hdu 1006 Tick and Tick

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

 #include <iostream>
 #include <stdio.h>
 #include <algorithm>
 #include <string.h>
 #include <cmath>
 using namespace std;
 double D;
 double sum;
 struct node
 {
     double l,r;
 };
 node ans[][];
 node solve(double a,double b)
 {
     node qu;
     if(a>)
     {
         qu.l=(D-b)/a;
         qu.r=(-D-b)/a;
     }
     else
     {
         qu.l=(-D-b)/a;
         qu.r=(D-b)/a;
     }
     if(qu.l<) qu.l=;
     if(qu.r>) qu.r=;
     if(qu.l>=qu.r) { qu.l=qu.r=;}
     return qu;
 }
 node mer_g(node a,node b)
 {
     node q;
     q.l=max(a.l,b.l);
     q.r=min(a.r,b.r);
     if(q.l>q.r) q.l=q.r=;
     return q;
 }
 int main()
 {
     int h,m;
     int i,j,k;
     double a1,a2,a3,b1,b2,b3;
     while(scanf("%lf",&D),D!=-)
     {
         sum=;
         node qu;
        for(h=;h<;h++)
          for(m=;m<;m++)
             {
                b1=m*;        a1=-5.9;
                b2=*h+(0.5-)*m; a2=1.0/-0.1;
                b3=*h+0.5*m; a3=1.0/-;
                ans[][]=solve(a1,b1);ans[][]=solve(-a1,-b1);
                ans[][]=solve(a2,b2);ans[][]=solve(-a2,-b2);
                ans[][]=solve(a3,b3);ans[][]=solve(-a3,-b3);
               for(i=;i<;i++)
                for(j=;j<;j++)
                 for(k=;k<;k++)
                  {
                    qu=mer_g(mer_g(ans[][i],ans[][j]),ans[][k]);
                    sum+=qu.r-qu.l;
                  }
             }
       printf("%.3lf\n",sum*/);
     }
     return ;
 }