Tick and Tick
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16508 Accepted Submission(s): 4004
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
Author
PAN, Minghao
AC代码：
#include <cstdio>
#include <cstdlib>
double D;
int H,M;
double S, Total;
int signal;
double minimum;
double maximum;
void get_next()
{
double HM,HS,MS; // HM是时针和分针的夹角, HS是时针和秒针的夹角, MS是分针和秒针的夹角
double HL,ML,SL; // HL是时针的位置, ML是分针的位置, SL是秒针的位置
double t1,t2,t3; // 临时值
SL = 6*S; // 秒针的位置是6度乘以当前的秒数
ML = 6*(M+S/60); // 分针的位置是6度乘以当前的分数 + 当前的秒数除以10
HL = 30*(H+M/60.0+S/3600); // 时针的位置是30度乘以当前的小时数 + 分钟数除以2 + 秒数除以120
HM = ML - HL; // 以分针在前, 算出分针和时针的夹角
HS = SL - HL; // 以秒针在前, 算出秒针和时针的夹角
MS = SL - ML; // 以秒针在前, 算出秒针和分针的夹角
while(HM+1e-6>=D) HM -= 360; // 如果时针和分针的夹角 >= D, 那么减360
while(HM+1e-6<D-360) HM += 360; // 如果时针和分针的夹角 < D-360, 那么加360
while(HS+1e-6>=D) HS -= 360; // 下面的操作是一样的, 目的是把每个夹角控制在 D-360 到 D之间
while(HS+1e-6<D-360) HS += 360; // 那样的话, 只要夹角小于-D, 他们就是Happy的
while(MS+1e-6>=D) MS -= 360; // 如果在-D和D之间, 就不happy
while(MS+1e-6<D-360) MS += 360;
if(HM>=-D||HS+1e-10>=-D||MS+1e-10>=-D) // 如果有任何一个夹角处于 -D到D之间
{
signal = 0; // 标记为Unhappy
t1 = (D - HM)/(1.0/10 - 1.0/120); // 时针和分针需要经过t1才变得happy
t2 = (D - HS)/(6-1.0/120); // 时针和秒针要经过t2才变得happy
t3 = (D - MS)/(6-1.0/10); // 分针和秒针要经过t3才变得happy
minimum = 0; // 下面是算t1, t2, t3的最大值
if(HM>=-D) minimum = minimum > t1? minimum : t1;
if(HS+1e-6>=-D) minimum = minimum > t2? minimum : t2;
if(MS+1e-6>=-D) minimum = minimum > t3? minimum : t3;
S += minimum; // 移动秒针到可能的happy处
while(S>=60) // 恢复正常的表示方式
{
S-=60;
M++;
}
while(M>=60) // 恢复正常的表示方式
{
M-=60;
H++;
}
return;
}
else
{
signal = 1; // 标记为happy
t1 = (-D - HM)/(1.0/10 - 1.0/120); // 时针和分针经过t1会变得unhappy
t2 = (-D - HS)/(6-1.0/120); // 时针和秒针经过t2会变得unhappy
t3 = (-D - MS)/(6-1.0/10); // 秒针和分针经过t3会变得unhappy
maximum = 100000; // 下面是算t1, t2, t3的最小值
maximum = maximum < t1? maximum : t1;
maximum = maximum < t2? maximum : t2;
maximum = maximum < t3? maximum : t3;
S += maximum; // 移动到unhappy处
while(S>=60) // 恢复正常的表示方式
{
S-=60;
M++;
}
while(M>=60) // 恢复正常的表示方式
{
M-=60;
H++;
}
return;
}
}
int main()
{
while(scanf("%lf",&D)&&D!=-1) // 读取数字
{
if(D==0) // 特殊处理
{
printf("100.000\n");
continue;
}
if(D>=120) // 特殊处理
{
printf("0.000\n");
continue;
}
H = M = 0; // 初始化时间
S = 0; // 初始化时间
Total = 0; // Happytime
while(1)
{
get_next(); // 得到下一个临界时间
if(H>=12) break; // 如果超过12小时, 就不管了, 跳出
if(signal) Total += maximum; // 如果从现在到下一个临界时间是Happy的, 就加上Maximum的Happy时间
}
printf("%.3lf\n",Total/432); // Total/432 % 就是总共happy的百分比
}
}
