Uva 10167 - Birthday Cake

Problem G. Birthday Cake

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

2
-20 20
-30 20
-10 -50
10 -5
0

Sample Output

0 1

题意：给你一2*n个点，让你找出一条直线AX+BY=0，使直线把这个蛋糕分为两个部分，每一边都有n个点。

由于A，B的范围在-500～500之间，所以直接枚举就可以了。

#include <stdio.h>
const int N = 110;

struct Node{
	int x;
	int y;
}node[N];

void solve(int n) {
	int less,more;		
	for(int i = -500; i <= 500; i++) {
		for(int j = -500; j <= 500; j++) {	
			less = more = 0;
			for(int k = 0; k < 2*n; k++) {
				if( i*node[k].x + j*node[k].y < 0) {
					less++;
				}
				if( i*node[k].x + j*node[k].y > 0) {
					more++;
				}
				if( i*node[k].x + j*node[k].y == 0) {
					break;
				}
				if( less == n && more == n) {
					printf("%d %d\n",i,j);
					return;
				}
			}
		}
	}
}

int main() {
	int n;
	while( scanf("%d",&n) != EOF && n) {
		for(int i = 0; i < 2*n; i++) {
			scanf("%d%d",&node[i].x,&node[i].y);	
		}
		solve( n );
	}
	return 0;
}