hdu 3980 Paint Chain 博弈+SG函数每次选连续M个

游戏策略博弈分析

最新推荐文章于 2023-07-11 00:59:15 发布

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本文介绍了一种基于博弈论的游戏策略分析方法，通过编程实现游戏中的最优策略选择过程，并以两个玩家轮流涂色珠子的游戏为例，展示了如何通过计算SG函数来判断胜者。

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

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1100;
//一条线每次取连续M个 SG函数
int sg[maxn],flag[maxn];
int calc_sg(int n,int m)
{
    for(int i=0;i<m;i++) sg[i]=0;
    for(int i=m;i<=n;i++)
    {
        memset(flag,0,sizeof(flag));
        for(int j=0;j+m<=i;j++)
        {
            flag[sg[j]^sg[i-j-m]]=1;
        }
        for(int j=0;;j++)
        {
            if(flag[j]==0)
            {
                sg[i]=j;break;
            }
        }
    }
}
int main()
{
    int ci,pl=1;scanf("%d",&ci);
    while(ci--)
    {
        int n,m;scanf("%d%d",&n,&m);
        if(n<m)
        {
            printf("Case #%d: abcdxyzk\n",pl++);
            continue;
        }
        calc_sg(n-m,m);
        if(sg[n-m]==0) printf("Case #%d: aekdycoin\n",pl++);
        else printf("Case #%d: abcdxyzk\n",pl++);
    }
    return 0;
}