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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
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思路：这题主要是求时钟和分钟夹角的大小，然后进行排序，而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)
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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[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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