HDU 5387 Clock

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

(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
 

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,以下同理
        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;
}