hdu1209 Clock

/*m点n分k秒时
时针从0点转过的角度a = [m + n/60 + k/3600]*30 
= 30m + n/2 + k/120(度) 

分针转过的角度b = [n/60 + k/3600]*360 
= 6n + k/10(度) */
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#define MAXN 10
using namespace std;

struct node 
{
	int h,m;
	double angle;
}times[MAXN];

int cmp(node x,node y)
{
	if(fabs(x.angle-y.angle)<1e-6)//注意浮点数大小比较
	{
		if(x.h==y.h)
			return x.m<y.m;
		else
			return x.h<y.h;
	}
	else
		return x.angle<y.angle;
}

int main()
{
	int test,i;
	scanf("%d",&test);
	while(test--)
	{
		for (i=0;i<5;++i)
		{
			scanf("%d:%d",×[i].h,×[i].m);
			if(times[i].h>12)
				times[i].angle=fabs(30.0*(times[i].h-12)+times[i].m/2.0-6.0*times[i].m);
			else
				times[i].angle=fabs(30.0*times[i].h+times[i].m/2.0-6.0*times[i].m);
			if(times[i].angle>180)
				times[i].angle=360-times[i].angle;
		}
		sort(times,times+5,cmp);
		printf("%02d:%02d\n",times[2].h,times[2].m);
	}
	return 0;
}