UESTC

Lucky Boy

Time Limit: 3000/1000MS (Java/Others)     Memory Limit: 65535/65535KB (Java/Others)

Submit Status

Recently, Lur have a good luck. He is also the cleverest boy in his school as he create the most popular computer game – Lucky Boy. The game is played by two players, aa and bb, in 22d planar .In the game Lucky Boy, there are nn different points on plane, each time one can remove one or multiple co-line points from the plane. The one who can firstly remove more than two points from the plane wins. The one who removes the last point on the plane can also win the game. You may assume that two players are both clever enough that they can always make the best choice. The winner is called Lucky Boy.

Given the n points, can you tell me who will be the Lucky Boy ? Note that player a will always the first one to remove points from the plane.

Input

The first line of each case is an integer n(0<n≤103)n(0<n≤103), following nn lines each contains two integers xx and y(0≤x,y≤108)y(0≤x,y≤108), describing the coordinates of each point. Ended byEOF.

Output

Output a is the lucky boy. in a single line if a win the game, otherwise you should output b is the lucky boy. in a single line.

Sample InputSample Output

30 01 12 230 01 12 3

a is the lucky boy.b is the lucky

Sample input and output

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <vector>using namespace std;#define ll long long#define ld long doubledouble a[1003][2];int n;int main(){    while(scanf("%d",&n) != EOF)    {        int sum = 0;        for(int i = 0; i < n; i++)        {            scanf("%lf%lf",&a[i][0],&a[i][1]);        }        int flag = 0;        for(int i = 0; i < n; i++)        {            for(int j = i+1; j < n; j++)            {                double a1 = a[j][1]-a[i][1],a2;                double b1 = a[i][0]-a[j][0],b2;                double c1 = a[j][0]*a[i][1] - a[j][1]*a[i][0],c2;                for(int k = j+1; k < n; k++)                {                    a2 = a[k][1]-a[i][1];                    b2 = a[i][0]-a[k][0];                    c2 = a[k][0]*a[i][1] - a[k][1]*a[i][0];                    if((a1*b2 == a2*b1) && a1*c2 == a2*c1)                    {                        flag = 1;                        break;                    }                }                if(flag)break;            }            if(flag)break;        }        if(flag || n%3 || n <= 2)printf("a is the lucky boy.\n");        else printf("b is the lucky boy.\n");    }    return 0;}