HDU 4664 Triangulation（2013多校6 1010题，博弈）

Triangulation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 96    Accepted Submission(s): 29

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, ..., n_N.

Sum of all N <= 10⁶.
1<=n_i<=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

 

Source

2013 Multi-University Training Contest 6

 

Recommend

zhuyuanchen520

 

 

 

 

这题一开始看错题目意思了。

 

导致连SG函数转移都写不出来。

 

其实这题看懂了就很好搞了。

 

每次加边，不能形成三角形，所以肯定不加共点的边，否则就是自杀。

 

x个点，转移后相当于 i    ,    x-i-2 .加的那两个点去掉了。

 

 

SG函数打表以后，很明显是要找规律。

发现周期是34.

而且周期要到后面才有周期。

 

所以前面打表，后面利用周期。

 

可以参考下oeis,发现这个是经典的问题。Sprague-Grundy values for Dawson's Chess

http://oeis.org/search?q=0%2C1%2C1%2C2%2C0%2C3%2C1%2C1%2C0%2C3%2C3%2C2%2C2%2C4%2C0%2C5%2C2%2C2%2C3%2C3%2C0&sort=&language=english&go=Search

 

Has period 34 with the only exceptions at n=0, 14, 16, 17, 31, 34 and 51.

 

 

然后胡搞下就过了

 

 

 /*
  * Author:  kuangbin
  * Created Time:  2013/8/8 11:54:23
  * File Name: 1010.cpp
  */
 #include <iostream>
 #include <cstdio>
 #include <cstdlib>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <string>
 #include <vector>
 #include <stack>
 #include <queue>
 #include <set>
 #include <time.h>
 using namespace std;
 const int MAXN = ;
 int sg[MAXN];
 bool vis[MAXN];
 int mex(int x)
 {

     if(sg[x]!=-)return sg[x];
     if(x == )return sg[x] = ;
     if(x == )return sg[x] = ;
     if(x == )return sg[x] = ;
     if(x == )return sg[x] = ;
     memset(vis,false,sizeof(vis));
     for(int i = ;i < x-;i++)
         vis[mex(i)^mex(x-i-)] = true;
     for(int i = ;;i++)
         if(!vis[i])
             return sg[x] = i;
 }

 int SG(int x)
 {
     if(x <= )return sg[x];
     else
     {
         x %= ;
         x += *;
         return sg[x];
     }
 }

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);
     memset(sg,-,sizeof(sg));
     for(int i = ;i <= ;i++)
     {
         sg[i] = mex(i);
     }
     int T;
     int n;
     int a;
     scanf("%d",&T);
     while(T--)
     {
         scanf("%d",&n);
         int sum = ;
         for(int i = ;i < n;i++)
         {
             scanf("%d",&a);
             sum ^= SG(a);
         }
         if(sum)printf("Carol\n");
         else printf("Dave\n");
     }
     return ;
 }