HDOJ 1209 Clock

排序函数很坑爹。

Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3411    Accepted Submission(s): 1048

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

300:00 01:00 02:00 03:00 04:0006:05 07:10 03:00 21:00 12:5511:05 12:05 13:05 14:05 15:05

 

Sample Output

02:0021:0014:05

 

Source

Asia 2003（Seoul）

 

Recommend

mcqsmall

 

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
 
#define eps 1e-6
 
using namespace std;
 
struct clock
{
    char st[6];
    int h,m;
    int time;
    double du;
}c[5];
 
bool cmp(clock a,clock b)
{
    if(a.du<b.du)
      return true;
     if(a.du>b.du)
        return false;
     if(a.du==b.du)
     {
         return a.time>b.time? false:true;
     }
}
 
double cnt(int h,int m)
{
    if(h>=12) h=h-12;
    double dh=h*30+m*0.5;
    double dm=m*6;
 
    double ma,mi;
    ma=max(dh,dm);
    mi=min(dh,dm);
 
    double ans=ma-mi;
    if(ans>=0&&ans<=180);
    else
        ans=360-ma+mi;
    return ans;
}
 
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        for(int i=0;i<5;i++)
        {
            cin>>c[i].st;
            c[i].h=(c[i].st[0]-'0')*10+(c[i].st[1]-'0');
            c[i].m=(c[i].st[3]-'0')*10+(c[i].st[4]-'0');
            c[i].time=c[i].h*60+c[i].m;
            c[i].du=cnt(c[i].h,c[i].m);
        }
 
        sort(c,c+5,cmp);
/*
        for(int i=0;i<5;i++)
            cout<<c[i].h<<":"<<c[i].m<<",,,,"<<c[i].time<<"..."<<c[i].du<<endl;
*/
        cout<<c[2].st<<endl;
    }
 
    return 0;
}