Tick and Tick

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16707    Accepted Submission(s): 4083

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

计算出每两个指针满足要求的角度所需的时间，及周期，然后按周期循环。

#include<iostream>
#include<stdio.h>
using namespace std;
double max(double a,double b,double c)
{
    double temp=(a>b)?a:b;
    return (temp>c)?temp:c;
}
double min(double a,double b,double c)
{
    double temp=(a<b)?a:b;
    return (temp<c)?temp:c;
}
int main()
{
    double wh=360.0/12/3600;
    double wm=360.0/60/60;
    double ws=360.0/60;
    double whm=wm-wh;
    double whs=ws-wh;
    double wms=ws-wm;
    //cout<<whm<<endl<<whs<<endl<<wms<<endl;
    double n;
    while(~scanf("%lf",&n)&&n!=-1)
    {
        double stahm=n/whm;
        double stahs=n/whs;
        double stams=n/wms;
        double endhm=(360-n)/whm;
        double endhs=(360-n)/whs;
        double endms=(360-n)/wms;
        double shm,shs,sms,ehm,ehs,ems;
        const double T_hm=43200.0/11,T_hs=43200.0/719,T_ms=3600.0/59;   //Ïà¶ÔÖÜÆÚ
        double sum=0;
        //cout<<"do"<<endl;
        for(shm=stahm,ehm=endhm; ehm<43200.000001; shm+=T_hm,ehm+=T_hm)
        {
            //cout<<shm<<endl;
            for(shs=stahs,ehs=endhs; ehs<43200.000001; shs+=T_hs,ehs+=T_hs)
            {
                if(ehm<shs) break;
                if(shm>ehs) continue;
                for(sms=stams,ems=endms; ems<43200.000001; sms+=T_ms,ems+=T_ms)
                {
                    if(ehm<sms||ehs<sms) break;
                    if(shm>ems||shs>ems) continue;
                    //cout<<"doing"<<endl;
                    double xsta=max(shm,shs,sms);
                    double xend=min(ehm,ehs,ems);
                    if(xsta<xend)
                        sum+=(xend-xsta);

                }
            }
        }
        printf("%.3lf\n",sum/432);

    }
    return 0;
}