：2019-02-28

本文共 2318 字，大约阅读时间需要 7 分钟。

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:

An integer K(1<=K<=2000) representing the total number of people;

K integer numbers(0s<=Si<=25s) representing the time consumed to buy a ticket for each person;

(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

题目大意：有一个买票的人Joe，今天要卖n张票出去，假如你是上帝，你知道今天买票的所有人买一次一张票和一次买两张票需要的时间。让你求出Joe最早回家的时间。

题目分析:

状态表示：f[i] //表示卖出第i张票时所需的最短时间

状态计算：有两种状态划分方法
1）当前的最后一个客人一次买了一张票，f[i]=f[i-1]+a[i]
2）当前最后一个客人一次买了两张票，f[i]=f[i-2]+b[i-1]
这两种划分取一个最小值即可

代码如下：

#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        #include 
         
          #include 
          
           #include 
           
            #define LL long longconst int N=1e4+5;using namespace std; int main(){ int t; //有多组测试数据 cin>>t; while(t--) { int n,a[N]={ 0},b[N]={ 0},f[N]={ 0}; cin>>n; for(int i=1;i<=n;i++) //a[i]存储卖出第i张票(一张)花的时间 cin>>a[i]; for(int i=1;i
            
             >b[i]; f[1]=a[1]; //只卖出第1张票的时间就是a[1] for(int i=2;i<=n;i++) //递推求解答案 { f[i]=min(f[i-1]+a[i],f[i-2]+b[i-1]); } int time=f[n],h,m,s; //将得到的秒数转化为时分秒 h=time/3600+8; //8点开始上班 time%=3600; m=time/60; time%=60; s=time; if(h<10) printf("0%d:",h); //当数值小于10时要手动加0 else printf("%d:",h); if(m<10) printf("0%d:",m); else printf("%d:",m); if(s<10) printf("0%d am\n",s); else printf("%d am\n",s); } return 0;}