sdnuoj 1094 clock

一道看着高大上其实很水很水的题

求时针分针角度,然后排序,输出中间的那个

1.找出时针分针和 12 的距离,然后一减就行了

2.如果角度大于180度,就取补角

3.常规知识,自己可以推的,分针 *1 = 6°,时针 *1 =  30°,分针动时针也会动,找好关系

4.用double 类型变量 因为有0.5°的出现

1094.Clock

Time Limit: 1000 MS    Memory Limit: 32768 KB
Total Submission(s): 120    Accepted Submission(s): 49

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
 
     
 
     
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
struct time{
    double h;
    double m;
    double hm;
    char s[6];
}tim[10010];
bool cmp(time a, time b)
{
    if(a.hm < b.hm)return 1;
    if(a.hm > b.hm)return 0;
    if(a.hm == b.hm&& a.h != b.h)return a.h < b.h;
    else return a.m < b.m;
}
int main()
{
    int t;
    int a, b;
    cin>>t;
    while(t--)
    {
        for(int i = 0; i < 5; i ++)
        {
            tim[i].h = 0;
            tim[i].m = 0;
            tim[i].hm = 0;
        }
        for(int i = 0; i < 5; i ++)
        {
            cin>>tim[i].s;
            a = (tim[i].s[0] - '0')*10 + (tim[i].s[1] - '0') * 1;
            b = (tim[i].s[3] - '0')*10 + (tim[i].s[4] - '0') * 1;
            tim[i].h = (a % 12) * 30 + b * 0.5;
            tim[i].m = b * 6;
            tim[i].hm = fabs(tim[i].h - tim[i].m);
            if(tim[i].hm > 180)tim[i].hm = 360 - tim[i].hm;
            //cout<
临近考试水一水,身心健康。。

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