HDU1006

Tick and Tick

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15570 Accepted Submission(s): 3677

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.

#include<iostream>
#include<cstdio>
using namespace std;
const double s_h = 719./120,s_m = 59./10,m_h =11./120;
const double tsh=43200./719,tsm =3600./59,tmh =43200./11;
double MAX(double a,double b,double c)
{
    double max=a;
    if (b>max)
        max=b;
    if(c>max)
        max=c;
    return max;
}
double MIN(double a,double b,double c)
{
    double min=a;
    if (b<min)
        min=b;
    if (c<min)
        min=c;
    return min;
}
int main()
{

    double N;
    while (~scanf("%lf",&N))
    {
        if (N==-1)
            break;
        double bsm,bsh,bmh,esm,esh,emh,begin,end,sum=0;
        bsm=N/s_m;
        bsh=N/s_h;
        bmh=N/m_h;
        esm=(360-N)/s_m;
        esh=(360-N)/s_h;
        emh=(360-N)/m_h;
        for(double bt3=bsh,et3=esh;et3<=43200.000001;bt3+=tsh,et3+=tsh)
        {
            for (double bt2=bmh,et2=emh;et2<=43200.000001;bt2+=tmh,et2+=tmh)
            {
                if (et2<bt3)
                    continue;
                if (bt2>et3)
                    break;
                for (double bt1=bsm,et1=esm;et1<43200.000001;bt1+=tsm,et1+=tsm)
                {
                    if (et1<bt2||et1<bt3)
                        continue;
                    if (bt1>et2||bt1>et3)
                        break;
                    begin=MAX(bt1,bt2,bt3);
                    end=MIN(et1,et2,et3);
                    sum+=(end-begin);
                }
            }
        }
        printf("%.3lf
",sum/432);
    }
    return 0;
}

View Code

Sample Input

Sample Output

000
000
251

1.数学题

2.思路要清晰

3.感谢discuss 0o恋蓝o0