bzoj1566

好题，这道题体现了换一个角度计数的思想

a1^2+a2^2+……ak^2=(变成第1种输出序列的操作序列数目)^2+(变成第2种输出序列的操作序列数目)^2……

脑洞大一点，这就相当于两个操作序列变成相同输出序列的对数（包括自己和自己）

于是dp即可……dp[i,j,k]表示输出到第i个，第一个操作序列在上面取了j个，第二个操作序列在上面取了k个，怎么弄大家都会……

 const mo=;

 var s1,s2:array[..] of char;
     p,m,n,i,j,k,j1,k1:longint;
     f:array[..,..,..] of longint;

 procedure reverse;
   var s:ansistring;
   begin
     readln(s);
     for i:= to n do
       s1[n-i+]:=s[i];
     s1[n+]:='$';
     readln(s);
     for i:= to m do
       s2[m-i+]:=s[i];
     s2[m+]:='#';
   end;

 begin
   readln(n,m);
   reverse;
   f[,,]:=;
   for i:= to m+n do
   begin
     p:=-p;
     fillchar(f[p],sizeof(f[p]),);
     for j:= to n do
       for k:= to n do
         if f[-p,j,k]>=mo then f[-p,j,k]:=f[-p,j,k] mod mo;
     for j:= to n do
       for k:= to n do
       begin
         j1:=i--j;
         k1:=i--k;
         if (j1>=) and (k1>=) and (j1<=m) and (k1<=m) then
         begin
           if s1[j+]=s1[k+] then inc(f[p,j+,k+],f[-p,j,k]);
           if s1[j+]=s2[k1+] then inc(f[p,j+,k],f[-p,j,k]);
           if s2[j1+]=s1[k+] then inc(f[p,j,k+],f[-p,j,k]);
           if s2[j1+]=s2[k1+] then inc(f[p,j,k],f[-p,j,k]);
         end;
       end;
   end;
   writeln(f[p,n,n] mod mo);
 end.