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");
}
}