Problem‘s Link:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=13&problem=1108&mosmsg=Submission+received+with+ID+14413715

http://luckycat.kshs.kh.edu.tw/homework/q10167.htm

analyse:

由于A，B的范围在-500～500之间，所以直接枚举就可以了。

Time complexity: O(n*n*m)

Source code: 

//  Memory   Time
//  1347K     0MS
//   by : Snarl_jsb
//   2014-10-24-20.52
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<vector>
#include<queue>
#include<stack>
#include<map>
#include<string>
#include<climits>
#include<cmath>
#define N 1000010
#define LL long long
using namespace std;
struct node
{
    int x,y;
};
node p[110];
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
//    freopen("C:\\Users\\ASUS\\Desktop\\cin.cpp","r",stdin);
//    freopen("C:\\Users\\ASUS\\Desktop\\cout.cpp","w",stdout);
    int n,i,j,k,x,y;
    while(cin>>n,n)
    {
        n<<=1;
        for(int i=0;i<n;++i)
        {
            cin>>x>>y;
            p[i].x=x,p[i].y=y;
        }
        int cnt;
        bool f=0;
        for(i=-500;i<=500;++i)
        {
            for(j=-500;j<=500;++j)
            {
                cnt=0;
                for(k=0;k<n;++k)
                {
                    if(p[k].x*i+p[k].y*j>0) cnt++;
                    if(p[k].x*i+p[k].y*j==0) break;
                }
                if(cnt==n>>1&&k==n)
                {
                    cout<<i<<" "<<j<<endl;
                    f=1;
                    break;
                }
            }
            if(f) break;
        }
    }
    return 0;
}