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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 (South Korea)
题意:
//给出 5 个时刻,按时钟的时针,分针夹角从小到大排序,
//输出中间的时刻。
代码例如以下:
#include <cstdio> #include <iostream> #include <algorithm> using namespace std; struct TIME { int h; int m; int angle; }a[7]; int cal(TIME TT) { if(TT.h > 12) { TT.h-=12; } int tt = abs((TT.h*60 + TT.m) - TT.m*12); //原式为:TT.h*30+(TT.m/60)*30-a.m*6; if(tt > 360) tt = 720 - tt; return tt; } bool cmp(TIME A, TIME B) { if(A.angle != B.angle) { return A.angle < B.angle; } else if(A.h != B.h) { return A.h < B.h; } else return A.m < B.m; } int main() { int t; scanf("%d",&t); while(t--) { for(int i = 0; i < 5; i++) { scanf("%d:%d",&a[i].h,&a[i].m); a[i].angle = cal(a[i]); } sort(a,a+5,cmp); printf("%02d:%02d\n",a[2].h,a[2].m); } return 0; }
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