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）
思路：首先要知道公式：（时间夹角） = | h * 30 - m * 5.5|
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define eps 1e-9
const int inf =
0x3f3f3f3f;
const int mod = 1e9+
7;
const int maxn =
8000 +
8;
int t;
struct node
{
int h, m;
double angle;
node() : angle(0){}
}ti[8];
bool cmp(node a, node b)
{
if(a.angle ==
b.angle)
return a.h <
b.h;
else if(a.angle == b.angle && a.h ==
b.h)
return a.m <
b.m;
return a.angle <
b.angle;
}
int main()
{
for(cin >> t; t--
; )
{
for(
int i =
0; i <
5; i++
)
{
scanf("%d:%d", &ti[i].h, &
ti[i].m);
ti[i].angle = fabs((ti[i].h %
12) *
30 -
5.5 * (ti[i].m %
60));
while(ti[i].angle >
180)ti[i].angle = fabs(
360 -
ti[i].angle);
}
sort(ti, ti +
5, cmp);
if(ti[
2].h <
10)
printf("0%d:", ti[
2].h);
else
printf("%d:", ti[
2].h);
if(ti[
2].m <
10)
printf("0%d\n", ti[
2].m);
else
printf("%d\n", ti[
2].m);
}
return 0;
}
转载于:https://www.cnblogs.com/RootVount/p/11448252.html
