ABC259 C - XX to XXX - 競プロ

ABC259 C - XX to XXX

https://atcoder.jp/contests/abc259/tasks/abc259_c

提出

code: python

s = input()

t = input()

sarray = [s0]

tarray = [t0]

for i in range(1, len(s)):

print(sarray)

if len(sarray) > 0 and sarray-1 == si and si == si-1:

sarray.pop(-1)

else:

sarray.append(si)

for i in range(1, len(t)):

print(tarray)

if len(tarray) > 0 and tarray-1 == ti and ti == ti-1:

tarray.pop(-1)

else:

tarray.append(ti)

# print(sarray)

print(tarray)

print("".join(sarray) == "".join(tarray))

解答

code: python

s = input()

t = input()

def rle(s):

cnt = 1

res = []

for i in range(1, len(s)):

if(si != si-1):

res.append((si-1, cnt))

cnt = 0

cnt += 1

res.append((s-1, cnt))

return res

sres = rle(s)

tres = rle(t)

if (len(sres) != len(tres)):

print("No")

exit()

ans = True

for i in range(len(sres)):

if (sresi0 != tresi0):

ans = False

if not (sresi1 == tresi1 or sresi1 < tresi1 and sresi1 >= 2): # 2つ以上なかったら間に入れられない

ans = False

print("Yes") if ans else print("No")

メモ

https://atcoder.jp/contests/abc259/editorial/4283

提出

TLE RE WA

code: python

s = list(input())

t = list(input())

# abbaac

# abbbbaaac

ends = []

while len(t) > 0 and len(s) > 0:

nows = s.pop(0)

nowt = t.pop(0)

if nows == nowt:

ends.append(nows)

continue

else:

if ends-1 == nowt and ends-2 == nowt:

t.pop(0)

ends.append(nows)

continue

else:

print("No")

exit()

print("Yes")