ACM HDU 1260 Tickets

Tickets

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 385    Accepted Submission(s): 192

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

 

Source

浙江工业大学第四届大学生程序设计竞赛

 

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JGShining

 

#include<stdio.h>
#include<algorithm>
using namespace std;
const int MAXN=2010;
int a[MAXN];//存放单个人买票的时间 
int b[MAXN];//存放两个相邻的人一起买票的时间
int dp[MAXN];//dp[i]表示前i个人买票需要的最少时间
int main()
{
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    int T,n,i,j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
           scanf("%d",&a[i]);
        for(i=2;i<=n;i++)
          scanf("%d",&b[i]);
        dp[0]=0;
        dp[1]=a[1];
        for(i=2;i<=n;i++)
          dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i]);
        int h=dp[n]/60/60+8;
        int m=dp[n]/60%60;
        int s=dp[n]%60;
        char hh[3],mm[3],ss[3];
        hh[0]=h/10+'0';hh[1]=h%10+'0';hh[2]='\0';
        mm[0]=m/10+'0';mm[1]=m%10+'0';mm[2]='\0';
        ss[0]=s/10+'0';ss[1]=s%10+'0';ss[2]='\0';
        if(h<=12) printf("%s:%s:%s am\n",hh,mm,ss);
        else printf("%s:%s:%s pm\n",hh,mm,ss);
    }    
    return 0;
}  

#include<stdio.h>
#include<algorithm>
#include<iomanip>
#include<iostream>
using namespace std;
const int MAXN=2010;
int a[MAXN];//存放单个人买票的时间 
int b[MAXN];//存放两个相邻的人一起买票的时间
int dp[MAXN];//dp[i]表示前i个人买票需要的最少时间
int main()
{
    //freopen("test.in","r",stdin);
    //freopen("test.out","w",stdout);
    int T,n,i,j;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(i=1;i<=n;i++)
           scanf("%d",&a[i]);
        for(i=2;i<=n;i++)
          scanf("%d",&b[i]);
        dp[0]=0;
        dp[1]=a[1];
        for(i=2;i<=n;i++)
          dp[i]=min(dp[i-1]+a[i],dp[i-2]+b[i]);
        int h=dp[n]/60/60+8;
        int m=dp[n]/60%60;
        int s=dp[n]%60;
        /*char hh[3],mm[3],ss[3];
        hh[0]=h/10+'0';hh[1]=h%10+'0';hh[2]='\0';
        mm[0]=m/10+'0';mm[1]=m%10+'0';mm[2]='\0';
        ss[0]=s/10+'0';ss[1]=s%10+'0';ss[2]='\0';
        if(h<=12) printf("%s:%s:%s am\n",hh,mm,ss);
        else printf("%s:%s:%s pm\n",hh,mm,ss);*/
        /*或者用下面的方法输出前面加iomanip头文件  */
        if(h<=12)
            cout<<setfill('0')<<setw(2)<<h<<":"<<setfill('0')<<setw(2)<<m<<":"<<setfill('0')<<setw(2)<<s<<" am"<<endl; 
        else 
           cout<<setfill('0')<<setw(2)<<h<<":"<<setfill('0')<<setw(2)<<m<<":"<<setfill('0')<<setw(2)<<s<<" pm"<<endl; 
        
        
        
       
    }    
    return 0;
}