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.

 

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(){
		
	}
}