hdoj1006--Tick and Tick

Problem Description

The three hands of the clock are rotating every second and meeting each other many times everyday. Finally, they get bored of this and each of them would like to stay away from the other two. A hand is happy if it is at least D degrees from any of the rest. You are to calculate how much time in a day that all the hands are happy.

 

Input

The input contains many test cases. Each of them has a single line with a real number D between 0 and 120, inclusively. The input is terminated with a D of -1.

 

Output

For each D, print in a single line the percentage of time in a day that all of the hands are happy, accurate up to 3 decimal places.

 

Sample Input

0

120

90

-1

 

Sample Output

100.000

0.000

6.251

 

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算法分析

秒针(S)： 1s 360/60= 6 度(degree)
分针(M)： 1s 360/60*60= 1/10度(degree)
时针(H)： 1s 360/60*60*12= 1/120度(degree)

秒针与分针的相对速度为：
V(M,S) = 6-1/10 = 59/10 d/s;
同理, 时针与秒针 V(H,S) = 6-1/120 =719/120 d/s;
V(H,M) = 1/10-1/120 = 11/120 d/s.

秒针与分针的相遇周期为
T(M,S)= 360/V(M,S)= 3600/59;
同理, T(H,S)= 360/V(H,S)=43200/719;
T(H,M)= 360/V(H,M)=43200/11;

在一天24小时内，后12个小时与前12个小时情况完全相同，故只取12个小时即可。
则12个小时总秒数
T = 12h * 60m* 60s = 43200s

秒针与分针的相遇次数为
N(M,S)=T/T(M,S) = 59*12 = 708
同理, N(H,S)=T/T(H,S) = 719
N(H,M)=T/T(H,M) = 11

由于周期性规律可知：
假设秒针与分针之间的角度用函数F(t)表示，则 F(t)为某一时刻t两针之间的角度。
F(t+n*T(M,S)) = F(t) , 其中n为自然数，
同理， F(t+n*T(H,S)) = F(t)
F(t+n*T(H,M)) = F(t)

所以，我们只要计算出第一个周期内的各指针之间的幸福时间区段，再加上若干个各自周期后，仍为幸福区段。
假设角度差至少为D时，秒针和分针才幸福，则在第一个周期T(M,S)内，
D<= V（M,S)*t <= 360-D

解得 t1=D/V(M,S)
t2=(360-D)/V(M,S);
根据周期性，
happy(t ms) =(t1+n*T(M,S),t2+n*T(M,S)) 0 <= n < N(M,S);
同理， happy(t hs ) =(t1+n*T(H,S),t2+n*T(H,S)) 0 <= n < N(H,S);
happy(t hm) =(t1+n*T(H,M),t2+n*T(H,M)) 0 <= n < N(H,M);

最后，求得三者的交集区间长度

L = happy(t ms) X happy(t hs) X happy(t hm) ,（X表示关系交）即可。

    #include <iostream>
    #include <iomanip>
    using namespace std;
    const int T = *, NMS = ,NHM = , NHS = ;  //相遇次数
    const double F = 0.0466631;                            //F为调节系数，使得各区间段为准确值
    const double hmlen = T*F/NHM,mslen = T*F/NMS,hslen = T*F/NHS;
    struct  interval
    {
        double low,high;
    };
    interval andset(interval S1,interval S2)
    {
        interval zone;
        zone.low  = S1.low > S2.low ? S1.low : S2.low;
        zone.high = S1.high < S2.high ? S1.high : S2.high;
        if( zone.low >= zone.high )
            zone.low = zone.high = 0.0;
        return zone;
    }
    int main()
    {
        int D=;
        while(cin>>D&&D!=-)
        {
            double len = 0.0;
            interval ms,hs,hm;  

            hm.low = hmlen*D/ - hmlen;
            hm.high = - hm.low -hmlen;  

            ms.low = mslen*D/ - mslen;
            ms.high = - ms.low -mslen;  

            hs.low = hslen*D/ - hslen;
            hs.high = - hs.low -hslen;  

            for(int i=,j=,k=;i<NHM;i++)
            {
                hm.low+=hmlen;
                hm.high+=hmlen;  

                for(;j<NMS;j++)
                {
                    ms.low+=mslen;
                    ms.high+=mslen;  

                    interval temp1 = andset(hm,ms);
                    if(temp1.low!=||temp1.high!=)
                    {
                        for(;k<NHS;k++)
                        {
                            hs.low+=hslen;
                            hs.high+=hslen;  

                            interval temp2 = andset(temp1,hs);
                            len+=temp2.high-temp2.low;
                            if(hs.high>=temp1.high)
                            {
                                hs.low-=hslen;
                                hs.high-=hslen;
                                break;
                            }
                        }
                    }
                    if(ms.high>=hm.high)
                    {
                         ms.low-=mslen;
                         ms.high-=mslen;
                         break;
                    }
                }
            }
            cout<<setprecision()<<fixed<<len/(*F)<<endl;
        }
        return ;
    }