HDU3980Paint Chain(尼姆博奕SG)

Paint Chain

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2051    Accepted Submission(s): 751

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

题意：一个有n个物品的环，每次必须拿m个，谁没的拿谁就输了，问谁赢。

思路：当第一个人拿了一次之后就变成一条链了，此时用SG函数求解，对于前i(i>.= m)个，我们取连续m个，就是sg(i-m)^sg(n-i)，这样递推求解。

整体值等于局部值的亦或。

#include<cstdio>#include<iostream>#include<algorithm>#include<queue>#include<stack>#include<cstring>#include<string>#include<vector>#include<cmath> #include<map>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long ll;const int maxn = 2e5+5;const int ff = 0x3f3f3f3f;int n,m;int sg[1005],vis[1005];int getsg(int x){mem(vis,0);for(int j = m;j<= x;j++)vis[sg[j-m]^sg[x-j]] = 1;for(int j = 0;;j++)if(!vis[j]){sg[x] = j;break;}return sg[x];}int main(){int t,cnt = 0;cin>>t;while(t--){scanf("%d %d",&n,&m);if(n< m){printf("Case #%d: abcdxyzk\n",++cnt);continue;}n-= m;mem(sg,-1);for(int i = 0;i<= n;i++)sg[i] = i< m?0:getsg(i);if(sg[n] == 0)printf("Case #%d: aekdycoin\n",++cnt);elseprintf("Case #%d: abcdxyzk\n",++cnt);}return 0;}