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



Sample input:
axyb
abyxb


Sample output:
axb

Submission status: Wrong Answer

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    R = lambda: list(map(int, r().split()))
    Rmap = lambda: map(int, r().split())
    Rfloat = lambda: list(map(float, r().split()))
    Rtuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main

    S = r()
    T = r()

    dp = [[0]*(len(T)+1) for _ in range(len(S)+1)]
    
    for si, s in enumerate(S):
        for ti in range(len(T)):
            dp[si+1][ti] = max(dp[si][ti], dp[si+1][ti])
            dp[si][ti+1] = max(dp[si][ti], dp[si][ti+1])
            if S[si] == T[ti]:
                dp[si+1][ti+1] = dp[si][ti] + 1 
            else:
                dp[si+1][ti+1] = max(dp[si+1][ti], dp[si][ti+1])

    ti = len(T)
    ans = []
    for si in range(len(S),0,-1):
        if ti == 0:
            break
        if dp[si][ti] == dp[si-1][ti-1]+1:
            ans.append(S[si-1])
            ti = ti-1
    print(*ans[::-1], sep='')
    
    # to finish code, use return instead of exit()
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Wrong Answer

Solution code:
import numpy as np
s1 = input()
s2 = input()

dp = [['' for j in range(len(s2))] for i in range(len(s1))]

idx = s2.find(s1[0])
if idx != -1:
    for j in range(idx, len(s2)):
        dp[0][j] =  s1[0]
idx = s1.find(s2[0])
if idx != -1:
    for j in range(idx, len(s1)):
        dp[j][0] =  s2[0]
for i in range(1, len(s1)):
  for j in range(len(s2)):
    if s1[i] == s2[j]:
      dp[i][j] = dp[i-1][j-1] + s1[i]
    else:
      dp[i][j] = dp[i-1][j] if len(dp[i-1][j])>len(dp[i][j-1]) else dp[i][j-1]
print(dp[len(s1)-1][len(s2)-1])

Submission status: Wrong Answer

Solution code:
s = input()
t = input()

s_len = len(s)
t_len = len(t)

dp = [[] * (t_len + 1)] * (s_len + 1)

for i in range(s_len + 1):
  dp[i] = [0] * (t_len + 1)


# 最長共通部分列問題（LCS）として計算
for i in range(s_len):
  for j in range(t_len):
    if s[i] == t[j]:
      # 文字が同じ場合はカウントアップ
      dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j] + 1)
    else:
      # 異なる場合は、1文字前のs or 1文字前のt から変化がないので、持ってくる
      dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j+1])
      dp[i+1][j+1] = max(dp[i+1][j+1], dp[i+1][j])
    

# 文字列生成
ans = ''
i = s_len
j = t_len

while(i > 0 and j > 0):
  if dp[i][j] == (dp[i-1][j-1]+1):
    ans = s[i-1] + ans
    i -= 1
    j -= 1
    
  elif dp[i][j] == dp[i-1][j]:
    i -= 1
  elif dp[i][j] == dp[i][j-1]:
    j -= 1

print(ans)
# for i in range(s_len+1):
#   print(dp[i])

Submission status: Wrong Answer

Solution code:
def lcs(str1, str2):
    str1_len = len(str1)
    str2_len = len(str2)

    result = [[0]*(str1_len + 1) for i in range(str2_len + 1)]

    for i in range(str1_len + 1):
        for j in range(str2_len + 1):
            if i == 0 or j == 0:
                result[j][i] = 0
            elif str1[i - 1] == str2[j - 1]:
                result[j][i] = 1 + result[j - 1][i - 1]
            else:
                result[j][i] = max(result[j - 1][i], result[j][i - 1])

    return result


def print_lcs(str1, str2):
    table = lcs(str1, str2)

    common = ''

    i = len(str1)
    j = len(str2)

    while i > 0 and j > 0:
        if str1[i - 1] == str2[j - 1]:
            common = str1[i - 1] + common
            j -= 1
            i -= 1
        elif table[j - 1][i] > table[j][i - 1]:
            j -= 1
        else:
            i -= 1

    return common if len(common) > 0 else 'The answer is (an empty string).'


