首页 > 代码库 > Tick and Tick------HDOJ杭电(解释不了,直接看代码)
Tick and Tick------HDOJ杭电(解释不了,直接看代码)
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
0 120 90 -1
Sample Output
100.000 0.000 6.251
#include <iostream> #include <iomanip> using namespace std; #define vs 6. #define vm 1./double(10) #define vh 1./double(120) int main() { double D; double T[3]= {(360./(vm-vh)),(360./(vs-vm)),(360./(vs-vh))}; ///时分 分秒 时秒 的相对周期 while(cin>>D && D!=-1) { double HS[3]= {(D/360.)*T[0],(D/360.)*T[1],(D/360.)*T[2]}; ///存储每对针的开始Happy时间 double HE[3]= {(360.-D)/360.*T[0],((360.-D)/360.*T[1]),((360.-D)/360.*T[2])}; ///存储每对针的结束Happy时间 double happyTime=0.,nextHS=HS[0],nextHE=min(HE[1],HE[2]); while(HS[1]<43200-(D/360.)*T[0] && HS[2]<43200-(D/360.)*T[0]) { nextHS= max(HS[0],max(HS[1],HS[2])); nextHE= min(HE[0],min(HE[1],HE[2])); happyTime += (nextHE-nextHS)>0.?nextHE-nextHS:0.; for(int i=0; i<3; i++) { HS[i]+=(nextHE-HE[i]<0.?HE[i]-nextHE:nextHE-HE[i])<1e-15?T[i]:0.; HE[i]+=(nextHE-HE[i]<0.?HE[i]-nextHE:nextHE-HE[i])<1e-15?T[i]:0.; } } double result= happyTime/432.; cout<<setiosflags(ios::fixed)<<setprecision(3)<<result<<endl; } return 0; }
此题由某大牛详细解答,传送门:大牛题解地址
Tick and Tick------HDOJ杭电(解释不了,直接看代码)
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。