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.
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.
time in a day that all of the hands are happy, accurate up to 3 decimal
places.
Sample Input
120
90
-1
Sample Output
100.000
0.000
6.251
题目的意思是：输入角度D，求符合这个角度的一天中的happy时间占一天时间的百分之多少.....
happy时间是指 时针，分针，秒针三者之间的角度大小满足>=D就是happy时间....
思路： 一天24小时，也就是时钟转两圈。那么就求12小时的happy时间在除以12*60*60
12小时的happy时间等于符合条件的每分钟的happy时间相加。
接下来就是想办法求1分钟中符合条件的happy时间
时针 分针 秒针
每秒角速度 1/120 1/10 6
一度所用时间 120 10 1/6
假设现在的时间为h小时，m分钟，s秒。以数字12为起点。
则hw时针角度为：（h+m/60+s/3600）*1/120*3600
mw分针角度为：（m+s/60）*1/10*60
sw秒针角度为：s*6
要是happy时间就要满足下面三个条件
D<=|hw-mw|<=360-D
D<=|hw-sw|<=360-D
D<=|mw-sw|<=360-D
求得一个形式为D<=|ax+b|<=360-D 这里的x是秒
就出来3个区间然后与[0.60]取交集得到h与m，m与s，h与s的三个两两指针的happy区间 最后在求这三个区间的交集就是符合条件的happy时间
代码如下：（详情 可以参考kuangbin的博客园，我也是看他的才知道的） http://www.cnblogs.com/kuangbin/archive/2011/12/04/2275470.html
#include <stdio.h>
#include <math.h>
#include<iostream>
#include <algorithm>
using namespace std;
struct qujian
{
double l,r;
};
double D; qujian solve(double a,double b) //解方程 D<=a*x+b<=360-D ,并且和 [0,60] 取交集
{
qujian p1;
if(a>)
{
p1.l=(D-b)/a;
p1.r=(-D-b)/a;
}
else
{
p1.l=(-D-b)/a;
p1.r=(D-b)/a;
}
if(p1.l<)
p1.l=;
if(p1.r>)
p1.r=;
if(p1.l>=p1.r)
p1.l=p1.r=;
return p1;
}
qujian compare(qujian a,qujian b)
{
qujian p;
p.l=max(a.l,b.l);
p.r=min(a.r,b.r);
if(p.l>p.r)
p.l=p.r=;
return p;
} double happytime(double h,double m)
{
double a,b;
int i,j,k;
qujian s[][],p2; //解方程D<=|hh-mm|<=360-D
a=1.0/120.0-0.1; //hh=30*h+m/2+s/120;
b=*h+m/2.0-*m;
s[][]=solve(a,b);
s[][]=solve(-a,-b); a=0.1-;
b=*m;
s[][]=solve(a,b);
s[][]=solve(-a,-b); a=1.0/-;
b=*h+m/2.0;
s[][]=solve(a,b);
s[][]=solve(-a,-b); double ans=;
for(i=; i<; i++)
{
for(j=; j<; j++)
for(k=; k<; k++)
{
p2=compare(compare(s[][i],s[][j]),s[][k]);
ans+=p2.r-p2.l;
}
}
return ans;
} int main()
{
while(scanf("%lf",&D))
{
if(D==-)
break;
double h,m,ans=;
for(h=; h<; h++)
{
for(m=; m<; m++)
ans+=happytime(h,m);
}
printf("%.3lf\n",ans*100.0/(**));
}
return ;
}