HDU 1006(时钟指针转角 **)

题意是说求出在一天中时针、分针、秒针之间距离均在 D 度以上的时间占比。

由于三针始终都在转动，所以要分别求出各个针之间的相对角速度，分别求出三针满足角度差的首次时间，再分别求出不满足角度差的最终时间，取这三个时间段的交集，也就是首次时间的最大值和最终时间的最小值之间的部分，要注意剪枝，去掉多余的部分，否则会超时的，

本人的代码也是借鉴了别人写出来的。代码如下：

 #include <bits/stdc++.h>
 using namespace std;
 const double hm = 11.0/;//时针分针相对角速度hm = m - h = 360/3600 - 360/(12*3600) = 11/120
 const double hs = 719.0/;//时针秒针相对角速度hs = s - h = 360/60 - 360/(12*3600) = 719/120
 const double ms = 59.0/;//分针秒针相对角速度ms = s - m = 360/60 - 360/3600 = 59/10
 const double thm = 43200.0/;//thm = 360/hm = 43200/11
 const double ths = 43200.0/;//ths = 360/hs = 43200/719
 const double tms = 3600.0/;//tms = 360/ms = 3600/59
 const double eps = 1e-;
 double max(double a,double b,double c)
 {
     if(a>b) return a>c?a:c;
     return b>c?b:c;
 }
 double min(double a,double b,double c)
 {
     if(a<b) return a<c?a:c;
     return b<c?b:c;
 }
 int main()
 {
     int deg;
     double from1,from2,from3,to1,to2,to3,x1,x2,x3,y1,y2,y3,st,ed,ans;
     while(scanf("%d",&deg))
     {
         if(deg==-) break;
         from1 = deg/hm;
         from2 = deg/hs;
         from3 = deg/ms;
         to1 = (-deg)/hm;
         to2 = (-deg)/hs;
         to3 = (-deg)/ms;
         ans = 0.0;
         for(x1 = from1, y1 = to1; y1 <= +eps; x1+=thm, y1+=thm)
             for(x2 = from2, y2 = to2; y2 <= +eps; x2+=ths, y2+=ths)
             {
                 if(x2>y1) break;
                 for(x3 = from3, y3 = to3; y3 <= +eps; x3+=tms, y3+=tms)
                 {
                     if(x3>y1 || x3>y2) break;
                     st = max(x1,x2,x3);
                     ed = min(y1,y2,y3);
                     if(st<ed) ans+=ed-st;
                 }
             }
         printf("%.3lf\n",ans*100.0/);//求出时间和的占比
     }
     return ;
 }