hdu1209（Clock）

點擊打開hdu1209

 

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：這題主要是求時鐘和分鐘夾角的大小，然後進行排序，而且還要注意一個小細節問題，當夾角相等時，時間小的放在前面。

夾角求法：分鐘旋轉一周要60分鐘，所以分鐘每分鐘旋轉360度除以60，為6度，而時鐘轉一周要12小時，一小時等於60分鐘，所以分鐘轉動的速度是時鐘轉動的12倍，即時鐘轉動速度為0,5度每分鐘。

從而得等公式r=h*30+0.5*m-m*6(h*30：代表時鐘准點的度數，而0.5*m：表示轉動m分鐘時，時鐘轉動的度數，二者相加即為時鐘的總的轉的角度，m*6則是分鐘轉動的角度)

 

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;itime[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(){
		
	}
}