杭电-1260 Tickets

最新推荐文章于 2022-07-22 01:11:50 发布

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最新推荐文章于 2022-07-22 01:11:50 发布

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Tickets

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3251 Accepted Submission(s): 1608

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

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

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题意就是一队人排队买票，先是输出n个数，代表每个人需要花费的时间；然后再输入n-1个数，代表两个人和在一起买票需要的时间；

动态规划方程：dp[i] = min(dp[i-1] + a[i], dp[i-2] + b[i]);

注意输出的时间，开始是8点；

#include<stdio.h> 
#include<string.h>
#include<algorithm>
using namespace std;
int a[2200];
int b[2200];
int dp[2200];
int main()  
{  
    int t;
	int n;  
    scanf("%d", &t);  
    while(t--)  
    {  
        scanf("%d", &n);  
        for(int i = 1; i <= n; i++)  
            scanf("%d", &a[i]);  
        for(int i = 2; i <= n; i++)  
            scanf("%d", &b[i]);  
        memset(dp, 0, sizeof(dp));  
        dp[1] = a[1];  
        for(int i = 2; i <= n; i++)  
            dp[i] = min(dp[i-1] + a[i], dp[i-2] + b[i]);  
        int t = 28800 + dp[n];  
        int h = t / 3600;  
        int m = t % 3600 / 60;  
        int s = (t % 3600) % 60;  
        if(t >= 43200)  
        {  
            if(h> 12)  
                h-= 12;  
            printf("%02d:%02d:%02d pm\n", h, m, s);  
        }  
        else  
            printf("%02d:%02d:%02d am\n", h, m, s);  
    }  
    return 0;  
}