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 ;
}