1006

/*

Description: to 
 * 
 * Author: Vincent
 * 
 * Date:2010/04/15
 * Contact:agilely@126.com 
 * Zju CS Lab
 
  
    
 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



输入一个角度  让时钟的一根指针远离其他任一根

输出一天中发生这种情况的百分比


首先是三个针角度（度数为单位）的函数：
其中t就是秒数  
秒针：360/60 * t = 6t
分针：(t%60)/10
时针：(t%3600)/120

a是题目给你的限制时间
 a <= | 6*t-(t%60)/10| <= 360-a
 a <= | 6*t-(t%3600)/120 | <= 360-a
 a <= | (t%60)/10-(t%3600)/120 | <= 360-a
时间只要满足上市就是happytime

由于double类型不好取mod
我就把时间分割成一分钟一分钟
循环个720次。。。。


然后就只计算上边三个式子的交集。。。
浮点型的交集好烦。。
*/
#include<stdio.h>
#include<stdlib.h>
#include<string>
double num[14];
int hash[13];
int cmp(const void *a,const void *b)
{
    return *(double *)a > *(double *)b ? 1 : -1;
}
void hh(double a ,double b)
{
    int i;
    double start,end;
    start = 0;
    end = a;
    for(i=0;i<13;i++)
        if(start <= num[i] && num[i+1] <= end)
            hash[i] ++;
    start = b;
    end = 60;
    for(i=0;i<13;i++)
        if(start <= num[i] && num[i+1] <= end)
            hash[i] ++;
}
void hhh(double a,double b)
{
    int i;
    if(b<0 || a>60)
        return ;
    for(i=0;i<13;i++)
        if(a <= num[i] && num[i+1] <= b)
            hash[i] ++;
}
double happytime(double ms,double hs,double a)
{
    //a <= | 6*t-(t+ms)/10 | <= 360-a
    
    //a <= | 6*t-(t+hs)/120 | <= 360-a
    
    //a <= | (t+ms)/10-(t+hs)/120 | <= 360-a
    
    //计算同时满足上边三个条件的t
    double aa,ab,ac,ad,ba,bb,bc,bd,ca,cb,cc,cd,sum;
    
    aa = (ms-a*10)/59;
    ab = (ms+a*10)/59;
    ac = (ms+10*a-3600)/59;
    ad = (ms+3600-10*a)/59;

    ba = (hs-12*ms-a*120)/11;
    bb = (hs-12*ms+a*120)/11;
    bc = (hs-12*ms+120*a-43200)/11;
    bd = (hs-12*ms+43200-120*a)/11;

    ca = (hs-a*120)/719;
    cb = (hs+a*120)/719;
    cc = (hs+120*a-43200)/719;
    cd = (hs+43200-120*a)/719;

    num[0] = 0;
    num[1] = 60;
    num[2] = aa;
    num[3] = ab;
    num[4] = ac;
    num[5] = ad;
    num[6] = ba;
    num[7] = bb;
    num[8] = bc;
    num[9] = bd;
    num[10] = ca;
    num[11] = cb;
    num[12] = cc;
    num[13] = cd;

    qsort(num,14,sizeof(num[0]),cmp);
    memset(hash,0,sizeof(hash));
    sum = 0;
    hh(aa,ab);
    hh(ba,bb);
    hh(ca,cb);
    hhh(ac,ad);
    hhh(bc,bd);
    hhh(cc,cd);
    for(int i=0;i<13;i++)
        if(hash[i]>=6)
            sum += num[i+1] - num[i];
    return sum;
}
int main()
{
    int i,start;
    double ans,a;
    while(scanf("%lf",&a),a!=-1)
    {
        ans = 0;
        start = 0;
        for(i=0;i<720;i++)
        {
            if(i==109)
                i = i;
            ans += happytime(start%3600,start,a);
            start += 60;
        }
        printf("%.3lf\n",ans*100/43200);
    }
    return 0;
}