hdu5387 六校 Clock（水题 模拟）

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=5387

Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 370    Accepted Submission(s): 255

Problem Description

Give a time.(hh:mm:ss)，you should answer the angle between any two of the minute.hour.second hand
Notice that the answer must be not more 180 and not less than 0

 

Input

There are  T (1≤T≤104)  test cases
for each case,one line include the time

0≤hh<24 , 0≤mm<60 , 0≤ss<60

 

Output

for each case,output there real number like A/B.(A and B are coprime).if it's an integer then just print it.describe the angle between hour and minute,hour and second hand,minute and second hand.

 

Sample Input

   
   
   
   

    
    
    
    4
00:00:00
06:00:00
12:54:55
04:40:00
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0 0 0 
180 180 0 
1391/24 1379/24 1/2 
100 140 120 

    
    
    
    
 
     
 
      Hint 
     
每行输出数据末尾均应带有空格 
    
    
    
    
     
   
   
   
   

 

一道比较水的题=，=，稍微用到了一点计算几何的东西

AC代码

#include<stdio.h>
#include<math.h>
#define eps 1e-8
int com(double m)
{
    if(fabs(m)<eps) return 0;
    if(m<0) return -1;
    return 1;
}
void print2(double k)
{
    for(int i=1;i<=360;i++)
    {
        if(com((i*k)-int(i*k+0.5))==0){
            if(i==1)
                printf("%d ",int(i*k));
            else
                printf("%d/%d ",int(i*k+0.5),i);
                return;
        }
    }
}
void print(double x,double y)
{
    double k;
    k=fabs(x-y);
    if(k>=180)
        k=360-k;
    print2(k);
}

int main()
{
    int hour,t;
    double h,m,s,minn,sec;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d:%lf:%lf",&hour,&minn,&sec);
        h=hour%12*30+minn*0.5+sec/120;
        m=minn*6+sec*0.1;
        s=sec*6;
        print(h,m);
        print(h,s);
        print(m,s);
        printf("\n");
    }
}