8个常见的硬币博弈的SG值规律

翻硬币游戏

 $N$枚硬币排成一排，有的正面朝上，有的反面朝上。我们从左开始对硬币按1 到N编号。

最右边那个硬币的必须是从正面翻到反面,.谁不能翻谁输。

局面的SG值为局面中每个正面朝上的棋子单一存在时的SG值的异或和

1.每次只能翻一个硬币,

显然,每个硬币的SG值为1

2.每次能翻转一个或两个(不用连续)

每个硬币的SG值为其从左到右的编号位置,从0开始

#include <bits/stdc++.h>
#define ll long long
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=1e5+10,maxm=2e6+10;
int casn,n,m,k,kase;
int sg[maxn];
int getmex(bool vis[]){
  int mex=0;
  while(vis[mex]) ++mex;
  return mex;
}
int getsg(int now){
  if(~sg[now]) return sg[now];
  bool vis[now+1]={0};
  rep(i,1,now-1) vis[getsg(i)]=1;
  vis[now]=1;
  return sg[now]=getmex(vis);
}
int main(){IO;
  memset(sg,-1,sizeof sg);
  cin>>n;
  getsg(n);
  showa(sg,1,n);
}

 3.每次只能翻转连续k枚硬币

每个硬币的SG值为0000...100000..1,其中每k个为一个周期,每个周期前k-1个为0

#include <bits/stdc++.h>
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=1e5+10,maxm=2e6+10;
int casn,n,m,k,kase;
int sg[maxn];
int getmex(bool vis[]){
  int mex=0;
  while(vis[mex]) ++mex;
  return mex;
}
int getsg(int now,int k){
  if(~sg[now]) return sg[now];
  bool vis[now+1]={0};
  int flag=0;
  rep(i,now-k+1,now-1){
    flag^=getsg(i,k);
  }
  vis[flag]=1;
  return sg[now]=getmex(vis);
}
int main(){IO;
  memset(sg,-1,sizeof sg);
  cin>>n>>k;
  rep(i,0,k-1) sg[i]=0;
  getsg(n,k);
  showa(sg,1,n);
}

4：每次翻动第X个硬币后，必须翻动其左侧最近K个硬币中的一个，除非X是小于等于3的

SG值为1,2,3,4..K,0,1,2,3,4..K,0每K+1个数字为一个周期

#include <bits/stdc++.h>
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=1e5+10,maxm=2e6+10;
int casn,n,m,k,kase;
int sg[maxn];
int getmex(bool vis[]){
  int mex=0;
  while(vis[mex]) ++mex;
  return mex;
}
int getsg(int now,int k){
  if(~sg[now]) return sg[now];
  bool vis[now+1]={0};
  rep(i,max(0,now-k),now-1){
    vis[getsg(i,k)]=1;
  }
  return sg[now]=getmex(vis);
}
int main(){
  memset(sg,-1,sizeof sg);
  cin>>n>>k;
  getsg(n,k);
  showa(sg,1,n);
}

 5.每次可以翻1,2,3..k个硬币

规律,sg[1]=1,其余值为从前几个中选出至多k个数字异或和的集合mex

#include <bits/stdc++.h>
#pragma GCC target ("popcnt")
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=1e5+10,maxm=2e6+10;
int casn,n,m,k,kase;
int sg[maxn];
int getmex(bool vis[]){
  int mex=0;
  while(vis[mex]) ++mex;
  return mex;
}
int getsg(int now,int k){
  if(~sg[now]) return sg[now];
  bool vis[now*200]={0};
  int st=(1<<(now-1))-1;
  rep(i,0,st){
    if(__builtin_popcount(i)<k){
      int x=0;
      for(int j=i;j;j&=j-1){
        x^=getsg(__builtin_ffs(j),k);
      }
      vis[x]=1;
    }
  }
  return sg[now]=getmex(vis);
}
int main(){
  memset(sg,-1,sizeof sg);
  cin>>n>>k;
  rep(i,1,n)getsg(i,k);
  showa(sg,1,n);
  return 0;
}

