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.
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.
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
Source
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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
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
1 /* 2 * Author: kuangbin 3 * Created Time: 2013/8/8 11:54:23 4 * File Name: 1010.cpp 5 */ 6 #include <iostream> 7 #include <cstdio> 8 #include <cstdlib> 9 #include <cstring> 10 #include <cmath> 11 #include <algorithm> 12 #include <string> 13 #include <vector> 14 #include <stack> 15 #include <queue> 16 #include <set> 17 #include <time.h> 18 using namespace std; 19 const int MAXN = 100010; 20 int sg[MAXN]; 21 bool vis[MAXN]; 22 int mex(int x) 23 { 24 25 if(sg[x]!=-1)return sg[x]; 26 if(x == 0)return sg[x] = 0; 27 if(x == 1)return sg[x] = 0; 28 if(x == 2)return sg[x] = 1; 29 if(x == 3)return sg[x] = 1; 30 memset(vis,false,sizeof(vis)); 31 for(int i = 0;i < x-1;i++) 32 vis[mex(i)^mex(x-i-2)] = true; 33 for(int i = 0;;i++) 34 if(!vis[i]) 35 return sg[x] = i; 36 } 37 38 int SG(int x) 39 { 40 if(x <= 200)return sg[x]; 41 else 42 { 43 x %= 34; 44 x += 4*34; 45 return sg[x]; 46 } 47 } 48 49 int main() 50 { 51 //freopen("in.txt","r",stdin); 52 //freopen("out.txt","w",stdout); 53 memset(sg,-1,sizeof(sg)); 54 for(int i = 0;i <= 1000;i++) 55 { 56 sg[i] = mex(i); 57 } 58 int T; 59 int n; 60 int a; 61 scanf("%d",&T); 62 while(T--) 63 { 64 scanf("%d",&n); 65 int sum = 0; 66 for(int i = 0;i < n;i++) 67 { 68 scanf("%d",&a); 69 sum ^= SG(a); 70 } 71 if(sum)printf("Carol "); 72 else printf("Dave "); 73 } 74 return 0; 75 }