hdu 5387

Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 311    Accepted Submission(s): 212

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

400:00:0006:00:0012:54:5504:40:00

 

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
 

 

题目大意：给出时间h： m： s  ，求出三个指针的角度

思       路 ：首先确认的一点是s转一圈走360下，跳一下就是1度    时针的角度要加上分针和秒针对应的角度 分针也是

                    因此需要算出三个针的角度再互相减去，同时保证数据在180 度以内 

代码如下：

/*踏实！！努力！！*/#include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<cctype>#include<functional>using namespace std;int gcd(int a,int b){    if(b==0)        return a;    return gcd(b,a%b);}void work(int t1,int t2){    int t=abs(t1-t2);    int k=120;    if(t>180*k) t=360*k-t;    int g=gcd(t,k);    if(k/g==1) printf("%d ",t/g);    else printf("%d/%d ",t/g,k/g);}int main(){    int t,h,m,s;    scanf("%d",&t);    while(t--){        scanf("%d:%d:%d",&h,&m,&s);        int hh,mm,ss;        if(h>=12) h-=12;        //先化为整数 全部乘以120        ss=s*720;          mm=m*720+s*12;        hh=h*3600+m*60+s;        work(hh,mm);        work(hh,ss);        work(mm,ss);        printf("\n");    }    return 0;}