 6.每次可以翻1,2,3..k个硬币

SG[1]=1,其余的SG值为,前面所有的SG值选出至多k个的所有组合的异或和集合的mex

#include <bits/stdc++.h>
#pragma GCC target ("popcnt")
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=1e5+10,maxm=2e6+10;
int casn,n,m,k,kase;
int sg[maxn];
int getmex(bool vis[]){
  int mex=0;
  while(vis[mex]) ++mex;
  return mex;
}
int getsg(int now,int k){
  if(~sg[now]) return sg[now];
  bool vis[now*1000]={0};
  int st=(1<<(now-1))-1;
  rep(i,0,st){
    if(__builtin_popcount(i)<k){
      int x=0;
      for(int j=i;j;j&=j-1){
        x^=getsg(__builtin_ffs(j),k);
      }
      vis[x]=1;
    }
  }
  return sg[now]=getmex(vis);
}
int main(){
  memset(sg,-1,sizeof sg);
  cin>>n>>k;
  rep(i,1,n)getsg(i,k);
  showa(sg,1,n);
  return 0;
}

 7.每次可以翻转连续的任意个硬币

sg: 1 2 1 4 1 2 1 8 1 2 1 4 1 2 1...

sg[x]为lowbit(x)

#include <bits/stdc++.h>
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=1e5+10,maxm=2e6+10;
int casn,n,m,k,kase;
int sg[maxn];
int getmex(bool vis[]){
  int mex=0;
  while(vis[mex]) ++mex;
  return mex;
}
int getsg(int now){
  if(~sg[now]) return sg[now];
  bool vis[now*10]={0};
  int flag=0;
  per(i,1,now-1){
    int x=0;
    rep(j,i,now-1) x^=getsg(j);
    vis[x]=1;
  }
  vis[0]=1;
  return sg[now]=getmex(vis);
}
inline int lb(int x) {return x&(-x);}
int main(){
  memset(sg,-1,sizeof sg);
  cin>>n;
  rep(i,1,n)getsg(i);
  showa(sg,1,n);
  cout<<"lb: ";rep(i,1,n) cout<<lb(i)<<' ';
}

8.每次必须翻转4个对称的硬币,其中最左与最右的硬币都必须是从正翻到反,且初始情况保证1和n为正

sg: 0 0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9

规律为max(0,(i-2)/2) 同时也是正多边形中不全等的对角线个数

注意此时,单个局面为首尾都是正面硬币

#include <bits/stdc++.h>
#define rep(ii,a,b) for(int ii=a;ii<=b;++ii)
#define per(ii,a,b) for(int ii=b;ii>=a;--ii)
#define show(x) cout<<#x<<"="<<x<<endl
#define show2(x,y) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<endl
#define show3(x,y,z) cout<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show4(w,x,y,z) cout<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define show5(v,w,x,y,z) cout<<#v<<"="<<v<<" "<<#w<<"="<<w<<" "<<#x<<"="<<x<<" "<<#y<<"="<<y<<" "<<#z<<"="<<z<<endl
#define showa(x,a,b) cout<<#x<<": ";rep(i,a,b) cout<<x[i]<<' ';cout<<endl
using namespace std;//head
const int maxn=1e5+10,maxm=2e6+10;
int casn,n,m,k,kase;
int sg[maxn];
int getmex(bool vis[]){
  int mex=0;
  while(vis[mex]) ++mex;
  return mex;
}
int getsg(int now){
  if(~sg[now]) return sg[now];
  bool vis[now*100]={0};
  rep(i,1,now){
    if(i>=now-i) break;
    int flag=getsg(now-2*i);
    vis[flag]=1;
  }
  return sg[now]=getmex(vis);
}
int main(){
  memset(sg,-1,sizeof sg);
  cin>>n;
  sg[1]=sg[2]=sg[3]=0;
  rep(i,1,n)getsg(i);
  showa(sg,1,n);
  cout<<"fx: ";rep(i,1,n) cout<<max(0,(i-2)/2)<<' ';
}