uva10167 Birthday Cake

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 mustn't in
[500;500]. Cherries are not allowed lying on the beeline. For each dataset there is at least one solution.
Input
The input le contains several scenarios. Each of them consists of 2 parts:
The rst 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 le
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

题目大意如下：平面直角坐标系中有一个以（0,0）为圆心，半径为100个单位长度的圆，圆内有一些点，求一条直线正好把所有点分成数量相等的两半，且没有点在直线上。

思路：枚举，外加一点初中知识。

来源：virtual judge

6714905

ksq2013

UVA

10167

Accepted

0

C++11 5.3.0

665

2016-08-02 17:26:10

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
using namespace std;
int n,chy[101][2];
bool judge(int A,int B)
{
	int cnt[2]={0,0};
	for(int i=1;i<=(n<<1);i++){
		int tmp=A*chy[i][0]+B*chy[i][1];
		if(!tmp)
			return false;
		cnt[tmp<0]++;
	}
	return cnt[0]==n&&cnt[1]==n;//保证两边确实都有n个cherries，没有任何一个cherry在直线上;
}
void solve()
{
	for(int i=-100;i<=100;i++)//圆的半径仅为100个单位长度;
		for(int j=-100;j<=100;j++){
			if(!j)continue;
			if(judge(i,j)){
				printf("%d %d
",i,j);
				return;
			}
		}
}
int main()
{
	while(scanf("%d",&n)&&n){
		for(int i=1;i<=(n<<1);i++)
			scanf("%d%d",&chy[i][0],&chy[i][1]);
		solve();
	}
	return 0;
}
/*
Sample input
2
-20 20
-30 20
-10 -50
10 -5
0
Sample output
0 1
*/