Brute force --- ultraviolet A 10167: birthday cake

 

Problem G. birthday cake

  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

 

Mean: 

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

Analyze:

Because the range of A and B is-500 ~ Between 500, so you can simply enumerate it.

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 longusing 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;}

　　

Brute force --- ultraviolet A 10167: birthday cake