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 = 100010;
int sg[MAXN];
bool vis[MAXN];
int mex(int x)
{

    if(sg[x]!=-1)return sg[x];
    if(x == 0)return sg[x] = 0;
    if(x == 1)return sg[x] = 0;
    if(x == 2)return sg[x] = 1;
    if(x == 3)return sg[x] = 1;
    memset(vis,false,sizeof(vis));
    for(int i = 0;i < x-1;i++)
        vis[mex(i)^mex(x-i-2)] = true;
    for(int i = 0;;i++)
        if(!vis[i])
            return sg[x] = i;
}

int SG(int x)
{
    if(x <= 200)return sg[x];
    else
    {
        x %= 34;
        x += 4*34;
        return sg[x];
    }
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    memset(sg,-1,sizeof(sg));
    for(int i = 0;i <= 1000;i++)
    {
        sg[i] = mex(i);
    }
    int T;
    int n;
    int a;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int sum = 0;
        for(int i = 0;i < n;i++)
        {
            scanf("%d",&a);
            sum ^= SG(a);
        }
        if(sum)printf("Carol\n");
        else printf("Dave\n");
    }
    return 0;
}