HDU 5387 Clock

Clock

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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
 

 暴力即可，我是先求出时针、分针、秒针的角度，化简为分母为120的分数，然后大角度减小角度，超过360度，减360，大于180度，用360减去原角度，通分输出即可。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <set>
#include <map>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <bitset>
#include <string>
#include <vector>
#include <iomanip>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
int main()
{
        int t;
        scanf("%d",&t);
        while(t--)
        {
                int a,b,c;
                scanf("%d:%d:%d",&a,&b,&c);
                if(a>12)
                {
                        a-=12;
                }
                int num1=(60*b+c+3600*a);
                int num2=(12*c+720*b);
                int num3=(720*c);
                int ans1;
                int ans2;
                int ans3;
                if(num1>num2)
                {
                        ans1=num1-num2;
                }
                else
                {
                        ans1=num2-num1;
                }
                //cout<<num1<<" "<<num2<<" "<<num3<<endl;
                if(num1>num3)
                {
                        ans2=num1-num3;
                }
                else
                {
                        ans2=num3-num1;
                }
                if(num2>num3)
                {
                        ans3=num2-num3;
                }
                else
                {
                        ans3=num3-num2;
                }
                if(ans1>360*120)
                {
                        ans1=-360*120+ans1;
                }
                if(ans2>360*120)
                {
                        ans2=-360*120+ans2;
                }
                if(ans3>360*120)
                {
                        ans3=-360*120+ans3;
                }
                //cout<<ans1<<" "<<ans2<<" "<<ans3<<endl;
                if(ans1>180*120)
                {
                        ans1=360*120-ans1;
                }
                if(ans2>180*120)
                {
                        ans2=360*120-ans2;
                }
                if(ans3>180*120)
                {
                        ans3=360*120-ans3;
                }
                //cout<<ans1<<" "<<ans2<<" "<<ans3<<endl;
                int gcd1=__gcd(ans1,120);
                int gcd2=__gcd(ans2,120);
                int gcd3=__gcd(ans3,120);
                if(gcd1==120)
                {
                        printf("%d ",ans1/120);
                }
                else
                {
                        printf("%d/%d ",ans1/gcd1,120/gcd1);
                }
                if(gcd2==120)
                {
                        printf("%d ",ans2/120);
                }
                else
                {
                        printf("%d/%d ",ans2/gcd2,120/gcd2);
                }
                if(gcd3==120)
                {
                        printf("%d ",ans3/120);
                }
                else
                {
                        printf("%d/%d ",ans3/gcd3,120/gcd3);
                }
                printf("\n");
        }
        return 0;
}