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)<<' ';
}