TnT/Multi_CodeNet4Repair huggingface dataset:
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
mport sys
sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))
from collections import deque
def resolve():
    S = deque(list(sr().split()), maxlen=1000000)
    Q = ir()
    rev = 1
    for i in range(Q):
        T = list(sr().split())
        if int(T[0]) == 1:
            rev *= -1
        else:
            x, F, C = T
            if int(F) == 1:
                if rev == 1:
                    S.appendleft(C)
                else:
                    S.append(C)
            else:
                if rev == 1:
                    S.append(C)
                else:
                    S.appendleft(C)
    if rev == -1:
        S.reverse()
    print(*S, sep='')
resolve()

import sys
sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))
from collections import deque
def resolve():
    S = deque(sr())
    Q = ir()
    rev = 1
    for i in range(Q):
        T = list(sr().split())
        if int(T[0]) == 1:
            rev *= -1
        else:
            x, F, C = T
            if int(F) == 1:
                if rev == 1:
                    S.appendleft(C)
                else:
                    S.append(C)
            else:
                if rev == 1:
                    S.append(C)
                else:
                    S.appendleft(C)
    if rev == -1:
        S.reverse()
    print(*S, sep='')
resolve()

Verdicts: Runtime Error

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
S=input()
Q=int(input())

guuki=0
ANS=S
ANS_front=""
ANS_back=""
for i in range(Q):
    A=list(input().split())
    if A[0]=="1":
        guuki+=1
    elif int(A[1])*guuki%2==1 or (A[1]=="2" and guuki%2==0):
        ANS_back+=A[2]
    else:
        ANS_front=A[2]+ANS_front
if guuki%2==1:
    ANS=ANS[::-1]
    ANS_back,ANS_front=ANS_front,ANS_back
print(ANS_front[::-1]+ANS+ANS_back)

S=input()
Q=int(input())

guuki=0
ANS=S
ANS_back=""
ANS_front=""
for i in range(Q):
    A=list(input().split())
    if A[0]=="1":
        guuki+=1
    elif int(A[1])*guuki%2==1 or (A[1]=="2" and guuki%2==0):
        ANS_back+=A[2]
    else:
        ANS_front+=A[2]
if guuki%2==1:
    ANS=ANS[::-1]
    ANS_back,ANS_front=ANS_front,ANS_back
print(ANS_front[::-1]+ANS+ANS_back)

Verdicts: Wrong Answer

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
def StrConv(S, T, F, C):
  ans = ''
  if T == '1':
    pass
    
  else:
    if F == '1':
      ans = C + S
    else:
      ans = S + C
  return ans
  

S = input()
Q = int(input())
dumy = '|' 
cnt = 0   


for i in range(Q):
  tmp = str(input())
  
  if tmp[0] == '1':
    T = tmp[0]
    F = '-'
    C = '-'
    cnt += 1
  else:
    T, F, C = tmp.split(' ')
  
  dumy = StrConv(dumy, T, F, C)


if (cnt % 2) == 0:
  ans = dumy[0:dumy.find('|')] + S + dumy[dumy.find('|')+1:]
else:
  ans = dumy[0:dumy.find('|')] + S[::-1] + dumy[dumy.find('|')+1:]
print(ans)



S = input()
Q = int(input())
cnt = 0   
head = ''
tail = ''


for i in range(Q):
  tmp = str(input())
  
  if tmp[0] == '1':
    cnt += 1
  else:
    T, F, C = tmp.split(' ')
    if cnt % 2 == 1:
      if F == '1':
        tail = tail + C
      else:
        head = C + head
    else:
      if F == '1':
        head = C + head
      else:
        tail = tail + C
    #cnt = 0
  #print('head=',head)
  #print('tail=',tail)

print('cnt=',cnt)
if cnt % 2 == 0:
  ans = head + S + tail
else:
  ans = tail[::-1] + S[::-1] + head[::-1]
print(ans)


def StrConv(S, T, F, C):
  ans = ''
  if T == '1':
    
  else:
    if F == '1':
      ans = C + S
    else:
      ans = S + C
  return ans
  

S = input()
Q = int(input())
dumy = '|' 
cnt = 0   


for i in range(Q):
  tmp = str(input())
  
  if tmp[0] == '1':
    T = tmp[0]
    F = '-'
    C = '-'
    cnt += 1
  else:
    T, F, C = tmp.split(' ')
  
  dumy = StrConv(dumy, T, F, C)


if (cnt % 2) == 0:
  ans = dumy[0:dumy.find('|')] + S + dumy[dumy.find('|')+1:]
else:
  ans = dumy[0:dumy.find('|')] + S[::-1] + dumy[dumy.find('|')+1:]
print(ans)



