Clock

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

Total Submission(s): 316 Accepted Submission(s): 215

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

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

/* ***********************************************

Author        :CKboss

Created Time  :2015年08月13日 星期四 22时23分29秒

File Name     :HDOJ5387.cpp

************************************************ */

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <string>

#include <cmath>

#include <cstdlib>

#include <vector>

#include <queue>

#include <set>

#include <map>

using namespace std;

int hh,mm,ss;

int alltime(int h,int m,int s)

{

	return h*3600+m*60+s;

}

const int mod=360*120;

int main()

{

	//freopen("in.txt","r",stdin);

	//freopen("out.txt","w",stdout);

	int T_T;

	scanf("%d",&T_T);

	while(T_T--)

	{

		scanf("%d:%d:%d",&hh,&mm,&ss);

		int at=alltime(hh,mm,ss);

		int hhdu=at%mod;

		int mmdu=12*at%mod;

		int ssdu=720*at%mod;

		int dur_hm=abs(mmdu-hhdu);

		if(dur_hm>mod/2) dur_hm=mod-dur_hm;

		int dur_hs=abs(hhdu-ssdu);

		if(dur_hs>mod/2) dur_hs=mod-dur_hs;

		int dur_ms=abs(mmdu-ssdu);

		if(dur_ms>mod/2) dur_ms=mod-dur_ms;

		int g1=__gcd(dur_hm,120);

		int g2=__gcd(dur_hs,120);

		int g3=__gcd(dur_ms,120);

		///print dur_hm

		if(g1==120) printf("%d ",dur_hm/g1);

		else printf("%d/%d ",dur_hm/g1,120/g1);

		///print dur_hs

		if(g2==120) printf("%d ",dur_hs/g2);

		else printf("%d/%d ",dur_hs/g2,120/g2);

		///print dur_hs

		if(g3==120) printf("%d \n",dur_ms/g3);

		else printf("%d/%d \n",dur_ms/g3,120/g3);

	}

    return 0;

}

巴特西

HDOJ 5387 Clock 水+模拟

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