str1 = input()
str2 = input()

print(print_lcs(str1, str2))

Submission status: Wrong Answer

Solution code:
X = input()
Y = input()
dp = [[0]*(len(Y)+1) for _ in range(len(X)+1)]
for i in range(len(X)):
  for j in range(len(Y)):
    if X[i]==Y[j]:
      dp[i+1][j+1] = dp[i][j]+1
    else:
      dp[i+1][j+1] = max(dp[i+1][j],dp[i][j+1])
      
print(dp[-1][-1])

Submission status: Wrong Answer

Solution code:
s=input()
t=input()
n=len(s)
m=len(t)
dp=[["" for _ in range(n+1)] for _ in range(2)]
for i in range(m):
    for j in range(n):
        if t[i] == s[j]:
            dp[1][j+1] = dp[0][j] + t[i]
        else:
            if len(dp[1][j]) >= len(dp[0][j+1]):
                dp[1][j+1] = dp[1][j]
            else:
                dp[1][j+1] = dp[0][j+1]
        #dp[i][j]=""
    #dp[i][n]=""
    dp[0]=dp[1]
print(dp[-1][-1])

Submission status: Wrong Answer

Solution code:
import sys
sys.setrecursionlimit(10**5)
s = input()
t = input()

#dp[i][j] => s[i], t[j]までの最長部分文字列

dp = [[0] * (len(t) + 1) for j in range(len(s) + 1)]
for i in range(1,len(s)+1):
    for j in range(1,len(t)+1):
        dp[i][j] = max(dp[i - 1][j],dp[i][j - 1])
        if s[i - 1] == t[j - 1]:
            dp[i][j] += 1

def make_ans(a, b):
    if a < 0 or b < 0:
        return 
    if s[a] == t[b]:
        make_ans(a-1,b-1)
        ans.append(s[a])
    else:
        if a == 0:
            make_ans(a, b-1)
        elif b == 0:
            make_ans(a - 1,b)
        else:
            if dp[a - 1][b] > dp[a][b - 1]:
                make_ans(a-1,b)
            else:
                make_ans(a,b-1)

ans = []
make_ans(len(s) - 1,len(t) - 1)
print("".join(ans))

Submission status: Wrong Answer

Solution code:
s=input()
t=input()
n=len(s)
m=len(t)
dp=[[0 for _ in range(n+1)] for _ in range(m+1)]
for i in range(m):
    for j in range(n):
        if t[i] == s[j]:
            dp[i+1][j+1] = dp[i][j] + 1
        else:
            dp[i+1][j+1] = max(dp[i+1][j],dp[i][j+1])
print(dp[-1][-1])

Submission status: Wrong Answer

Solution code:
s = input()
t = input()

#dp[i][j] => s[i], t[j]までの最長部分文字列

dp = [[""] * (len(t) + 1) for j in range(len(s) + 1)]
for i in range(1,len(s)+1):
    for j in range(1,len(t)+1):
        if s[i - 1] == t[j - 1]:
            dp[i][j] = dp[i - 1][j] + s[i - 1]
        else:
            c, d = dp[i][j-1], dp[i-1][j]
            #上か左の長いほうを入れる
            if len(c) < len(d):
                dp[i][j] = d
            else:
                dp[i][j] = c
print(dp[-1][-1])

Submission status: Wrong Answer

Solution code:
s = input()
t = input()

#dp[i][j] => s[i], t[j]までの最長部分文字列

dp = [[""] * (len(t) + 1) for j in range(len(s) + 1)]
for i in range(1,len(s)+1):
    for j in range(1,len(t)+1):
        if s[i - 1] == t[j - 1]:
            dp[i][j] = (dp[i - 1][j] + s[i - 1])
        else:
            #上か左の長いほうを入れる
            if len(dp[i][j-1]) < len(dp[i-1][j]):
                dp[i][j] = dp[i - 1][j]
            else:
                dp[i][j] = dp[i][j - 1]
print(dp[-1][-1])

Submission status: Wrong Answer

Solution code:
s = input()
t = input()
 
