HDU1209：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

 

 

注意这题的排序是按照时针与分针的夹角大小来排序的

#include <stdio.h>
#include <algorithm>
#include <math.h>
using namespace std;

struct Node
{
    int h,m;
    double r;
}clock[10000];

int cmp(Node x,Node y)
{
    if(x.r!=y.r)
    return x.r<y.r;
    return x.h<y.h;
}

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

    return 0;
}