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=1024523;

var s1,s2:array[0..600] of char;
    p,m,n,i,j,k,j1,k1:longint;
    f:array[0..1,0..510,0..510] of longint;

procedure reverse;
  var s:ansistring;
  begin
    readln(s);
    for i:=1 to n do
      s1[n-i+1]:=s[i];
    s1[n+1]:='$';
    readln(s);
    for i:=1 to m do
      s2[m-i+1]:=s[i];
    s2[m+1]:='#';
  end;

begin
  readln(n,m);
  reverse;
  f[0,0,0]:=1;
  for i:=1 to m+n do
  begin
    p:=1-p;
    fillchar(f[p],sizeof(f[p]),0);
    for j:=0 to n do
      for k:=0 to n do
        if f[1-p,j,k]>=mo then f[1-p,j,k]:=f[1-p,j,k] mod mo;
    for j:=0 to n do
      for k:=0 to n do
      begin
        j1:=i-1-j;
        k1:=i-1-k;
        if (j1>=0) and (k1>=0) and (j1<=m) and (k1<=m) then
        begin
          if s1[j+1]=s1[k+1] then inc(f[p,j+1,k+1],f[1-p,j,k]);
          if s1[j+1]=s2[k1+1] then inc(f[p,j+1,k],f[1-p,j,k]);
          if s2[j1+1]=s1[k+1] then inc(f[p,j,k+1],f[1-p,j,k]);
          if s2[j1+1]=s2[k1+1] then inc(f[p,j,k],f[1-p,j,k]);
        end;
      end;
  end;
  writeln(f[p,n,n] mod mo);
end.