HDU1006 ：Tick and Tick 数学+枚举

题意：现在有一个普通的时钟，有时针、分针、秒针，(秒针是连续走，而不是一秒一秒的走)给你一个度数D(小于等于120°)。求：24小时中，三个指针超过这个度数的时间占所有时间的百分比是多少。保留小数点后三位。

import java.text.DecimalFormat;
import java.util.Scanner;

public class Main {
	// 角速度差
	static final double s_m = 59.0 / 10, s_h = 719.0 / 120, m_h = 11.0 / 120;
	// 相差360°的时间
	static final double tsm = 3600.0 / 59, tsh = 43200.0 / 719,
			tmh = 43200.0 / 11;

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		DecimalFormat df = new DecimalFormat("0.000");
		double D, v = 43200.0, eq = 1e-7;
		while (in.hasNext()) {
			D = in.nextDouble();
			if (D == -1)
				break;
			double bsm, bsh, bmh, esm, esh, emh, Begin, End, ans = 0;
			// 计算各自第一次满足条件的时间
			bsm = D / s_m;
			bsh = D / s_h;
			bmh = D / m_h;
			// 计算各自第一次不满足条件的时间
			esm = (360 - D) / s_m;
			esh = (360 - D) / s_h;
			emh = (360 - D) / m_h;
			for (double b1 = bsm, e1 = esm; e1 - v < eq; b1 += tsm, e1 += tsm) {
				for (double b2 = bsh, e2 = esh; e2 - v < eq; b2 += tsh, e2 += tsh) {
					if (e2 < b1)
						continue;
					if (b2 > e1)
						break;
					for (double b3 = bmh, e3 = emh; e3 - v < eq; b3 += tmh, e3 += tmh) {
						if (e3 < b1 || e3 < b2)
							continue;
						if (b3 > e1 || b3 > e2)
							break;
						Begin = Max(b1, b2, b3);// 开始时间最晚
						End = Min(e1, e2, e3);// 结束时间最早
						ans += (End - Begin);
					}
				}
			}
			System.out.println(df.format(ans / 432));
		}
	}

	private static double Max(double a, double b, double c) {
		return Math.max(a, Math.max(b, c));
	}

	private static double Min(double a, double b, double c) {
		return Math.min(a, Math.min(b, c));
	}
}