hdu-3980-nim博弈/sg函数

Paint Chain

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2465    Accepted Submission(s): 880

Problem Description

Aekdycoin and abcdxyzk are playing a game. They get a circle chain with some beads. Initially none of the beads is painted. They take turns to paint the chain. In Each turn one player must paint a unpainted beads. Whoever is unable to paint in his turn lose the game. Aekdycoin will take the first move.

Now, they thought this game is too simple, and they want to change some rules. In each turn one player must select a certain number of consecutive unpainted beads to paint. The other rules is The same as the original. Who will win under the rules ?You may assume that both of them are so clever.

 

Input

First line contains T, the number of test cases. Following T line contain 2 integer N, M, indicate the chain has N beads, and each turn one player must paint M consecutive beads. (1 <= N, M <= 1000)

 

Output

For each case, print "Case #idx: " first where idx is the case number start from 1, and the name of the winner.

 

Sample Input

2 3 1 4 2

 

Sample Output

Case #1: aekdycoin Case #2: abcdxyzk

 

Author

jayi

 

Source

2011 Multi-University Training Contest 14 - Host by FZU

 

        一串长度为n的圆形珠子，每次可以找连续的m个白色珠子染色，不能操作者输，问先手状态。

　　 先计算出线性状态下的游戏的sg函数，再计算与真实状态的sg函数时判断sg[n-m]=sg[n-m]==0?1:0。

       由于子状态只有一种。

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int sg[1010];
 4 bool vis[1010];
 5 int main(){
 6     int t,n,m,i,j,k;
 7     cin>>t;
 8     for(int cas=1;cas<=t;++cas){
 9         cin>>n>>m;
10         printf("Case #%d: ",cas);
11         if(n<m){
12             puts("abcdxyzk");
13             continue;
14         }
15         for(i=0;i<m;++i) sg[i]=0;
16         for(i=m;i<=n;++i){
17             memset(vis,0,sizeof(vis));
18             for(j=1;j+m-1<=i;j++){
19                 vis[sg[j-1]^sg[i-m-j+1]]=1;
20             }
21             for(j=0;;j++){
22                 if(!vis[j]){
23                     sg[i]=j;
24                     break;
25                 }
26             }
27         }
28         for(i=0;i<=n;++i){
29             if(sg[i]==0) sg[i]=1;
30             else sg[i]=0;
31         }
32         sg[n-m]?puts("aekdycoin"):puts("abcdxyzk");
33     }
34     return 0;
35 }