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Paint Chain
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：有一个带有n个珠子的圆链，起初圆链上的珠子都未被染色。规定两人轮流染色，每人每次只能挑选m个连续的未被染色的珠子进行染色，Aekdycoin先来，最后谁先不能染色谁则输掉了这场比赛，问最后谁会赢
思路：由于f[]数组不好求，所以要使用dfs版的SG函数
Aekdycoin先挑m个染色，则这个环就变成了一条链，那么接下来就类似于尼姆博弈了，此时abcdxyzk变成了先手
对于每一步有以下情况（现在珠子数为n-m）:
前面空0个，开始染m个，则子游戏为(0,m)，(n-m-m,m)
前面空1个，开始染m个，则子游戏为(1,m)，(n-m-m-1,m)
前面空2个，开始染m个，则子游戏为(2,m)，(n-m-m-2,m)
……
前面空n-m-m个，开始染m个，则子游戏为(n-m-m,m)，(0,m)
对于每一种情况的游戏，它的结果等于其所有子游戏结果的异或。
这样即可求出谁赢
代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn=1e3+10;
int sg[maxn];
int n,m;
int SG_dfs(int x)
{
if(x<m)
return sg[x]=0;
if(sg[x]!=-1)
return sg[x];
bool vis[maxn]={
false};
for(int i=0;i<=x-m;++i)
vis[SG_dfs(i)^SG_dfs(x-m-i)]=true;
for(int j=0;;++j)
if(!vis[j])
{
sg[x]=j;
break;
}
return sg[x];
}
int main()
{
int t,k=0;
scanf("%d",&t);
while(++k<=t)
{
memset(sg,-1,sizeof(sg));
scanf("%d%d",&n,&m);
if(n<m)
{
printf("Case #%d: abcdxyzk\n",k);
continue;
}
SG_dfs(n-m);
if(sg[n-m])
printf("Case #%d: abcdxyzk\n",k);
else
printf("Case #%d: aekdycoin\n",k);
}
return 0;
}
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