hdu 4664 Triangulation 经典博弈

题意：

N个平面，每个平面有ni个点，组成凸多边形，两个人玩游戏，划线，他们可以划任意一个平面的两个点，有以下要求：两个人划得线不能交叉，不要划已经划过的线，如果一个平面被划了一个空心的三角形，那么这个平面就不能继续划线了。Carol先来，两个人轮着画，谁没线划了就输了，问你最后谁赢。

思路：

sg函数暴力求法：

一个平面上连接点时，不能连接已经有边的顶点，因为对方只需要再连接一次就可以组成一个三角形了。又所有的边不能相交，因此每连接一条边，就相当于把整个平面上的点划分成了两个部分，在接下来的游戏中，只能单独在两部分里面进行，相当于将一个游戏划分成了两个游戏。因此当前状态x的sg函数值就是两个子游戏的Nim和了。

即：sg(x) = mex{ sg(i)^sg(x-i-2) }

这样复杂度很高。但是这道题目有规律，在大数据的范围下，会出现循环，循环长度为34，因此只需要小数据暴力，大

数据打表就哦了 .........ＳＧ函数-------->>>点击打开链接

 

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, ..., n_N.

Sum of all N <= 10 ⁶.
1<=n _i<=10 ⁹.

 

Output

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

 

Sample Input

 21222 2 

 

Sample Output

 CarolDave 

 

---

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int sg[] = {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,            0,3,1,1,0,3,3,2,2,4,4,5,5,9,3,3,0,1,1,3,0,2,1,1,0,4,5,3,7,4,8,1,1,2,            0,3,1,1,0,3,3,2,2,4,4,5,5,9,3,3,0,1,1,3,0,2,1,1,0,4,5,3,7,4,8,1,1,2,            0,3,1,1,0,3,3,2,2,4,4,5,5,9,3,3,0,1,1,3,0,2,1,1,0,4,5,3,7,4,8,1,1,2,            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           };int main(){    int t;    scanf("%d",&t);    while(t--)    {      int n,m,t;      int ans=0;      scanf("%d",&n);      for(int i=0;i<n;i++)      {         scanf("%d",&m);         if(m<=150)          t=sg[m];          else          {            m-=150;            m%=34;            m+=150;            t=sg[m];          }          ans^=t;      }      if(ans==0)        printf("Dave\n");        else        printf("Carol\n");    }    return 0;}

打表函数：

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<stdio.h>#include<string.h>using  namespace  std;int vis[10011],sg[10011];void init(){    sg[0]=0;    sg[1]=0;    sg[2]=1;    for(int i=3; i<=10000; i++)//查找前300的    {        memset(vis,0,sizeof(vis));        for(int j=0; j<=(i-2); j++)  //找到他前面的可以划分的进行划分标记            vis[sg[j]^sg[i-2-j]]=1;        for(int j=0;; j++) //找到前面未出现的第一个整数            if(!vis[j])            {                sg[i]=j;                break;            }    }    for(int i=0; i<=300; i++)    {        printf("%d ",sg[i]);        if(i%50==0)            printf("\n");    }}int  main(){    freopen("stdout.txt","w",stdout);    init();}

说实话会了也感觉不容易，第一次接触博弈看了好多博客，终于搞懂了。。。。