LibreOJ#6436. 「PKUSC2018」神仙的遊戲

鏈接

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題解

不重疊的時候就是個帶有通配符的字符串匹配，很裸

但是當前綴和後綴重合的時候，情況就不太一樣了

可以去想一想當年初學KMP的時候做的一類循環串題，最後會發現：我要檢查$i$是不是$border$，就等價於檢驗$n-i$是不是循環節

檢驗的過程就用多項式卷積即可

代碼

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct EasyMath
{
    ll prime[maxn], phi[maxn], mu[maxn];
    bool mark[maxn];
    ll fastpow(ll a, ll b, ll c)
    {
        ll t(a%c), ans(1ll);
        for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
        return ans;
    }
    void exgcd(ll a, ll b, ll &x, ll &y)
    {
        if(!b){x=1,y=0;return;}
        ll xx, yy;
        exgcd(b,a%b,xx,yy);
        x=yy, y=xx-a/b*yy;
    }
    ll inv(ll x, ll p)  //p是素數
    {return fastpow(x%p,p-2,p);}
    ll inv2(ll a, ll p)
    {
        ll x, y;
        exgcd(a,p,x,y);
        return (x+p)%p;
    }
    void shai(ll N)
    {
        ll i, j;
        for(i=2;i<=N;i++)mark[i]=false;
        *prime=0;
        phi[1]=mu[1]=1;
        for(i=2;i<=N;i++)
        {
            if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
            for(j=1;j<=*prime and i*prime[j]<=N;j++)
            {
                mark[i*prime[j]]=true;
                if(i%prime[j]==0)
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    ll CRT(vector<ll> a, vector<ll> m) //要求模數兩兩互質
    {
        ll M=1, ans=0, n=a.size(), i;
        for(i=0;i<n;i++)M*=m[i];
        for(i=0;i<n;i++)(ans+=a[i]*(M/m[i])%M*inv2(M/m[i],m[i]))%=M;
        return ans;
    }
}em;
#define mod 998244353ll
struct NTT
{
		ll n;
		vector<ll> R;
		void init(ll bound)    //bound是積多項式的最高次冪
		{
			ll L(0);
			for(n=1;n<=bound;n<<=1,L++);
			R.resize(n);
			for(ll i=0;i<n;i++)R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
		}
		void ntt(vector<ll>& a, int opt)
		{
			ll i, j, k, wn, w, x, y, inv(em.fastpow(n,mod-2,mod));
			for(i=0;i<n;i++)if(i>R[i])swap(a[i],a[R[i]]);
			for(i=1;i<n;i<<=1)
			{
				if(opt==1)wn=em.fastpow(3,(mod-1)/(i<<1),mod);
				else wn=em.fastpow(3,(mod-1-(mod-1)/(i<<1)),mod);
				for(j=0;j<n;j+=i<<1)
					for(w=1,k=0;k<i;k++,w=w*wn%mod)
					{
						x=a[k+j], y=a[k+j+i]*w%mod;
						a[k+j]=(x+y)%mod, a[k+j+i]=(x-y)%mod;
					}
			}
			if(opt==-1)for(i=0;i<n;i++)(a[i]*=inv)%=mod;
		}
};
struct Poly
{
    vector<ll> v;
    ll n;	//n是最高次項的次數
    Poly(ll N){v.resize(N+1);n=N;}
    Poly(const Poly& p){v=p.v; n=p.n;}
    void resize(ll N){n=N; v.resize(N+1);}
    ll& operator[](ll id){return v[id];}
    void show()
    {
        printf("n=%lld\n",n);
        ll i; rep(i,0,n-1)printf("%lldx^%lld + ",(v[i]+mod)%mod,i);
        printf("%lldx^%lld\n",(v[n]+mod)%mod,n);
    }
};
Poly operator+(Poly A, Poly B)
{
	ll i;
    Poly C(max(A.n,B.n));
	A.resize(C.n), B.resize(C.n);
    rep(i,0,C.n)C[i]=(A[i]+B[i])%mod;
    return C;
}
Poly operator*(ll x, Poly A)
{
    x%=mod;
    ll i; rep(i,0,A.n)(A[i]*=x)%=mod;
    return A;
}
Poly operator*(Poly A, Poly B)
{
    NTT ntt;
    ll i, n=A.n+B.n;
    ntt.init(A.n+B.n);
    A.resize(ntt.n-1), B.resize(ntt.n-1);
    ntt.ntt(A.v,1), ntt.ntt(B.v,1);
    Poly C(ntt.n-1);
    rep(i,0,C.n)C[i]=(A[i]*B[i])%mod;
    ntt.ntt(C.v,-1);
    C.resize(n);
    return C;
}
int main()
{
	ios::sync_with_stdio(0);
	string s; cin>>s;
	ll n=s.length(), i, ans=0, j;
	Poly A(n-1), B(n-1);
	rep(i,0,n-1)
	{
		A[i] = (s[i]=='0');
		B[i] = (s[n-1-i]=='1');
	}
	auto P = A*B;
	rep(i,1,n-1)
	{
			bool flag = true;
			auto L = n-i;
			for(j = n%L ? n%L : L; j<=n; j+=L)
			{
					if(P[j-1] or P[2*n-1-j])flag=false;
			}
			if(flag)ans ^= i*i;
	}
	ans ^= n*n;
	cout<<ans;
	return 0;
}