HDU1209 ZOJ2680 UVALive2807 Clock【排序】

Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6451    Accepted Submission(s): 2027

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）

问题链接：HDU1209 ZOJ2680 UVALive2807 Clock

问题简述：

  输入有多个样例，每个样例有5个时间（小时：分钟），按时间的夹锐角从小到大进行排序，若夹角相等则按时间顺序。输出排序后位于中间的时间。

问题分析：

  夹角求法：分针60分钟旋转一周，分针每1分钟旋转360/60=6度；时针12小时旋转一周，每1小时等于60分钟，分针转动速度是时针的12倍，即时针转动速度为每分钟6/12=0.5度。得等公式a=h*30+m/2-m*6。其中：

  h*30表示时针准点度数；

  m/2表示分针转动m分钟所转动的度数；

  m*6表示分针转动的度数。

程序说明：（略）

题记：（略）

参考链接：（略）

AC的C++语言程序如下：

/* HDU1209 ZOJ2680 UVALive2807 Clock */

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <math.h>

using namespace std;

const int N = 5;
struct Node {
    int h, m;
    double a;
} clock2[N];

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

int main()
{
    int t;
    scanf("%d", &t);
    while(t--) {
        for(int i = 0; i < N; i++) {
            scanf("%d:%d", &clock2[i].h, &clock2[i].m);

            if(clock2[i].h > 12)
                clock2[i].a = fabs(30.0 * (clock2[i].h - 12) + clock2[i].m / 2.0 -6.0 * clock2[i].m);
            else
                clock2[i].a = fabs(30.0 * clock2[i].h + clock2[i].m / 2.0 - 6.0 * clock2[i].m);
            if(clock2[i].a > 180)
                clock2[i].a = 360 - clock2[i].a;
        }

        sort(clock2, clock2 + N, cmp);

        printf("%02d:%02d\n", clock2[2].h, clock2[2].m);
    }

    return 0;
}