hdu 5387 Clock 2015多校联合训练赛#8

Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 328    Accepted Submission(s): 225

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 
    
    
    
    
     
   
   
   
   

 

Source

2015 Multi-University Training Contest 8

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;


int gcd(int a,int b){
    if(b == 0) return a;
    return gcd(b,a%b);
}
void work(int a,int b){
    int u = 120*360;
    int f = ((a-b)%u+u)%u;
    f = min(f,u-f);
    u = 120;
    int g = gcd(f,u);
    u /= g;
    f /= g;
    if(u == 1) printf("%d ",f);
    else printf("%d/%d ",f,u);
}
int main(){
    int t,h,m,s;
    scanf("%d",&t);
    while(t--){
        scanf("%d:%d:%d",&h,&m,&s);
        h = h*60*60+m*60+s;
        m = m*12*60+s*12;
        s = s*12*60;
        work(h,m);
        work(h,s);
        work(m,s);
        printf("\n");
    }
    return 0;
}