Clock
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 422 Accepted Submission(s): 294
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
Author
SXYZ
Source
2015 Multi-University Training Contest 8
题意：给你一个时刻，求出时针与分针，分针与秒针，秒针与时针的夹角。
题解：分别求出时针，分针，秒针与12整点的夹角，求的时候把夹角*120，再求两两的夹角除以120.
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#define ll int
using namespace std;
ll gcd(ll a,ll b) {
return b==0?a:gcd(b,a%b);
}
int main() {
//freopen("test.in","r",stdin);
int t;
cin>>t;
while(t--) {
int h,m,s;
scanf("%d:%d:%d",&h,&m,&s);
if(h>=12)h-=12;
///ah=h*30+0.5*m+s/120 120为分母的公倍数
int ah=h*30*120+m*60+s;///时针与12正点的夹角*120
///放大120倍是她为整数,下面同理
int am=m*6*120+s*12;
int as=6*s*120;
int gd=120;
int a1,a2,a3;
///时-分
a1=ah>am?ah-am:am-ah;
a1=a1>180*120?360*120-a1:a1;
///分-秒
a2=am>as?am-as:as-am;
a2=a2>180*120?360*120-a2:a2;
///秒-时
a3=as>ah?as-ah:ah-as;
a3=a3>180*120?360*120-a3:a3;
if(a1%gd==0)
printf("%d ",a1/gd);
else
printf("%d/%d ",a1/gcd(a1,gd),gd/gcd(a1,gd));
if(a3%gd==0)
printf("%d ",a3/gd);
else
printf("%d/%d ",a3/gcd(a3,gd),gd/gcd(a3,gd));
if(a2%gd==0)
printf("%d ",a2/gd);
else
printf("%d/%d ",a2/gcd(a2,gd),gd/gcd(a2,gd));
cout<<endl;
}
return 0;
}
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