[杭電]Tick and Tick

題目：http://acm.hdu.edu.cn/showproblem.php?pid=1006

思路：這題參考了網上的答案。http://www.cnblogs.com/Lyush/archive/2011/11/29/2266925.html

改進： 改變了三層for循環的次序，講循環次數自小的放在了最外面。AC耗時從93MS降到了78MS.           

代碼：

#include<iostream>
#include <iomanip>
using namespace std;

inline double Max(double a, double b, double c) {
	double tmp = a > b ? a : b;
	tmp = tmp > c ? tmp : c;
	return tmp;
}
inline double Min(double a, double b, double c) {
	double tmp = a < b ? a : b;
	tmp = tmp < c ? tmp : c;
	return tmp;
}

int main() {
	double ss, mm, hh, sm, mh, sh, t_sm, t_mh, t_sh;
	ss = 6.0, mm = 0.1, hh = 0.1 / 12.0;
	sm = 6.0 - 0.1;
	mh = 0.1 - 0.1 / 12.0;
	sh = 6.0 - 0.1 / 12.0;   //relative velocity 
	t_sm = 360.0 / sm;
	t_mh = 360.0 / mh;
	t_sh = 360.0 / sh;      //relative cycle
	int D;
	double m[3], n[3], x[3], y[3];
	while (~scanf("%d", &D)) {
		if (D == -1) break;

		x[0] = D / mh;            //begin time of meeting of hour and minute
		y[0] = (360.0 - D) / mh;  //end time of meeting of hour and minute

		x[1] = D / sm;
		y[1] = (360.0 - D) / sm;

		x[2] = D / sh;	
		y[2] = (360.0 - D) / sh;
		double st, ed;
		double ans = 0;

		// m[i] and n[i] stands for begin time and end time in this cycle. 
		for (m[0] = x[0], n[0] = y[0]; m[0] <= 43200; m[0] += t_mh , n[0] += t_mh )
		{
			for (m[1] = x[1], n[1] = y[1]; m[1] <= 43200; m[1] += t_sm, n[1] += t_sm)
			{
				if (m[0] > n[1]) continue;
				if (n[0] < m[1]) break;
				for (m[2] = x[2], n[2] = y[2]; m[2] <= 43200; m[2] += t_sh, n[2] += t_sh)
				{
					if (n[2] < m[1] || n[2] < m[0]) continue;
					if (m[2] > n[0] || m[2] > n[1]) break;
					st = Max(m[0], m[1], m[2]);
					ed = Min(n[0], n[1], n[2]);
					if (ed > st) ans += ed - st;
				}
			}
		}
		cout << setiosflags(ios::fixed) << setprecision(3) << ans *100.0/43200<< endl;
	}
}