传送门

思路

首先通过各种手玩/找规律/严谨证明，发现当\(n-i\)为border当且仅当对于任意\(k\in[0,i)\)，模\(i\)余\(k\)的位置没有同时出现0和1。

换句话说，拿出任意一个1的位置\(x\)，一个0的位置\(y\)，那么对于\(|x-y|\)的所有约数\(i\)，\(n-i\)均不合法。

考虑用NTT优化这个过程：记两个多项式\(A(x),B(x)\)。若\(s_i=0\)则\([x^i]A(x)=1\)；若\(s_i=1\)则\([x^{n-i}]B(x)=1\)。然后把\(A\)和\(B\)卷积起来即可。

代码

#include<bits/stdc++.h>
clock_t t=clock();
namespace my_std{
    using namespace std;
    #define pii pair<int,int>
    #define fir first
    #define sec second
    #define MP make_pair
    #define rep(i,x,y) for (int i=(x);i<=(y);i++)
    #define drep(i,x,y) for (int i=(x);i>=(y);i--)
    #define go(x) for (int i=head[x];i;i=edge[i].nxt)
    #define templ template<typename T>
    #define sz 4004040
    #define mod 998244353ll
    typedef long long ll;
    typedef double db;
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    templ inline T rnd(T l,T r) {return uniform_int_distribution<T>(l,r)(rng);}
    templ inline bool chkmax(T &x,T y){return x<y?x=y,1:0;}
    templ inline bool chkmin(T &x,T y){return x>y?x=y,1:0;}
    templ inline void read(T& t)
    {
        t=0;char f=0,ch=getchar();double d=0.1;
        while(ch>'9'||ch<'0') f|=(ch=='-'),ch=getchar();
        while(ch<='9'&&ch>='0') t=t*10+ch-48,ch=getchar();
        if(ch=='.'){ch=getchar();while(ch<='9'&&ch>='0') t+=d*(ch^48),d*=0.1,ch=getchar();}
        t=(f?-t:t);
    }
    template<typename T,typename... Args>inline void read(T& t,Args&... args){read(t); read(args...);}
    char __sr[1<<21],__z[20];int __C=-1,__zz=0;
    inline void Ot(){fwrite(__sr,1,__C+1,stdout),__C=-1;}
    inline void print(register int x)
    {
        if(__C>1<<20)Ot();if(x<0)__sr[++__C]='-',x=-x;
        while(__z[++__zz]=x%10+48,x/=10);
        while(__sr[++__C]=__z[__zz],--__zz);__sr[++__C]='\n';
    }
    void file()
    {
        #ifdef NTFOrz
        freopen("a.in","r",stdin);
        #endif
    }
    inline void chktime()
    {
        #ifndef ONLINE_JUDGE
        cout<<(clock()-t)/1000.0<<'\n';
        #endif
    }
    #ifdef mod
    ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x%mod) if (y&1) ret=ret*x%mod;return ret;}
    ll inv(ll x){return ksm(x,mod-2);}
    #else
    ll ksm(ll x,int y){ll ret=1;for (;y;y>>=1,x=x*x) if (y&1) ret=ret*x;return ret;}
    #endif
//  inline ll mul(ll a,ll b){ll d=(ll)(a*(double)b/mod+0.5);ll ret=a*b-d*mod;if (ret<0) ret+=mod;return ret;}
}
using namespace my_std;

int r[sz],limit;
void NTT_init(int n)
{
    limit=1;int l=-1;
    while (limit<=n+n) limit<<=1,++l;
    rep(i,0,limit-1) r[i]=(r[i>>1]>>1)|((i&1)<<l);
}
void NTT(ll *a,int type)
{
    rep(i,0,limit-1) if (i<r[i]) swap(a[i],a[r[i]]);
    for (int mid=1;mid<limit;mid<<=1)
    {
        ll Wn=ksm(3,(mod-1)/mid>>1);if (type==-1) Wn=inv(Wn);
        for (int len=mid<<1,j=0;j<limit;j+=len)
        {
            ll w=1;
            for (int k=0;k<mid;k++,w=w*Wn%mod)
            {
                ll x=a[j+k],y=a[j+k+mid]*w%mod;
                a[j+k]=(x+y)%mod;a[j+k+mid]=(x-y+mod)%mod;
            }
        }
    }
    if (type==1) return;
    ll I=inv(limit);
    rep(i,0,limit-1) a[i]=a[i]*I%mod;
}

int n;
char s[sz];

ll tmp1[sz],tmp2[sz],a[sz];
ll ans;

int main()
{
    file();
    cin>>(s+1);n=strlen(s+1);
    rep(i,1,n) if (s[i]=='0') tmp1[i]=1;
    rep(i,1,n) if (s[i]=='1') tmp2[n-i]=1;
    NTT_init(n);
    NTT(tmp1,1);NTT(tmp2,1);
    rep(i,0,limit-1) tmp1[i]=tmp1[i]*tmp2[i]%mod;
    NTT(tmp1,-1);
    rep(i,1,n+n) a[i]=tmp1[i];
    rep(i,1,n-1)
    {
        bool flg=1;
        for (int j=i;j<n;j+=i) flg&=(a[n-j]==0&&a[n+j]==0);
        if (flg) ans^=1ll*(n-i)*(n-i);
    }
    ans^=1ll*n*n;
    cout<<ans;
    return 0;
}