RT.

写法是二分长度，然后算出需要的长度，维护每个字母后缀出现次数、位置。

WA 31 个点。

代码极其丑陋/kk

int n;
i128 m,k,suf[100010][26],id[26][100010],cnt[26];
string s,t;
bool check(i128 x)
{
//	cout<<" "<<(int)x<<endl;
	i128 c=0;
	for(int i=0;i<k;i++)
	{
//		cout<<" "<<(int)c<<endl;
		i128 p=t[i]-'a';
		i128 d=suf[c%m][p];
//		cout<<" "<<i<<" "<<(int)d<<" "<<(int)c<<endl;
		if(d>=x)
		{
			i128 l=c/m*m;
			c=l+id[p][d-x+1]+1;	
			if(c>n*m)return 0;
		}
		else
		{
//			cout<<"  "<<(int)c<<endl;
			c=c/m*m+m;
//			cout<<"  "<<(int)c<<endl;
//			if(c>n*m)return 0;
			int px=x;
			px-=d;
			i128 ca=suf[0][p];
			if(!ca)return 0; 
			i128 lt=px/ca;
//			cout<<"   "<<(int)lt<<endl;
			if(px%ca==0)
			{
				c+=m*(lt-1);
//				c-=m;
//				cout<<"   "<<(int)c<<" "<<(int)(m*(lt-1))<<endl;
				if(c>n*m||lt-1>n)return 0;
//				cout<<"   "<<(int)c<<" "<<(int)id[p][1]<<endl;
				c+=id[p][1]+1;
				if(c>n*m)return 0;
				continue;
			}
			c+=m*lt;
			if(c>n*m||lt>n)return 0;
			c+=id[p][ca-px%ca];
			if(c>n*m)return 0;
			c++;
			if(c>n*m)return 0;
		}
//		cout<<"  "<<(int)x<<" "<<(int)i<<" "<<(int)c<<endl;
	}
	return c<=n*m;
}
void main_solve(int test_id)
{
	cin>>n>>s>>t;
	m=s.size();
	k=t.size();
	int l=1,r=n*m/k,ans=0;
	for(int i=m-1;i>=0;i--)
	{
		for(int j=0;j<26;j++)
			suf[i][j]=suf[i+1][j]+(s[i]-'a'==j);
		id[s[i]-'a'][++cnt[s[i]-'a']]=i;
	}
	while(l<=r) 
	{
		int mid=l+r>>1;
		if(check(mid))ans=mid,l=mid+1;
		else r=mid-1;
	}
	cout<<ans;
	return;
}