UVa 10167 Birthday Cake (枚举)

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本文探讨了一种算法，用于解决将含有多个樱桃的蛋糕公平切割为两半的问题，确保每一半都包含相同数量的樱桃。通过枚举法求解，确保切割线通过蛋糕中心并使两边形状相等。

10167 - Birthday Cake

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=107&page=show_problem&problem=1108

Background

Lucy and Lily are twins. Today is their birthday. Mother buys a birthday cake for them.Now we put the cake onto a Descartes coordinate. Its center is at (0,0), and the cake's length of radius is 100.

There are 2N (N is a integer, 1<=N<=50) cherries on the cake. Mother wants to cut the cake into two halves with a knife (of course a beeline). The twins would like to be treated fairly, that means, the shape of the two halves must be the same (that means the beeline must go through the center of the cake) , and each half must have N cherrie(s). Can you help her?

Note: the coordinate of a cherry (x , y) are two integers. You must give the line as form two integers A,B(stands for Ax+By=0), each number in the range [-500,500]. Cherries are not allowed lying on the beeline. For each dataset there is at least one solution.

Input

The input file contains several scenarios. Each of them consists of 2 parts: The first part consists of a line with a number N, the second part consists of 2N lines, each line has two number, meaning (x,y) .There is only one space between two border numbers. The input file is ended with N=0.

Output

For each scenario, print a line containing two numbers A and B. There should be a space between them. If there are many solutions, you can only print one of them.

Sample Input

Sample Output

0 1

枚举A,B即可。

完整代码：

/*0.015s*/

#include <cstdio>
const int maxn = 30;///不需要500那么大，30足矣

int chx[101], chy[101];

int main()
{
	int n, cnt, a, b, i;
	bool flag;
	while (scanf("%d", &n), n)
	{
		for (i = 0; i < n * 2; ++i)
			scanf("%d%d", &chx[i], &chy[i]);
		flag = false;
		for (a = -maxn; a <= maxn; ++a)
		{
			for (b = -maxn; b <= maxn; ++b)
			{
				cnt = 0;
				for (i = 0; i < n * 2; ++i)
				{
					if (chx[i] * a + chy[i] * b == 0)
					{
						cnt = 0;
						break;
					}
					else if (chx[i] * a + chy[i] * b > 0) ++cnt;
				}
				if (cnt == n)
				{
					printf("%d %d\n", a, b);
					flag = true;
					break;
				}
			}
			if (flag) break;
		}
	}
	return 0;
}