原题链接 Tick and Tick 题意 计算时针.分针.秒针24小时之内三个指针之间相差大于等于n度一天内所占百分比. 思路 每隔12小时时针.分针.秒针全部指向0,那么只需要计算12小时内的百分比就行. 秒针转动速度 6度/s 分针转动速度 1/10度/s 时针转动速度 1/120度/s 那么秒针与分针的相对速度为 59/10度/s 秒针与时针的相对速度为 719/120度/s 分针和时针的相对速度为 11/120度/s 计算两个指针相差一度需要的时间: 秒针和分针 sm=10/59s/度…

Tick and Tick Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10194    Accepted Submission(s): 2859 Problem Description The three hands of the clock are rotating every second and meeting each ot…

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 ot…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=1006 Tick and Tick Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 22203    Accepted Submission(s): 5877 Problem Description The three hands of the…

Tick and Tick Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 19764    Accepted Submission(s): 5164 Problem Description The three hands of the clock are rotating every second and meeting each ot…

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 th…

hdu1006 标签(空格分隔): 暴力枚举 好久没有打题了,退队了有好几个月了,从心底不依赖那个人了,原来以为的爱情戏原来都只是我的独角戏.之前的我有时候好希望有个人出现,告诉自己去哪里,做什么,哪种仅仅是渴望被爱的感觉都很美好,大概是因为离开了acm也不再依赖他所以现在我自己告诉自己,自己要去哪里,做什么,突然感觉很踏实,也不那么累了.现在我想把之前没有做好的做好,虽然我已经失去了最好的时机,但还是要在最忙的时间里抽出来做一些曾经有所遗憾的事情. 不多说了,上题! 开始推了半天公式,发现这个…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=1006 题意: 24小时中,三个指针之间超过某个度数的时间占所有时间的百分比是多少. 思路:主要是物理和数学的综合 难在相对速度的运用秒钟的速度 s=6°/s,分针是1/10°/s,时针是1/120°/s 所以相对速度 s_m=59/10°/s,s_h=719/120°/s,m_h=11/120°/s 所以相差一度所需要的时间 sm=10/59 s/°,sh=120/719 s/°,mh=120/11 s…

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 de…

先缩短一半的时间:早上的12个小时和下午的12小时对时钟是一样的,因为时钟12小时与0小时的三针位置相同.接着就是了解到每次所有的针从有重合到再次有重合至多有一段连续的段符合三针分离度大于n.所以只要枚举每个重合到重合的时间段,然后求出每段中符合条件的时间段即可. 由于枚举的是重合到重合,所以先求出时针和分针,分针和秒针,时针和秒针之间的相对角速度和再重合周期.通过相对角速度hm,hs,ms,我们可以求出从重合到分离n度所需的时间n/hm,n/hs,n/ms,以及到再重合前n度所需的时间(360…