题目：在一个矩形中摆放圆，要求每行或每列的圆是相切的，问最多能放多少个。

分析：计算几何，数论。首先计算矩形摆放，然后计算交叉摆放（每行相同和相邻行差1个）求最大即可。

说明：╮(╯▽╰)╭当成最大摆放WA好几次。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>

using namespace std;

//题目方式摆放 
int getmin(double x, double y)
{
	int a = (int)x,m = 0;
	int b = (int)(2.0*(y-1.0)/sqrt(3.0))+1;
	if (m < a*b-b/2)
		m = a*b-b/2;
		a = (int)x;
	if (a+0.5 <= x && m < a*b)
		m = a*b;
	return m;
}

//最大数目摆放 
int getmax(double x, double y)
{
	int a = (int)x,m = 0;
	//n+1,n,n+1方式 
	while (a > 1 && y >= 1) {
		double d = (x-1.0)/(a-1.0);
		if (d >= 2) break;
		double r = sqrt(1.0-0.25*d*d);
		if (r < 0.5) r = 0.5;
		int b = (int)((y-1.0)/r)+1;
		if (m < a*b-b/2)
			m = a*b-b/2;
		a --;
	}
	//n,n,n方式
	a = (int)x;
	while (a > 0 && y >= 1) {
		double d = 2*(x-1.0)/(2*a-1.0);
		if (d >= 2) break;
		if (d >= 1) {
			double r = sqrt(1.0-0.25*d*d);
			if (r < 0.5) r = 0.5;
			int b = (int)((y-1.0)/r)+1;
			if (m < a*b)
				m = a*b;
		}
		a --;
	}
	return m;
}

int main()
{
	double x,y;
	while (cin >> x >> y) {
		int n = (int)x*(int)y,m = 0,p,q;
		
		p = getmin(x, y);
		if (p > m) m = p;
		q = getmin(y, x);
		if (q > m) m = q;
		
		if (m > n && x >= 1 && y >= 1)
			cout << m << " skew" << endl;
		else cout << n << " grid" << endl;
	}
	return 0;
}

UVa 121 - Pipe Fitters