杭电1209Clock

Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5296    Accepted Submission(s): 1645

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）

 

求出时针和分针的夹角的绝对值，排序后输出角度为中间的那个时间，要求如果夹角一样，则时间比较小的排在前边：

附代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;struct node{int hour;int minute;double angle;}t[10];int cmp(node a,node b){if(a.angle==b.angle)return a.hour*60+a.minute<b.hour*60+b.minute;return a.angle<b.angle ;}int i,j,k,l,m,n;int main(){while(~scanf("%d",&k)){while(k--){for(i=0;i<5;i++){scanf("%d:%d",&t[i].hour,&t[i].minute);int hour=t[i].hour;if(hour>=12)hour-=12;t[i].angle=fabs(hour*30+t[i].minute*1.00/2-t[i].minute*1.00*6);t[i].angle=min(360-t[i].angle,t[i].angle);}sort(t,t+5,cmp);printf("%02d:%02d\n",t[2].hour,t[2].minute);//被double惯坏了，错了两边后发现输出的是.2d，虽然结果相同.}}return 0;}