hdu 3863 No Gambling

题目链接：点这里。

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~

【题意】

给你一个散点图，每次每个人只能选择两个相邻属于自己颜色的点，然后链接起来，并且不能和之前的连线相交，问你谁最先得到一条从最左边到最右边的折线。

【分析】

感觉和博弈还是有点不一样啊，因为两个人的可行操作不完全一样。而且还有一些神奇的推理，反正结果比较简单：先手一定必胜，其实结果还是还比较容易理解的，因为不管如何后手一定会先一步堵住先手的。这样会造成后手慢一步，所以一定会造成后手慢一点，这就造成了先手一定成功。

【代码】

#include <bits/stdc++.h>int main() {    int a;    while(scanf("%d",&a),a+1) puts("I bet on Oregon Maple~");    return 0;}

【续】

没错，答案就是这么简单。