Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

Source

Asia 2003（Seoul）

这个题 刚开始没思路，试想小时是从1---23两个指针都在变化，角度如何确定？看了看网上竟然没有多少相关资料！干脆自己想，试想可以借助12点，把小时和分钟和12点的夹角

之差即为两个表针的夹角，这样很容易表示结构体中的angle，但是有一个问题，如果小时大于12呢？这时候确定度数就要对12取余，这个过程中不能变动结构体中的数据成员，

否则影响angle相同下的hour排序。注意度数是实型，故wa一次！

Code：

#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;

struct clock
{
	int hour;
	int minute;
    double angle;//
}time[5];

bool cmp(clock V,clock X)//结构体多级排序
{
	if(V.angle!=X.angle)
		return V.angle<X.angle;
	else if((V.angle==X.angle)&&(V.hour==X.hour))
       return V.minute<X.minute;
	else
		return V.hour<X.hour;
}
int main()
{
	char ch;int T;
	cin>>T;

	while(T--)
	{
		for(int j=0;j<5;j++)
		{
			  cin>>time[j].hour>>ch>>time[j].minute;

			 double anglehour=( time[j].hour%12+time[j].minute*1.0/60)*30;
			  double anglemin=time[j].minute*6;
			  time[j].angle=abs(anglehour-anglemin)>180?(360-abs(anglehour-anglemin)):abs(anglehour-anglemin);
			}
			sort(time,time+5,cmp);
			printf("%02d:%02d\n",time[2].hour,time[2].minute);
	}
	return 0;
}