【LOJ #6436】「PKUSC2018」神仙的游戏（NTT）

传送门

考虑对于一个长度为$k$的$border$
那么任意$i,j$满足$i\equiv j(\mod n-k)$要满足$s_i=s_j$
考虑不满足的话一定是有$i,j$满足$s_i\not=s_j$
变下形是$(n-k)|i+(-j)$
考虑类似卷积做字符串匹配的方法
把$\sum_{i}[s[i]==0]x^i$和$\sum_{j}[s[j]==1]x^{n-j}$做卷积
然后调和级数看所有$p*(x-k)$的位置是否有值即可

#include<bits/stdc++.h>
using namespace std;
#define cs const
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define pb push_back
cs int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ib==ob)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ib==ob)?EOF:*ib++;
}
inline int read(){
	char ch=gc();
	int res=0;bool f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
inline int readstring(char *s){
	int top=0;
	char ch=gc();
	while(isspace(ch))ch=gc();
	while(!isspace(ch)&&ch!=EOF)s[++top]=ch,ch=gc();
	return top;
}
template<class tp>inline void chemx(tp &a,tp b){a<b?a=b:0;}
template<class tp>inline void chemn(tp &a,tp b){a>b?a=b:0;}
cs int mod=998244353,G=3;
inline int add(int a,int b){return (a+b)>=mod?(a+b-mod):(a+b);}
inline void Add(int &a,int b){a=(a+b)>=mod?(a+b-mod):(a+b);}
inline int dec(int a,int b){a-=b;return a+(a>>31&mod);}
inline void Dec(int &a,int b){a-=b,a+=a>>31&mod;}
inline int mul(int a,int b){static ll r;r=1ll*a*b;return r>=mod?r%mod:r;}
inline void Mul(int &a,int b){static ll r;r=1ll*a*b;a=r>=mod?r%mod:r;}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,Mul(a,a))(b&1)&&(Mul(res,a),1);return res;}
inline int Inv(int x){return ksm(x,mod-2);}
cs int C=21;
#define poly vector<int>
poly w[C+1];
int rev[(1<<C)|5];
inline void init_rev(int lim){
	for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1));
}
inline void init_w(){
	for(int i=1;i<=C;i++)w[i].resize(1<<(i-1));
	int wn=ksm(G,(mod-1)/(1<<C));
	w[C][0]=1;
	for(int i=1,lim=(1<<(C-1));i<lim;i++)
	w[C][i]=mul(w[C][i-1],wn);
	for(int i=C-1;i;i--)
	for(int j=0,lim=1<<(i-1);j<lim;j++)
	w[i][j]=w[i+1][j<<1];
}
inline void ntt(int *f,int lim,int kd){
	for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);
	for(int mid=1,l=1,a0,a1;mid<lim;mid<<=1,l++)
	for(int i=0;i<lim;i+=(mid<<1))
	for(int j=0;j<mid;j++)
	a0=f[i+j],a1=mul(f[i+j+mid],w[l][j]),f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1);
	if(kd==-1){
		reverse(f+1,f+lim);
		for(int i=0,iv=Inv(lim);i<lim;i++)Mul(f[i],iv);
	}
}
inline void pmul(int *a,int *b,int deg){
	int lim=1;
	while(lim<deg)lim<<=1;
	init_rev(lim);
	ntt(a,lim,1),ntt(b,lim,1);
	for(int i=0;i<lim;i++)Mul(a[i],b[i]);
	ntt(a,lim,-1);
}
cs int N=500005;
char s[N];
int n,A[(1<<C)+5],B[(1<<C)+5],c[(1<<C)+5];
int main(){
#ifdef Stargazer
freopen("lx.in","r",stdin);
#endif
	init_w();
	n=readstring(s);
	for(int i=1;i<=n;i++)if(s[i]=='0')A[i]=1;else if(s[i]=='1')B[n-i]=1;
	pmul(A,B,(n+1)*2);
	for(int i=1;i<=(n+1)*2;i++)c[abs(i-n)]+=A[i];
	ll ret=0;
	for(int i=1;i<=n;i++){
		int flag=1;
		for(int j=0;j<=n;j+=i)flag&=(!c[j]);
		ret^=1ll*(n-i)*(n-i)*flag;
	}
	cout<<(ret^(1ll*n*n))<<'\n';
	return 0;
}