hdu1209(Clock)
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Problem Description
There is an analog clock with two hands: an hour hand and a minute hand. The two hands form an angle. The angle is measured as the smallest angle between the two hands. The angle between the two hands has a measure that is greater
than or equal to 0 and less than or equal to 180 degrees.
Given a sequence of five distinct times written in the format hh : mm , where hh are two digits representing full hours (00 <= hh <= 23) and mm are two digits representing minutes (00 <= mm <= 59) , you are to write a program that finds the median, that is, the third element of the sorted sequence of times in a nondecreasing order of their associated angles. Ties are broken in such a way that an earlier time precedes a later time.
For example, suppose you are given a sequence (06:05, 07:10, 03:00, 21:00, 12:55) of times. Because the sorted sequence is (12:55, 03:00, 21:00, 06:05, 07:10), you are to report 21:00.
Given a sequence of five distinct times written in the format hh : mm , where hh are two digits representing full hours (00 <= hh <= 23) and mm are two digits representing minutes (00 <= mm <= 59) , you are to write a program that finds the median, that is, the third element of the sorted sequence of times in a nondecreasing order of their associated angles. Ties are broken in such a way that an earlier time precedes a later time.
For example, suppose you are given a sequence (06:05, 07:10, 03:00, 21:00, 12:55) of times. Because the sorted sequence is (12:55, 03:00, 21:00, 06:05, 07:10), you are to report 21:00.
Input
The input consists of T test cases. The number of test cases (T) is given on the first line of the input file. Each test case is given on a single line, which contains a sequence of five distinct times, where times are given in the
format hh : mm and are separated by a single space.
Output
Print exactly one line for each test case. The line is to contain the median in the format hh : mm of the times given. The following shows sample input and output for three test cases.
Sample Input
3 00:00 01:00 02:00 03:00 04:00 06:05 07:10 03:00 21:00 12:55 11:05 12:05 13:05 14:05 15:05
Sample Output
02:00 21:00 14:05
思路:这题主要是求时钟和分钟夹角的大小。然后进行排序。并且还要注意一个小细节问题。当夹角相等时,时间小的放在前面。
夹角求法:分钟旋转一周要60分钟,所以分钟每分钟旋转360度除以60,为6度,而时钟转一周要12小时,一小时等于60分钟,所以分钟转动的速度是时钟转动的12倍。即时钟转动速度为0,5度每分钟。
从而得等公式r=h*30+0.5*m-m*6(h*30:代表时钟准点的度数。而0.5*m:表示转动m分钟时,时钟转动的度数,二者相加即为时钟的总的转的角度。m*6则是分钟转动的角度)
import java.util.Scanner; public class P1209 { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int t=sc.nextInt(); while(t-->0){ String s; String[] str; int h,m; Time[] time=new Time[5]; for(int i=0;i<5;i++){ s=sc.next(); str=s.split(":"); h=Integer.parseInt(str[0]); m=Integer.parseInt(str[1]); time[i]=new Time(h,m); } sort(time); System.out.printf("%02d:%02d",time[2].h,time[2].m); System.out.println(); } } private static void sort(Time[] time) { for(int i=0;i<time.length-1;i++){ for(int j=0;j<time.length-1-i;j++){ if(time[j].r>time[j+1].r){ swap(time,j,j+1); }else if(time[j].r==time[j+1].r){ if(time[j].h>time[j+1].h){ swap(time,j,j+1); } } } } } private static void swap(Time[] time, int j, int i) { Time t=new Time(); t=time[j]; time[j]=time[i]; time[i]=t; } } class Time{ public int h; public int m; public double r; public Time(int h,int m){ this.h=h; this.m=m; setR(); } private void setR() { this.r=Math.abs(h%12*30.0+m*0.5-m*6.0); if(this.r>180){ this.r=360-this.r; } } public Time(){ } }
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