【HDU 5387】Clock

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$$(1leq T leq 10^4)$ test cases
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

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 }