HDU 1006(时钟指针转角 **)

Posted 2020-12-31 taskr212

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：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const double hm = 11.0/120;//时针分针相对角速度hm = m - h = 360/3600 - 360/(12*3600) = 11/120
 4 const double hs = 719.0/120;//时针秒针相对角速度hs = s - h = 360/60 - 360/(12*3600) = 719/120
 5 const double ms = 59.0/10;//分针秒针相对角速度ms = s - m = 360/60 - 360/3600 = 59/10
 6 const double thm = 43200.0/11;//thm = 360/hm = 43200/11
 7 const double ths = 43200.0/719;//ths = 360/hs = 43200/719
 8 const double tms = 3600.0/59;//tms = 360/ms = 3600/59
 9 const double eps = 1e-6;
10 double max(double a,double b,double c)
11 {
12     if(a>b) return a>c?a:c;
13     return b>c?b:c;
14 }
15 double min(double a,double b,double c)
16 {
17     if(a<b) return a<c?a:c;
18     return b<c?b:c;
19 }
20 int main()
21 {
22     int deg;
23     double from1,from2,from3,to1,to2,to3,x1,x2,x3,y1,y2,y3,st,ed,ans;
24     while(scanf("%d",&deg))
25     {
26         if(deg==-1) break;
27         from1 = deg/hm;
28         from2 = deg/hs;
29         from3 = deg/ms;
30         to1 = (360-deg)/hm;
31         to2 = (360-deg)/hs;
32         to3 = (360-deg)/ms;
33         ans = 0.0;
34         for(x1 = from1, y1 = to1; y1 <= 43200+eps; x1+=thm, y1+=thm)
35             for(x2 = from2, y2 = to2; y2 <= 43200+eps; x2+=ths, y2+=ths)
36             {
37                 if(x2>y1) break;
38                 for(x3 = from3, y3 = to3; y3 <= 43200+eps; x3+=tms, y3+=tms)
39                 {
40                     if(x3>y1 || x3>y2) break;
41                     st = max(x1,x2,x3);
42                     ed = min(y1,y2,y3);
43                     if(st<ed) ans+=ed-st;
44                 }
45             }
46         printf("%.3lf
",ans*100.0/43200);//求出时间和的占比
47     }
48     return 0;
49 }

View Code