bzoj3530

比较恶心的题目
不难发现是在自动机上做数位dp
注意要考虑前导0，题目中给出的233是幸运数，20233不是
为此我非常猥琐的写了一个四维dp，用记忆化搜索实现

 const mo=;
 var dp:array[..,..,..,..] of longint;
     trie:array[..,''..''] of longint;
     q,f:array[..] of longint;
     can:array[..] of boolean;
     ans,i,n,m,t,j,k,l:longint;
     s,ss:ansistring;
     ch:char;

 procedure ac;
   var h,r,i,x,y:longint;
       c:char;
   begin
     h:=;
     r:=;
     for c:='' to '' do
       if trie[,c]> then
       begin
         inc(r);
         q[r]:=trie[,c];
       end;

     while h<=r do
     begin
       x:=q[h];
       for c:='' to '' do
         if trie[x,c]> then
         begin
           y:=trie[x,c];
           inc(r);
           q[r]:=y;
           j:=f[x];
           while (j>) and (trie[j,c]=) do j:=f[j];
           f[y]:=trie[j,c];
           if can[trie[j,c]] then can[y]:=true; //注意
         end;
       inc(h);
     end;
   end;

 function calc(p,z,i,j:longint):longint; //p表示当前位置上的数是否需要小于N当前位上的数
                                         //z表示假如当前位出现0，是否是多余的0
   var k,q,zz:longint;
       s,c,e:char;
   begin
     if can[j] then exit();
     if i=m then exit();
     if dp[p,z,i,j]<>- then exit(dp[p,z,i,j]);
     dp[p,z,i,j]:=;
     if p= then e:=ss[i+] else e:='';
     for c:='' to e do
     begin
       if (p=) and (c=e) then q:=
       else q:=;
       k:=j;
       while k>- do
       begin
         if trie[k,c]> then break;
         k:=f[k];
       end;
       if k=- then k:= else k:=trie[k,c];
       if (z=) and (c='') then  //开头多余的0不考虑
       begin
         zz:=;
         k:=;  //多余的0不在自动机上匹配
       end
       else zz:=;
       dp[p,z,i,j]:=(dp[p,z,i,j]+calc(q,zz,i+,k)) mod mo;
     end;
     exit(dp[p,z,i,j]);
   end;

 begin
   read(ch);
   while ch='' do read(ch);
   readln(ss);
   ss:=ch+ss;
   m:=length(ss);
   f[]:=-;
   readln(n);
   for i:= to n do
   begin
     readln(s);
     j:=;
     l:=length(s);
     for k:= to l do
     begin
       if trie[j,s[k]]= then
       begin
         inc(t);
         trie[j,s[k]]:=t;
       end;
       j:=trie[j,s[k]];
     end;
     can[j]:=true;
   end;
   ac;
   fillchar(dp,sizeof(dp),);
   ans:=(calc(,,,)-+mo) mod mo;
   writeln(ans);
 end.