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): 18663    Accepted Submission(s): 4760


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

 

Author

PAN, Minghao

 

Source

ZJCPC2004

 

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题意是说时分秒三个的夹角都相差D时，每一个针都会开心，这题我是看了http://blog.csdn.net/a402630999/article/details/8130308的题解，才有点理解，涉及到角速度和区间求值。。。

#include<iostream>
#include<math.h>
#include<string>
using namespace std;
double max(double a,double b,double c)
{
if(a>=b&&a>=c) return a;
else if(b>c&&b>a) return b;
else return c;
}
double min(double a,double b,double c)
{
if(a<=b&&a<=c) return a;
else if(b<c&&b<a) return b;
else return c;
}
int main()
{// 白天12个小时和晚上12个小时的概率相同
double ss,mm,hh,sm,mh,sh,t_sm,t_mh,t_sh;
ss=6.0;//角速度
mm=0.1;
hh=0.1/12.0;
sm=ss-mm;//相对速度
mh=mm-hh;
sh=ss-hh;
t_sm=360.0/sm;//相对周期
t_mh=360.0/mh;
t_sh=360.0/sh;
int d;
double m[3],n[3],x[3],y[3];
while(cin>>d)
{
if(d==-1) break;
x[0]=d/sm;//第一次满足条件的时间
x[1]=d/mh;
x[2]=d/sh;
y[0]=(360.0-d)/sm;//第一个不满足条件的时间
y[1]=(360.0-d)/mh;
y[2]=(360.0-d)/sh;
double st,ed,ans=0;
for(m[0]=x[0],n[0]=y[0];m[0]<=43200;m[0]+=t_sm,n[0]+=t_sm)
{
for(m[1]=x[1],n[1]=y[1];m[1]<=43200;m[1]+=t_mh,n[1]+=t_mh)
{
if(m[0]>n[1]) continue;//h满足条件的大于s不满足条件的 继续
if (n[0]<m[1]) break;
for(m[2]=x[2],n[2]=y[2];m[2]<=43200;m[2]+=t_sh,n[2]+=t_sh)
{
if(n[2]<m[1]||n[2]<m[0]) continue;
if(m[2]>n[0]||m[2]>n[1]) break;
st=max(m[0],m[1],m[2]);
ed=min(n[0],n[1],n[2]);
if(ed>st) ans+=ed-st;
}
}
}
printf("%.3lf\n",ans/432.0);
}
}