HDU1006 tick and tick

Tick and Tick

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21216    Accepted Submission(s): 5600

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

012090-1

 

Sample Output

100.0000.0006.251

 

Author

PAN, Minghao

 

Source

ZJCPC2004

 

Recommend

JGShining

题目大意是时针分针秒针待在一起感觉不开心，要间隔至少D°才能感到开心，求他们一天之内开心的时间占得百分比。

一开始是使用的散点来求和求百分比，也就是一秒一秒的计算，然后除以43200【12小时的秒数】，但是出现了精度问题，也就是输入90的时候输出了6.225

在经过反复思量之后觉得思路应该是没有问题的，那么问题就应该出现在除法的精度上面。时针分针秒针应该按照连续的转动来计算而不是散点。

据说如果化为毫秒可以满足要求，但是会超时

然后翻了大神的代码，知道应该使用角速度来计算，然后再度开始重新编写

题目的整体思路大约是计算出来时针，分针，秒针的角速度，然后求相对速度，求得满足最大距离大于degree的区间，最后求交集

先贴一下原来的代码，WA的

#include<iostream>
#include<cstdio>
#include<cmath>
#include<iomanip>
using namespace std;
int main()
{
    int n;
    float h,m,s,hm,ms,sh,sum;
    while(scanf("%d",&n)&&n!=-1)
    {
        sum = 0;
        for(int i = 0; i < 12; i ++)
        {
            for(int j = 0; j < 60; j ++)
            {
                for(int k = 0; k < 60; k ++)
                {
                    s = k*6;
                    m = j*6+k*0.1;
                    h = i*30+j*0.5+k*0.5/60;
                    hm = fabs(h-m);
                    ms = fabs(m-s);
                    sh = fabs(s-h);
                    hm = min(fabs(h-m),360-fabs(h-m));
                    ms = min(fabs(m-s),360-fabs(m-s));
                    sh = min(fabs(s-h),360-fabs(s-h));
                    if(hm >= n && ms >= n && sh >= n)
                        sum++;
                }
            }
        }
        sum = sum/432.0;
        cout << fixed << setprecision(3) << sum << endl;
    }
    return 0;
}

然后贴一下成功AC了的代码

#include<iostream>
#include<cstdio>
#include<iomanip>
#include<cmath>
using namespace std;
struct interval
{
    double b;
    double e;
};
int main()
{
    double ans,h,m,s;
    interval hm[11],hs[719],ms[709];
    double degree;
    while(1)
    {
    	ans = 0;
    	scanf("%lf",°ree);
		if(degree == -1.0)break;
        for(int i = 0; i < 11; i ++)
        {
            hm[i].b = (360*i+degree)/11.0*120;
            hm[i].e = (360*i+360-degree)/11.0*120;
        }
        for(int j = 0; j < 708; j ++)
        {
            ms[j].b = (360*j+degree)/5.9;
            ms[j].e = (360*j+360-degree)/5.9;
        }
        for(int k = 0; k < 719; k ++)
        {
            hs[k].b = (360*k+degree)/719.0*120;
            hs[k].e = (360*k+360-degree)/719.0*120;
        }
        for(int i = 0; i < 11; i ++)
        {
            for(int j = 0; j < 708; j ++)
            {
                if(hm[i].e > ms[j].b && hm[i].b < ms[j].e){
                for(int k = 0; k < 719; k ++)
                {
                    if(ms[j].e > hs[k].b && ms[j].b < hs[k].e)
                    {
                    	if(hm[i].b < hs[k].e && hm[i].e > hs[k].b)
                    	{
                    		double maxs = min(min(hs[k].e,ms[j].e),hm[i].e);
                    		double mins = max(max(hs[k].b,ms[j].b),hm[i].b);
                    		//cout << maxs << " " << mins << endl;
                    		ans = ans + maxs - mins;
						}
					}
                }
            	}
            }
        }
        cout << fixed << setprecision(3) << ans / 432.0 << endl;
    }
    return 0;
}

贴一些参考过的博客，致谢

语言分别是

C++

java