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…

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): 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…

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…

Tick and Tick Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 20120    Accepted Submission(s): 5262 Problem Description The three hands of the clock are rotating every second and meeting each ot…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1006 题目大意:钟表有时.分.秒3根指针.当任意两根指针间夹角大于等于n°时,就说他们是happy的,问一天有百分之多少时间是happy的. 关键思想:两根指针两根指针地考虑,从重合到重合中有且仅有一段连续时间这两根指针是happy的.共有3种组合(时分.分秒.时秒),所以若以时间为横轴,夹角为纵轴,图像为三个连续三角形.另y大于等于n,得到的图像求符合区间的长度. 代码如下: //多个连续三角区域…

一開始思考的时候认为好难的题目,由于感觉非常多情况.不知道从何入手. 想通了就不难了. 能够转化为一个利用速度建立不等式.然后解不等式的问题. 建立速度,路程,时间的模型例如以下: /*************************************************************************** * * * 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…

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…