对作文库中的串建出广义SAM，然后显然可以二分答案，二分之后考虑暴力dp，设f[i]为前i位最长匹配长度，显然有f[i]=max(f[i-1],f[j]+i-j) (i-j>=l&&j+1~i能在作文库中匹配)。在SAM上跑并记录匹配长度，单调队列优化dp即可。

#include
     
      using namespace std;#define ll long long#define N 2200010char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}int gcd(int n,int m){return m==0?n:gcd(m,n%m);}int read(){	int x=0,f=1;char c=getchar();	while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}	while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();	return x*f;}int n,m,son[N][2],fail[N],len[N],f[N],q[N],cnt=1,last;char s[N];int ins(int c,int p){    if (son[p][c])    {        int q=son[p][c];        if (len[p]+1==len[q]) return q;        else        {            int y=++cnt;            len[y]=len[p]+1;            memcpy(son[y],son[q],sizeof(son[q]));            fail[y]=fail[q],fail[q]=y;            while (son[p][c]==q) son[p][c]=y,p=fail[p];            return y;        }    }    else    {        int x=++cnt;len[x]=len[p]+1;        while (!son[p][c]&&p) son[p][c]=x,p=fail[p];        if (!p) fail[x]=1;        else        {            int q=son[p][c];            if (len[p]+1==len[q]) fail[x]=q;            else            {                int y=++cnt;                len[y]=len[p]+1;                memcpy(son[y],son[q],sizeof(son[q]));                fail[y]=fail[q],fail[x]=fail[q]=y;                while (son[p][c]==q) son[p][c]=y,p=fail[p];            }        }        return x;    }}bool check(int k,int n){	int x=1,l=0;int head=0,tail=0;	for (int i=1;i<=n;i++)	{		while (!son[x][s[i]-'0']&&x) x=fail[x],l=len[x];		if (!x) x=1,l=0;		else x=son[x][s[i]-'0'],l++;		f[i]=f[i-1];		if (i>=k)		{			while (head<=tail&&f[q[tail]]-q[tail]<=f[i-k]-(i-k)) tail--;			q[++tail]=i-k;		}		if (l>=k)		{			while (q[head]
      
       =n*9;}int main(){#ifndef ONLINE_JUDGE	freopen("a.in","r",stdin);	freopen("a.out","w",stdout);	const char LL[]="%I64d\n";#else	const char LL[]="%lld\n";#endif	n=read(),m=read();	for (int i=1;i<=m;i++)	{		scanf("%s",s+1);int _=strlen(s+1);last=1;		for (int i=1;i<=_;i++) last=ins(s[i]-'0',last);	}	for (int i=1;i<=n;i++)	{		scanf("%s",s+1);int _=strlen(s+1);		int l=1,r=_,ans=0;		while (l<=r)		{			int mid=l+r>>1;			if (check(mid,_)) ans=mid,l=mid+1;			else r=mid-1;		}		printf("%d\n",ans);	}	return 0;}