HDU 3980 Paint Chain(sg関数)

Paint Chain
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 804    Accepted Submission(s): 292
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
 
标题:1周にN個の硬貨があり、2人のプレイヤーが交代で取り、毎回M個の連続硬貨を取ることができ、最後の1つが取れないのはマイナスである.

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int N=1010;

int n,m,sg[N];

int mex(int x){
    if(sg[x]!=-1)
        return sg[x];
    int re[N];
    memset(re,0,sizeof(re));
    for(int i=0;i<=x-m-i;i++)
        re[mex(i)^mex(x-m-i)]=1;
    int i=0;
    while(re[i]!=0)
        i++;
    return sg[x]=i;
}

int main(){

    //freopen("input.txt","r",stdin);

    int t,cases=0;
    scanf("%d",&t);
    while(t--){
        memset(sg,-1,sizeof(sg));
        printf("Case #%d: ",++cases);
        scanf("%d%d",&n,&m);
        if(n<m){
            puts("abcdxyzk");
            continue;
        }
        if(m==1){
            if(n&1)
                puts("aekdycoin");
            else
                puts("abcdxyzk");
            continue;
        }
        if(!mex(n-m))
            puts("aekdycoin");
        else
            puts("abcdxyzk");
    }
    return 0;
}