hdu while .cn tsm ret name cas begin script

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 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

Source ZJCPC2004 分析：

三個指針走的角速度：

秒針速度S = 6°/s，分針速度M = (1/10)°/s，時針速度H = (1/120)°/s

這三個指針兩兩之間的相對速度差為：

秒時相差S_H = (719/120)°/s，秒分相差S_M = (59/10)°/s，分時相差M_H = (120/11)°/s

相差一度需要的時間為

秒時相差SH = (120/719)s/度，秒分相差SM = (10/59)s/度，分時相差MH = (120/11)s/度

相差360°需要的時間為

秒時相差tSH = 43200.0/719，秒分相差tSM = 3600.0/59，分時相差tMH = 43200.0/11

算出兩兩指針在43200s（12小時）內滿足條件時間的區間的交集。

代碼：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<iostream>

using namespace std;
///秒針角速度 6度/s, 分針0.1度/s,時針1/120度/s
const double sh = 719.0/120 , sm = 59.0/10, mh = 11.0/120;///三個指針兩兩相對角速度。
const double tsh = 43200.0 / 719, tsm = 3600.0 / 59, tmh = 43200.0 / 11;///三個指針兩兩相差360度所需的時間。
///將相對角速度變成周期。（即兩針間需要多久出現夾角的循環）
/// 同樣可求得三個周期的最小公倍數為 43200 秒，即12小時，

double max1(double a, double b, double c)
{
return max(a, max(b, c));
}
double min1(double a, double b, double c)
{
return min(a, min(b, c));
}
int main(void)
{
int D;

while(scanf("%d", &D), D != -1)
{
double bsh, bsm, bmh, esh, esm, emh, total, beginn, endd;
///第一次滿足條件的時間。(開始時間）
bsh = D / sh;
bsm = D / sm;
bmh = D / mh;
///第一次出現不滿足條件的時間。（結束時間）
esh = (360 - D) / sh;
esm = (360 - D) / sm;
emh = (360 - D) / mh;

total = 0;

for(double bt1 = bsh, et1 = esh; et1<= 43200.000001; bt1 += tsh, et1 += tsh)
{
for(double bt2 = bsm, et2 = esm; et2 <= 43200.000001; bt2 += tsm, et2 += tsm)
{
///判斷是否有交集。
if(bt2 > et1)
break;

if(et2 < bt1)
continue;

for(double bt3 = bmh, et3 = emh; et3 <= 43200.000001; bt3 += tmh, et3 += tmh)
{
if(bt3 > et1 || bt3 > et2)
break;

if(et3 < bt1 || et3 < bt2)
continue;

beginn = max1(bt1, bt2, bt3);
endd = min1(et1, et2, et3);

total += (endd - beginn);
}

}
}

printf("%.3f\n", total / 432);

}
return 0;
}

HDU 1006 模擬