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 ",sum*100/43200); 67 } 68 return 0; 69 }