Clock

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

 

#include <stdio.h>
#include <math.h>
#include <algorithm>
using std::sort;
struct Node{
	int h, m;
	double val;
} a[5];

bool cmp(Node x, Node y){
	if(x.val < y.val) return 1;
	if(x.val == y.val && x.h < y.h) return 1;
	if(x.val == y.val && x.h == y.h && x.m < y.m) return 1;
	return 0;
}

int main(){
	int t, h, m;
	double val;
	scanf("%d", &t);
	while(t--){
		for(int i = 0; i < 5; ++i){
			scanf("%d:%d", &a[i].h, &a[i].m);
			h = a[i].h; m = a[i].m;
			if(h > 12) h -= 12;
			val = fabs((h + m / 60.0) * 30.0 - m * 6.0);
			if(val > 180) val = 360 - val;
			a[i].val = val;
		}
		sort(a, a + 5, cmp);
		printf("%02d:%02d\n", a[2].h, a[2].m);
	}
	return 0;
}