#dp[i][j] => s[i], t[j]までの最長部分文字列
 
dp = [[""] *(len(t) + 1) for j in range(len(s) + 1)]
for i in range(1,len(s)+1):
    for j in range(1,len(t)+1):
        if s[i - 1] == t[j - 1]:
            dp[i][j] = dp[i - 1][j] + s[i - 1]
        else:
            #上か左の長いほうを入れる
            if len(dp[i][j-1]) < len(dp[i-1][j]):
                dp[i][j] = dp[i - 1][j]
            else:
                dp[i][j] = dp[i][j - 1]
print(dp[-1][-1])

Submission status: Wrong Answer

Solution code:
s = input()
t = input()

#dp[i][j] => s[i], t[j]までの最長部分文字列

dp = [[""] *(len(t) + 1) for j in range(len(s) + 1)]
for i in range(1,len(s)+1):
    for j in range(1,len(t)+1):
        if s[i - 1] == t[j - 1]:
            dp[i][j] = "".join(dp[i - 1][j] + s[i - 1])
        else:
            #上か左の長いほうを入れる
            dp[i][j] = dp[i - 1][j]
print(dp[-1][-1])

Submission status: Wrong Answer

Solution code:

s = input()
t = input()

#dp[i][j] => s[i], t[j]までの最長部分文字列

dp = [[""] *(len(t) + 1) for j in range(len(s) + 1)]
for i in range(len(s)):
    print(dp[i])
for i in range(1,len(s)+1):
    for j in range(1,len(t)+1):
        if s[i - 1] == t[j - 1]:
            dp[i][j] = "".join(dp[i - 1][j] + s[i - 1])
        else:
            #上か左の長いほうを入れる
            if len(dp[i][j-1]) < len(dp[i-1][j]):
                dp[i][j] = dp[i - 1][j]
            else:
                dp[i][j] = dp[i][j - 1]
print(dp[-1][-1])

Submission status: Wrong Answer

Solution code:
s = " "+input()
t = " "+input()
record = [[0 for _ in range(len(t))] for _ in range(len(s))]
for i in range(1,len(s)):
    for j in range(1,len(t)):
        if t[j] == s[i]:
            record[i][j] = record[i-1][j-1] + 1
        else:
            record[i][j] = max(record[i-1][j-1],record[i][j-1])
print(record[len(s)-1][len(t)-1])

Submission status: Wrong Answer

Solution code:
import sys
# input処理を高速化する
input = sys.stdin.readline

# coding: utf-8
# Your code here!
s=list(input())
t=list(input())

def mmax(a,b):
    if len(a)>=len(b):
        return a
    else:
        return b

dp=[[""]*(len(t)+1) for i in range(len(s)+1)]

for i in range(1,len(s)+1):
    flag=True
    for j in range(1,len(t)+1):
        if s[i-1]==t[j-1] and flag:
            dp[i][j]=mmax(dp[i-1][j]+s[i-1],dp[i][j-1]+t[j-1])
            flag=False
        else:
            dp[i][j]=mmax(dp[i-1][j],dp[i][j-1])
            if dp[i][j]==dp[i-1][j]:
                flag=True

print(dp[-1][-1])

Submission status: Wrong Answer

Solution code:
# coding: utf-8
# Your code here!
s=list(input())
t=list(input())

def mmax(a,b):
    if len(a)>=len(b):
        return a
    else:
        return b

dp=[[""]*(len(t)+1) for i in range(len(s)+1)]

for i in range(1,len(s)+1):
    flag=True
    for j in range(1,len(t)+1):
        if s[i-1]==t[j-1] and flag:
            dp[i][j]=mmax(dp[i-1][j]+s[i-1],dp[i][j-1]+t[j-1])
            flag=False
        else:
            dp[i][j]=mmax(dp[i-1][j],dp[i][j-1])
            if dp[i][j]==dp[i-1][j]:
                flag=True

print(dp[-1][-1])

