hdu 4664 Triangulation 博弈论。。。。我把这题称作是花样打表。。受不了了，各种坑

Triangulation

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

Total Submission(s): 633 Accepted Submission(s): 248



Problem Description
There are n points in a plane, and they form a convex set.

No, you are wrong. This is not a computational geometry problem.

Carol and Dave are playing a game with this points. (Why not Alice and Bob? Well, perhaps they are bored. ) Starting from no edges, the two players play in turn by drawing one edge in each move. Carol plays first. An edge means a line segment connecting two
different points. The edges they draw cannot have common points.

To make this problem a bit easier for some of you, they are simutaneously playing on N planes. In each turn, the player select a plane and makes move in it. If a player cannot move in any of the planes, s/he loses.

Given N and all n's, determine which player will win.



Input
First line, number of test cases, T.

Following are 2*T lines. For every two lines, the first line is N; the second line contains N numbers, n1, ..., nN.

Sum of all N <= 106.

1<=ni<=109.


Output
T lines. If Carol wins the corresponding game, print 'Carol' (without quotes;) otherwise, print 'Dave' (without quotes.)


Sample Input

2
1
2
2
2 2

Sample Output

Carol
Dave

这题意说的很笼统，，给你们一篇博客看吧，http://www.xuebuyuan.com/2010248.html
说的很详细的
这题的规律还真不好找，，出这题的真特么蛋疼。
前100个有几个数总是操蛋的不等于，，所以规律从后一百个找，比较方便和正确。
SG打表找规律代码（不是杭电提交代码）：

#include <stdio.h>
#include <string.h>
#define MAX 150
int sg[MAX] ;
int main()
{
	sg[2] = 1 ;
	bool visited[MAX] ;
	for(int i = 0 ; i < MAX ; ++i)
	{
		memset(visited,false,sizeof(visited)) ;
		for(int j = 0 ; j <= i-2 ; ++j)
		{
			visited[sg[i-2-j]^sg[j]] = true ;
		}
		for(int j = 0 ; j <= MAX ; ++j)
		{
			if(!visited[j])
			{
				sg[i] = j;
				printf("%d,",j);
				break ;
			}
		}
	}
	return 0 ;
}

下面是我的花样打表代码（杭电提交代码）：

#include <stdio.h>
int sg1[]={0,0,1,1,2,0,3,1,1,0,3,3,2,2,4,0,5,2,2,3,3,0,1,1,3,0,2,1,1,0,4,5,2,7,4,0,1,1,2,0,
			3,1,1,0,3,3,2,2,4,4,5,5,2,3,3,0,1,1,3,0,2,1,1,0,4,5,3,7,4,8,1,1,2,0,3,1,1,0,3,3,
			2,2,4,4,5,5,9,3,3,0,1,1,3,0,2,1,1,0,4,5} ;
int sg2[] = {3,7,4,8,1,1,2,0,3,1,1,0,3,3,2,2,4,4,5,5,9,3,3,0,1,1,3,0,2,1,1,0,4,5};
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		int ans = 0 ;
		for(int i = 0 ; i < n ; ++i)
		{
			int temp ;
			scanf("%d",&temp);
			if(temp>=100)
			{
				ans ^= sg2[(temp-100)%34] ;
			}
			else
			{
				ans ^= sg1[temp] ;
			}
		}
		if(ans)
		{
			puts("Carol") ;
		}
		else
		{
			puts("Dave");
		}
	}
	
	return 0 ;
}

与君共勉