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.
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
Recommend
mcqsmall
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <cmath> 5 6 #define eps 1e-6 7 8 using namespace std; 9 10 struct clock 11 { 12 char st[6]; 13 int h,m; 14 int time; 15 double du; 16 }c[5]; 17 18 bool cmp(clock a,clock b) 19 { 20 if(a.du<b.du) 21 return true; 22 if(a.du>b.du) 23 return false; 24 if(a.du==b.du) 25 { 26 return a.time>b.time? false:true; 27 } 28 } 29 30 double cnt(int h,int m) 31 { 32 if(h>=12) h=h-12; 33 double dh=h*30+m*0.5; 34 double dm=m*6; 35 36 double ma,mi; 37 ma=max(dh,dm); 38 mi=min(dh,dm); 39 40 double ans=ma-mi; 41 if(ans>=0&&ans<=180); 42 else 43 ans=360-ma+mi; 44 return ans; 45 } 46 47 int main() 48 { 49 int t; 50 cin>>t; 51 while(t--) 52 { 53 for(int i=0;i<5;i++) 54 { 55 cin>>c[i].st; 56 c[i].h=(c[i].st[0]-'0')*10+(c[i].st[1]-'0'); 57 c[i].m=(c[i].st[3]-'0')*10+(c[i].st[4]-'0'); 58 c[i].time=c[i].h*60+c[i].m; 59 c[i].du=cnt(c[i].h,c[i].m); 60 } 61 62 sort(c,c+5,cmp); 63 /* 64 for(int i=0;i<5;i++) 65 cout<<c[i].h<<":"<<c[i].m<<",,,,"<<c[i].time<<"..."<<c[i].du<<endl; 66 */ 67 cout<<c[2].st<<endl; 68 } 69 70 return 0; 71 }