2013
12-04

# Clock

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.

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.

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.

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

02:00
21:00
14:05

mark：题目很简单，角度也很好算：时针每小时走30度，每分钟走0.5度。分针每分钟走6度。做差后取绝对值，再和180比一下，如果大于180，用360减。

# include <stdio.h>
# include <stdlib.h>

typedef struct TIME{
int hh, mm ;
int ang ;
} TIME ;

int cmp(const void *a, const void *b)
{
TIME *p = (TIME*)a, *q = (TIME*)b ;
if (p->ang != q->ang)
return p->ang - q->ang ;

if (p->hh != q->hh) return p->hh - q->hh ;
return p->mm - q->mm ;
}

int abs(int x){return x<0?-x:x;}

int main ()
{
int i, T ;
TIME time[5] ;
scanf ("%d", &T) ;
while (T--)
{
for(i = 0 ; i < 5 ; i++)
{
scanf ("%d:%d", &time[i].hh, &time[i].mm) ;
time[i].ang = abs((time[i].hh%12)*60+time[i].mm - time[i].mm*12) ;
if (time[i].ang > 360) time[i].ang = 720 - time[i].ang ;
}
qsort(time, 5, sizeof(time[0]), cmp) ;
printf ("%02d:%02d\n", time[2].hh, time[2].mm) ;
}
return 0 ;
}