Tick and Tick------HDOJ杭州电（无法解释，直接看代码）

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

#include <iostream>
#include <iomanip>
using namespace std;
#define vs 6.
#define vm 1./double(10)
#define vh 1./double(120)

int main()
{
    double D;
    double T[3]= {(360./(vm-vh)),(360./(vs-vm)),(360./(vs-vh))}; ///时分 分秒 时秒 的相对周期
    while(cin>>D && D!=-1)
    {
        double HS[3]= {(D/360.)*T[0],(D/360.)*T[1],(D/360.)*T[2]}; ///存储每对针的開始Happy时间
        double HE[3]= {(360.-D)/360.*T[0],((360.-D)/360.*T[1]),((360.-D)/360.*T[2])}; ///存储每对针的结束Happy时间
        double happyTime=0.,nextHS=HS[0],nextHE=min(HE[1],HE[2]);
        while(HS[1]<43200-(D/360.)*T[0] && HS[2]<43200-(D/360.)*T[0])
        {
            nextHS= max(HS[0],max(HS[1],HS[2]));
            nextHE= min(HE[0],min(HE[1],HE[2]));
            happyTime += (nextHE-nextHS)>0.?nextHE-nextHS:0.;
            for(int i=0; i<3; i++)
            {
                HS[i]+=(nextHE-HE[i]<0.?HE[i]-nextHE:nextHE-HE[i])<1e-15?T[i]:0.;
                HE[i]+=(nextHE-HE[i]<0.?
HE[i]-nextHE:nextHE-HE[i])<1e-15?T[i]:0.;
            }
        }
        double result= happyTime/432.;
        cout<<setiosflags(ios::fixed)<<setprecision(3)<<result<<endl;
    }
    return 0;
}

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