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$$(1leq T leq 10^4)$ test cases
for each case,one line include the time
$0leq hh<24$,$0leq mm<60$,$0leq ss<60$
for each case,one line include the time
$0leq hh<24$,$0leq mm<60$,$0leq ss<60$
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：
1 #include<stdio.h> 2 int t,h,m,s,tol,up[4],down[4],g; 3 int gcd(int a,int b) 4 {return b?gcd(b,a%b):a;} 5 int main() 6 { 7 scanf("%d",&t); 8 while(t--) 9 { 10 scanf("%d:%d:%d",&h,&m,&s); 11 tol=h*3600+m*60+s; 12 up[0]=(11*tol)%(43200); 13 up[1]=(719*tol)%(43200); 14 up[2]=(59*tol)%(3600); 15 down[0]=down[1]=120;down[2]=10; 16 for(int i=0; i<3; i++) 17 { 18 g=gcd(up[i],down[i]); 19 up[i]/=g;down[i]/=g; 20 if(up[i]/down[i]>179)up[i]=down[i]*360-up[i]; 21 if(down[i]==1)printf("%d ",up[i]); 22 else printf("%d/%d ",up[i],down[i]); 23 } 24 printf(" "); 25 } 26 return 0; 27 }