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
Notice that the answer must be not more 180 and not less than 0
Input
There are $T$$(1\leq T \leq 10^4)$ test cases
for each case,one line include the time
for each case,one line include the time
$0\leq hh<24$,$0\leq mm<60$,$0\leq ss<60$
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
题意：
求给定时间点的时针、分针、秒针的夹角（分数、角度制）
分析：
tol=总秒数，时针1/120 °/s，分针1/10 °/s，秒针 6 °/s，
分针和时针转过的角度差（tol/10-tol/120）=11*tol/120；
秒针和时针转过的角度差（6*tol-tol/120）=719*tol/120；
秒针和分针转过的角度差（6*tol-tol/10）=59*tol/10；
然后前两个角度差的分子%（120*360）即%43200，后面一个%3600，这样就<360了，就是我们要求的夹角或其补角了。
然后判断夹角up/down 是否大于180度，刚开始我让它>180就360减去它，但是 这里是整除，也就是180多一点点的整除了变成180，所以要写成>=180或>179时就让360减去它。
接下来约分输出。
代码：
#include<stdio.h>
int t,h,m,s,tol,up[],down[],g;
int gcd(int a,int b)
{return b?gcd(b,a%b):a;}
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d:%d:%d",&h,&m,&s);
tol=h*+m*+s;
up[]=(*tol)%();
up[]=(*tol)%();
up[]=(*tol)%();
down[]=down[]=;down[]=;
for(int i=; i<; i++)
{
g=gcd(up[i],down[i]);
up[i]/=g;down[i]/=g;
if(up[i]/down[i]>)up[i]=down[i]*-up[i];
if(down[i]==)printf("%d ",up[i]);
else printf("%d/%d ",up[i],down[i]);
}
printf("\n");
}
return ;
}