No Gambling

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65568/32768 K (Java/Others)

Problem Description

One day, Flyvan introduced a new game to his two friends, Oregon Maple and Grape Skin. The game is quite simple. Given an N-sized grids, like the figure A shown below (as N = 4). The blue points are the places the first player can
choose, and the red points are the places the second player can choose.

In the game, the two players take turns to choose two points to get connected by a stick. The two chosen points’ distance should be exactly one-unit length. The first player’s goal is to create a ‘bridge’ that connects a most left point and a most right point.
The second player’s goal is to create a ‘bridge’ that connects a most top point and a most bottom point. Figure B shows a possible result (the first player won). In addition, the stick shouldn’t get crossed.

Now Flyvan will give the number N, and his two friends will play the game. Both of the two players will choose the best strategy. You can bet on one player, and if he wins the game, you’ll get twice money you bet~

Since you are a talented programmer, you surely won’t just do gambling. Please write a program to find out the player who you should bet on. As Oregon Maple is elder, he will always play first.

Input

Each line of the input is an integer N (2 <= N <= 270000), which indicated the number Flyvan chose. The end-of-file is denoted by a single line containing the number -1.

Output

If you think the first player will win, please output “I bet on Oregon Maple~”, else please output “I bet on Grape Skin~”.

Sample Input

2
-1

Sample Output

I bet on Oregon Maple~

Source

题意：

第一个人从上往下连红点

第二个人从左往右连蓝点

连线得多的人胜

这题就是英文看不懂 ，知道题意了就可以知道两个人开始所面临的状态是一样的，所以先手必胜

#include<iostream>
using namespace std;
int main()
{
    int n;
    while(cin>>n)
    {
        if(n==-1)break;
        cout<<"I bet on Oregon Maple~"<<endl;
    }
    return 0;
}