Problem:
Score : 100 points

Problem Statement

You are given strings s and t.
Find one longest string that is a subsequence of both s and t.

Notes

A subsequence of a string x is the string obtained by removing zero or more characters from x and concatenating the remaining characters without changing the order.

Constraints

s and t are strings consisting of lowercase English letters.

1 \leq |s|, |t| \leq 3000

Input

Input is given from Standard Input in the following format:

s
t

Output

Print one longest string that is a subsequence of both s and t.
If there are multiple such strings, any of them will be accepted.

Sample Input 1

axyb
abyxb

Sample Output 1

axb

The answer is axb or ayb; either will be accepted.

Sample Input 2

aa
xayaz

Sample Output 2

aa

Sample Input 3

a
z

Sample Output 3

The answer is (an empty string).

Sample Input 4

abracadabra
avadakedavra

Sample Output 4

aaadara

Solution in LuaJIT (2.0.4) (judge verdict: Time Limit Exceeded)

local s = io.read()
local t = io.read()
local n, m = #s, #t
local ret = {}
local len = {}
for i = 1, n * m do ret[i], len[i] = "", 0 end

local str_s = s:sub(1, 1)
if t:sub(1, 1) == str_s then
  ret[1] = str_s
  len[1] = 1
end
for i_t = 2, m do
  if t:sub(i_t, i_t) == str_s then
    ret[i_t] = str_s
    len[i_t] = 1
  else
    ret[i_t] = ret[i_t - 1]
    len[i_t] = len[i_t - 1]
  end
end
for i_s = 2, n do
  str_s = s:sub(i_s, i_s)
  if t:sub(1, 1) == str_s then
    ret[(i_s - 1) * m + 1] = str_s
  end
  for i_t = 2, m do
    local idx = (i_s - 1) * m + i_t
    if len[idx - 1] < len[idx - m] then
      ret[idx] = ret[idx - m]
      len[idx] = len[idx - m]
    else
      ret[idx] = ret[idx - 1]
      len[idx] = len[idx - 1]
    end
    if t:sub(i_t, i_t) == str_s then
      local cand = ret[idx - m - 1] .. str_s
      if len[idx] < #cand then
        ret[idx] = cand
        len[idx] = #cand
      end
    end
  end
end
print(ret[n * m])