HUT 排序训练赛 G - Clock

Clock

Time Limit: 1000MS

Memory Limit: 32768KB

64bit IO Format: %I64d & %I64u

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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

 

Source

Asia 2003（Seoul）

【题目来源】

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=41455#problem/G

【题目大意】

输入5个时间点，要求计算时针与分针的夹角，然后比较夹角大小，输出第三大的那个时间点。

【细节】

注意这道题不需要考虑去重，也就是说直接比较输出就可。

还有就是角度相同的时候要比较时针，时针相同时比较分针。

角度最好定义为double型。

当时间大于或等于12时，与12相减然后取绝对值。

【解题思路】首先还是定义一个结构体用来存储输入的数据，然后就是输入，接着计算角度，结构体排序，输出。

【角度计算方法】

大家已经认识了钟表。钟表上的分针、时针在不停息地转动着，两针有时相互重合，有时相互垂直，有时又成一条直线，而求时针、分针形成的各种不同位置所需的时间，就构成了饶有兴趣的时钟问题。

[基础知识]

（1）周角是360°，钟面上有12个大格，每个大格是360°÷12=30°；有60个小格，每个小格是360°÷60=6°。

（2）时针每小时走一个大格（30°），所以时针每分钟走30°÷60=0.5°；分针每小时走60个小格，所以分针每分钟走6°.

一般来说，已知钟面的时刻求时针和分针所成夹角的度数（小于或等于180°的角度），可以找出时针（整时刻）和分针（当前时刻）之间相差的大格或小格数。求出相应度数以后，再减去时针所走的度数（用分针数乘以0.5°）

下面是我的AC代码：

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>

#define eps 1e-6

using namespace std;

struct clock
{
    char st[];
    int h,m;
    int time;
    double du;
}c[];

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>=) h=h-;
    double dh=h*+m*0.5;
    double dm=m*;

    double ma,mi;
    ma=max(dh,dm);
    mi=min(dh,dm);

    double ans=ma-mi;
    if(ans>=&&ans<=);
    else
        ans=-ma+mi;
    return ans;
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        for(int i=;i<;i++)
        {
            cin>>c[i].st;
            c[i].h=(c[i].st[]-'')*+(c[i].st[]-'');
            c[i].m=(c[i].st[]-'')*+(c[i].st[]-'');
            c[i].time=c[i].h*+c[i].m;
            c[i].du=cnt(c[i].h,c[i].m);
        }

        sort(c,c+,cmp);
        cout<<c[].st<<endl;
    }

    return ;
}