Submission status: Wrong Answer

Solution code:
s = input()
t = input()
dp = [[""]*(len(t)+1) for i in range(len(s)+1)]
for i in range(1, len(s)+1) :
    for j in range(1, len(t)+1) :
        if s[i-1] == t[j-1] :
            dp[i][j] = dp[i-1][j] + t[j-1]
        elif len(dp[i-1][j]) >= len(dp[i][j-1]) :
            dp[i][j] = dp[i-1][j]
        else :
            dp[i][j] = dp[i][j-1]

print(dp[len(s)][len(t)])

Submission status: Wrong Answer

Solution code:
#EDP-F
import sys

S = sys.stdin.readline().rstrip()
T = sys.stdin.readline().rstrip()
sn, tn = len(S), len(T)
dp = [[0 for j in range(tn)] for i in range(sn)]

#方針:その時点での最長数を入れる
for j in range(tn):
    dp[0][j] = dp[0][j-1]
    if S[0]==T[j]:
        dp[0][j] = 1
for i in range(sn):
    dp[i][0] = dp[i-1][0]
    if S[i]==T[0]:
        dp[i][0] = 1
        

for i in range(1, sn):
    for j in range(1, tn):
        if S[i] == T[j]:
            dp[i][j] = dp[i-1][j-1]+1
        else:
            dp[i][j] = max(dp[i-1][j], dp[i][j-1])

#pprint.pprint(dp)

i,j= sn-1, tn-1
ans =''
while True:
    if i ==0 or j==0:
        if dp[i][j]==1:
            ans+= S[0]
        print(ans[::-1])
        break
    if dp[i-1][j] == dp[i][j]:
        i-=1
    elif dp[i][j-1] == dp[i][j]:
        j-=1
    elif dp[i-1][j-1] +1 == dp[i][j]:
        ans += S[i]
        i-=1
        j-=1

Submission status: Wrong Answer

Solution code:
s=input()
t=input()
i,j=0,0
ans=''
while True:
    if s[i]==t[j]:
        ans+=s[i]
        i+=1
        j+=1
    else:
        if len(s) - i >= len(t) - j:
            i+=1
        else:
            j+=1
    if i==len(s) or j==len(t):
            break
print(ans if ans else  ' ')

Submission status: Wrong Answer

Solution code:
s=input()
t=input()
i,j=0,0
ans=''
while True:
    if s[i]==t[j]:
        ans+=s[i]
        i+=1
        j+=1
        if i==len(s) or j==len(t):
            break
    else:
        if len(s) - i >= len(t) - j:
            i+=1
            if i==len(s):
                break
        else:
            j+=1
            if j==len(t):
                break
print(ans if ans else ' ')

Submission status: Wrong Answer

Solution code:
#EDP-F
import sys
import pprint

S = sys.stdin.readline().rstrip()
T = sys.stdin.readline().rstrip()
sn, tn = len(S), len(T)
dp = [[0 for j in range(tn)] for i in range(sn)]

#方針:その時点での最長数を入れる
for j in range(tn):
    dp[0][j] = dp[0][j-1]
    if S[0]==T[j]:
        dp[0][j] = 1


for i in range(1, sn):
    for j in range(tn):
        if j ==0:
            dp[i][j] = dp[i-1][j]
            if S[i]==T[j]:
                dp[i][j]=1
        else:
            if S[i] == T[j]:
                dp[i][j] = dp[i-1][j-1]+1
            else:
                dp[i][j] = max(dp[i-1][j], dp[i][j-1])

#pprint.pprint(dp)

i,j= sn-1, tn-1
ans =''
while True:
    if i ==0 or j==0:
        if dp[i][j]==1:
            ans+= S[0]
        print(ans[::-1])
        break
    if dp[i-1][j] == dp[i][j]:
        i-=1
    elif dp[i][j-1] == dp[i][j]:
        j-=1
    elif dp[i-1][j-1] +1 == dp[i][j]:
        ans += S[i]
        i-=1
        j-=1

Submission status: Wrong Answer

Solution code:
#EDP-F
import sys
import pprint

S = sys.stdin.readline().rstrip()
T = sys.stdin.readline().rstrip()
sn, tn = len(S), len(T)
dp = [[0 for j in range(tn)] for i in range(sn)]

