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 Python (3.4.3) (judge verdict: Time Limit Exceeded)

import sys
readline = sys.stdin.readline

S = readline().rstrip()
T = readline().rstrip()

# dp[i][j] = Sをi文字目、Tをj文字目まで進めたときの最長
# S[i] == S[j]のとき
# dp[i][j] = dp[i - 1][j - 1] + 1
# S[i] != S[j]のとき
# dp[i][j] = max(dp[i - 1][j],dp[i][j - 1])

# その際、自分がどこから来たか？を記録する
# dp[i - 1][j] > dp[i][j - 1]であれば
# route[i][j] = (i,j)を記録

dp = [[0] * (len(T) + 1) for i in range(len(S) + 1)]

for i in range(len(S)):
  for j in range(len(T)):
    if S[i] == T[j]:
      dp[i + 1][j + 1] = dp[i][j] + 1
    else:
      if dp[i][j + 1] > dp[i + 1][j]:
        dp[i + 1][j + 1] = dp[i][j + 1]
      else:
        dp[i + 1][j + 1] = dp[i + 1][j]

# dpテーブルの数字を見て復元していく
ans = ""
x = len(S)
y = len(T)
while x != 0 and y != 0:
  if dp[x][y] == dp[x - 1][y]: # x - 1から遷移してきた
    x -= 1
  elif dp[x][y] == dp[x][y - 1]: # y - 1から遷移してきた
    y -= 1
  else:
    ans += S[x - 1]
    x -= 1
    y -= 1

print(ans[::-1])