hdu 1006 Tick and Tick

Posted 2020-10-05 极速快码

Tick and Tick

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19764    Accepted Submission(s): 5164

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

 

 

题解：直接就枚举就好了

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <algorithm>
 4 #include <string.h>
 5 #include <cmath>
 6 using namespace std;
 7 double D;
 8 double sum;
 9 struct node
10 {
11     double l,r;
12 };
13 node ans[3][2];
14 node solve(double a,double b)
15 {
16     node qu;
17     if(a>0)
18     {
19         qu.l=(D-b)/a;
20         qu.r=(360-D-b)/a;
21     }
22     else
23     {
24         qu.l=(360-D-b)/a;
25         qu.r=(D-b)/a;
26     }
27     if(qu.l<0) qu.l=0;
28     if(qu.r>60) qu.r=60;
29     if(qu.l>=qu.r) { qu.l=qu.r=0;}
30     return qu;
31 }
32 node mer_g(node a,node b)
33 {
34     node q;
35     q.l=max(a.l,b.l);
36     q.r=min(a.r,b.r);
37     if(q.l>q.r) q.l=q.r=0;
38     return q;
39 }
40 int main()
41 {
42     int h,m;
43     int i,j,k;
44     double a1,a2,a3,b1,b2,b3;
45     while(scanf("%lf",&D),D!=-1)
46     {
47         sum=0;
48         node qu;
49        for(h=0;h<12;h++)
50          for(m=0;m<60;m++)
51             {
52                b1=m*6;        a1=-5.9;
53                b2=30*h+(0.5-6)*m; a2=1.0/120-0.1;
54                b3=30*h+0.5*m; a3=1.0/120-6;
55                ans[0][0]=solve(a1,b1);ans[0][1]=solve(-a1,-b1);
56                ans[1][0]=solve(a2,b2);ans[1][1]=solve(-a2,-b2);
57                ans[2][0]=solve(a3,b3);ans[2][1]=solve(-a3,-b3);
58               for(i=0;i<2;i++)
59                for(j=0;j<2;j++)
60                 for(k=0;k<2;k++)
61                  {
62                    qu=mer_g(mer_g(ans[0][i],ans[1][j]),ans[2][k]);
63                    sum+=qu.r-qu.l;
64                  }
65             }
66       printf("%.3lf\n",sum*100/43200);
67     }
68     return 0;
69 }