#方針:その時点での最長数を入れる
for j in range(tn):
    dp[0][j] = dp[0][j-1]
    if S[0]==T[j]:
        dp[0][j] = 1


for i in range(1, sn):
    for j in range(tn):
        if j ==0:
            dp[i][j] = dp[i-1][j]
        else:
            if S[i] == T[j]:
                dp[i][j] = dp[i-1][j-1]+1
            else:
                dp[i][j] = max(dp[i-1][j], dp[i][j-1])

#pprint.pprint(dp)

i,j= sn-1, tn-1
ans =''
while True:
    if i ==0 or j==0:
        if dp[i][j]==1:
            ans+= S[0]
        print(ans[::-1])
        break
    if dp[i-1][j] == dp[i][j]:
        i-=1
    elif dp[i][j-1] == dp[i][j]:
        j-=1
    elif dp[i-1][j-1] +1 == dp[i][j]:
        ans += S[i]
        i-=1
        j-=1

Submission status: Wrong Answer

Solution code:
#EDP-F
import sys
import pprint

S = sys.stdin.readline().rstrip()
T = sys.stdin.readline().rstrip()
sn, tn = len(S), len(T)
dp = [[0 for j in range(tn)] for i in range(sn)]

#方針:その時点での最長数を入れる
for j in range(tn):
    dp[0][j] = dp[0][j-1]
    if S[0]==T[j]:
        dp[0][j] = 1


for i in range(1, sn):
    for j in range(tn):
        if j ==0:
            dp[i][j] = dp[i-1][j]
        else:
            if S[i] == T[j]:
                dp[i][j] = dp[i-1][j-1]+1
            else:
                dp[i][j] = max(dp[i-1][j], dp[i][j-1])

#pprint.pprint(dp)

i,j= sn-1, tn-1
ans =''
while True:
    if i ==0:
        if dp[i][j]==1:
            ans+= S[0]
        print(ans[::-1])
        break
    if dp[i-1][j] == dp[i][j]:
        i-=1
    elif dp[i][j-1] == dp[i][j]:
        j-=1
    elif dp[i-1][j-1] +1 == dp[i][j]:
        ans += S[i]
        i-=1
        j-=1

Submission status: Wrong Answer

Solution code:
s=str(input())
t=str(input())
n=len(s)
m=len(t)
dp = [['' for i in range(m+1)] for j in range(n+1)]
for i in range(n):
    for j in range(m):
        if s[i]==t[j]:
            #print(s[i])
            dp[i+1][j+1]=dp[i][j]+s[i]
        elif len(dp[i][j+1])!=0 and len(dp[i+1][j])==len(dp[i][j+1]):
            #print(dp[i][j+1],dp[i+1][j])
            dp[i+1][j+1]=min(dp[i][j+1],dp[i+1][j])
            #print(dp[i+1][j+1])
        else:
            print(dp[i][j+1],dp[i+1][j])
            dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j])
            print(dp[i+1][j+1])
print(dp[n][m])

Submission status: Wrong Answer

Solution code:
"""
【最長共通部分列問題】
2つの文字列s,tが与えられたときの2つの共通部分文字列の長さの最大値とその文字列
!!制約!!
1<=n,m<=1000
"""
#-------------------------------------------------------------------------------

#入力,準備
s=input()
t=input()
n=len(s)
m=len(t)
dp=[[0 for i in range(m+1)] for j in range(n+1)]


#メイン

#dp[i][j]:=s_1~s_iとt_1~t_jに対するLCSの長さ

for i in range(n):
    for j in range(m):
        if s[i]==t[j]:
            dp[i+1][j+1]=dp[i][j]+1 #同じならくっつけられる
        else:
            dp[i+1][j+1]=max(dp[i][j+1],dp[i+1][j])
print(dp[n][m])

#dpを逆からたどって文字列を復元
ans = ''
while(dp[n][m]>0):
  if dp[n][m] == dp[n-1][m-1] +1 and dp[n][m] == dp[n-1][m] +1 and dp[n][m] == dp[n][m-1] +1:
    ans += s[n-1]
    n -= 1
    m -= 1
  elif dp[n-1][m] > dp[n][m-1]:
    n -= 1
  else:
    m -= 1
print(''.join(reversed(list(ans))))