Bzoj1566 [noi2009] pipe beads

This is a very complicated thinking DP question.

Why is the same as the two sequences? In YY, it is equivalent to getting two people together, and the final sequence is the same.

Then the water goes:

F [I, j, k] indicates that the first bead was obtained, and the first person took the J beads in the pipe No. 1, when the second person took K beads in the No. 1 pipeline, they took the number of equal sequences.

So it's really water !!! (For the equation, please make up your mind or read the program)

Then I am so confused about the programming ability, one hour ah, Grandpa...

 

 1 /************************************************************** 2     Problem: 1566 3     User: rausen 4     Language: Pascal 5     Result: Accepted 6     Time:11920 ms 7     Memory:3592 kb 8 ****************************************************************/ 9  10 const prime = 1024523;11  12 var13   s1, s2 : array[0..600] of char;14   m, n, i, j, k, j1, k1 : longint;15   p, q : longint;16   f : array[0..1, 0..600, 0..600] of longint;17  18 procedure reverse;19 var20   s : ansistring;21   i : longint;22  23 begin24   readln(s);25   for i := 1 to m do26     s1[m - i + 1] := s[i];27   readln(s);28   for i := 1 to n do29     s2[n - i + 1] := s[i];30 end;31  32 begin33   readln(m, n);34   reverse;35   f[0, 1, 1] := 1;36   for i := 1 to m + n + 1 do begin37     p := i mod 2;38     q := p xor 1;39     fillchar(f[p], sizeof(f[p]), 0);40     for j := 1 to m + 1 do41       for k := 1 to m + 1 do42         while f[q, j, k] >= prime do43           dec(f[q, j, k], prime);44     for j := 1 to m + 1 do45       for k := 1 to m + 1 do begin46         j1 := i - j + 1;47         k1 := i - k + 1;48         if (j1 > 0) and (k1 > 0) and (j1 <= n + 1) and (k1 <= n + 1) then begin49           if s1[j] = s1[k] then inc(f[p, j + 1, k + 1], f[q, j, k]);50           if s1[j] = s2[k1] then inc(f[p, j + 1, k], f[q, j ,k]);51           if s2[j1] = s1[k] then inc(f[p, j, k + 1], f[q, j ,k]);52           if s2[j1] = s2[k1] then inc(f[p, j, k], f[q, j, k]);53         end;54       end;55   end;56   writeln(f[q, m + 1, m + 1]);57 end.

View code

(P.s. The above signature was previously played, so it was written by Pascal)

Bzoj1566 [noi2009] pipe beads