hdu-Tick and Tick

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本文针对HDU在线评测平台上的TickandTick问题，详细解析了如何通过计算时钟三针之间的角度来确定“快乐”状态的时间百分比。文章提供了具体的数学模型和C语言实现代码。

hdu-Tick and Tick解题报告

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

Sample Output

100.000
0.000
6.251

题目大意：

输入一个非负整数D，当时针、分针、秒针，之中任意两者之间的夹角大于D，则会happy。现在要求总的happy的时间占的百分比（基于12个小时），保留三位小数。

解题思路：

1、每秒钟的移动速度(多少度)，（s：秒针，m：分针，h：时针）： s=6 /s m=(1/10) /s h=(1/120) /s

2、三种针之间相对速度： s_m=(59/10) /s s_h=(719/120) /s m_h=(11/120) /s

3、移动一度所需要的时间： s_m=(10/59) s s_h=(120/719) s m_h=(120/11) s

4、周期： ts_m=(3600/59) s ts_h=(43200/719) s tm_h=(43200/11) s

所以，假设输入的D=n；则有：

n*s_m + k1*ts_m < T < ts_m - n*s_m + k1*ts_m

n *s_h + k2*ts_h < T < ts_h - n*s_h + k2*ts_h

n* m_h + k3*tm_h <T < tm_h - n*m_h +k3*tm_h

当T同时满足这三个式子时，便是happy的时间；

#include<stdio.h>
#include<math.h>
int main()
{
    int t;
    double n,sm,sh,mh,tsm,tsh,tmh,fsm,fsh,fmh,esm,esh,emh,b1,e1,b2,e2,b3,e3,min,max,sum;
    sm=10.0/59.0;
    sh=120.0/719.0;
    mh=120.0/11.0;
    tsm=3600.0/59.0;
    tsh=43200.0/719.0;
    tmh=43200.0/11.0;
    while(~scanf("%lf",&n))
    {
        if(n<0)break;
        sum=0;
        fsm=n*sm;esm=tsm-fsm;
        fsh=n*sh;esh=tsh-fsh;
        fmh=n*mh;emh=tmh-fmh;
        for(b3=fmh,e3=emh;e3<=43200;b3+=tmh,e3+=tmh)//找出满足三个式子的T，并且求和得到sum
        {
            for(b2=fsh,e2=esh;e2<=43200;b2+=tsh,e2+=tsh)
            {
                if(e2<b3)
                    continue;
                if(b2>e3)
                    break;
                for(t=0,b1=fsm,e1=esm;e1<=43200;t=t+1,b1=fsm+t*tsm,e1=esm+t*tsm)
                {
                    if(e1<b2||e1<b3)
                        continue;
                    if(b1>e2||b1>e3)
                        break;
                    max=b1>b2?b1:b2;
                    max=max>b3?max:b3;
                    min=e1<e2?e1:e2;
                    min=min<e3?min:e3;
                    sum+=min-max;
                }
            }
        }
        printf("%.3lf\n",sum/432.0);//因为要求百分比，所以sum应该除以12个小时即43200秒，再乘以100
    }
    return 0;
}