Clock
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5870 Accepted Submission(s):
1872
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
题意：n个测试案例，每一个测试案例5个时间，
每一个时间的时针和分针都有一个相对应的角度，
按角度排序，角度排序排在中间的那个时间
{注意： 角度相同的时候 要按时间来排序}
#include <stdlib.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
#define N 1000
struct SS
{
int hh,mm;
double r;
}f[N];
int cmp(SS a,SS b)
{
if(a.r!=b.r)
return a.r<b.r;
if(a.r==b.r&&a.hh!=b.hh)
return a.hh<b.hh;
}
int main()
{ //freopen("1.txt","r",stdin);
int i,test,m,n;
cin>>test;
while(test--)
{
for(i=0;i<5;i++)
scanf("%d:%d",&f[i].hh,&f[i].mm);
for(i=0;i<5;i++)
{
if(f[i].hh>12)
{
f[i].r=fabs(30.0*(f[i].hh-12)+f[i].mm/2.0-6.0*f[i].mm);
}
else
{
f[i].r=fabs(30.0*f[i].hh+f[i].mm/2.0-6.0*f[i].mm);
}
if(f[i].r>180)
f[i].r=360-f[i].r;
}
sort(f,f+5,cmp);
printf("%02d:%02d\n",f[2].hh,f[2].mm);
}
return 0;
}