S = input()
Q = int(input())
cnt = 0   
head = ''
tail = ''


for i in range(Q):
  tmp = str(input())
  
  if tmp[0] == '1':
    cnt += 1
  else:
    T, F, C = tmp.split(' ')
    if cnt % 2 == 1:
      if F == '1':
        tail = tail + C
      else:
        head = C + head
    else:
      if F == '1':
        head = C + head
      else:
        tail = tail + C
    #cnt = 0
  #print('head=',head)
  #print('tail=',tail)

#print('cnt=',cnt)
if cnt % 2 == 0:
  ans = head + S + tail
else:
  ans = tail[::-1] + S[::-1] + head[::-1]
print(ans)


Verdicts: Wrong Answer

Wrong Answer

Runtime Error

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
s = input()
q = int(input())
for i in range(2, q):
    txt3 = input().split()
    if txt3[0] == '1':
        s = s[::-1]
    else:
        if txt3[1] == '1':
            s = txt3[2] + s
        else:
            s = s + txt3[2]
print(s)


def ABC158D(txt):
    txt2 = txt.split("\n")
    s = txt2[0]
    q = int(txt2[1])
    for i in range(2, len(txt2)):
        txt3 = txt2[i].split()
        if txt3[0] == '1':
            s = s[::-1]
        else:
            if txt3[1] == '1':
                s = txt3[2] + s
            else:
                s = s + txt3[2]
    print(s)
                
ABC158D(txt)


def ABC158D():
    s = input()
    q = int(input())
    for i in range(2, q):
        txt3 = input().split()
        print(f"txt3 = {txt3}")
        if txt3[0] == '1':
            s = s[::-1]
        else:
            if txt3[1] == '1':
                s = txt3[2] + s
            else:
                s = s + txt3[2]
    print(s)

ABC158D()



s = input()
q = int(input())
ht = 1
count = 0
for i in range(2, q):
    txt3 = input().split()
    if txt3[0] == '1':
        ht = 3 - ht
        count += 1
    else:
        if int(txt3[1]) == ht:
            s = txt3[2] + s
        else:
            s = s + txt3[2]
if count % 2 == 1:
    s = s[::-1]
print(s)


print(1)

def ABC158D():
    txt2 = txt.split("\n")
    s = txt2[0]
    q = int(txt2[1])
    for i in range(2, len(txt2)):
        txt3 = txt2[i].split()
        if txt3[0] == '1':
            s = s[::-1]
        else:
            if txt3[1] == '1':
                s = txt3[2] + s
            else:
                s = s + txt3[2]
    print(s)
                
ABC158D()


def ABC158D():
    txt2 = txt.split("\n")
    s = txt2[0]
    q = int(txt2[1])
    for i in range(2, len(txt2)):
        txt3 = txt2[i].split()
        if txt3[0] == '1':
            s = s[::-1]
        else:
            if txt3[1] == '1':
                s = txt3[2] + s
            else:
                s = s + txt3[2]
    print(s)

f = open("foot/bar.dat")
txt = f.read()
f.close()
ABC158D(txt)


# ABC158D

s = input()
q = int(input())
ht = 1
head = ''
tail = ''
count = 0
for i in range(q):
    txt3 = input().split()
    if txt3[0] == '1':
        ht = 3 - ht
        count += 1
    else:
        if int(txt3[1]) == ht:
            head = txt3[2] + head
        else:
            tail = tail + txt3[2]
s = head + s + tail
if count % 2 == 1:
    s = s[::-1]
print(s)


Verdicts: Wrong Answer

Runtime Error

Runtime Error

Wrong Answer

Wrong Answer

Runtime Error

Runtime Error

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
s=list(input())
q=int(input())
rev=0
for i in range(q):
    t=list(input().split())
    if t[0]=="1":
        rev=rev+1
    else:
        if t[1]=="1":
            if rev%2==0:
                s.appendleft(t[2])
            else:
                s.append(t[2])
        else:
            if rev%2==0:
                s.append(t[2])
            else:
                s.appendleft(t[2])
if rev%2==0:
    print("".join(s))
else:
    s.reverse()
    print("".join(s))



from collections import deque
s=deque(list(input()))
q=int(input())
rev=0
for i in range(q):
  t=list(input().split(" "))
  if t[0]=="1":
    rev=rev+1
  else:
    if t[1]=="1":
      if rev%2==0:
        s.appendleft(t[2])
      else:
        s.append(t[2])
    else:
      if rev%2==0:
        s.append(t[2])
      else:
        s.appendleft(t[2])
if rev%2==0:
  print("".join(s))
else:
  s.reverse()
  print("".join(s))

Verdicts: Runtime Error

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
from collections import deque
S=deque(input())
Q=int(input())
for i in range(Q):
  N=list(input().split())
  if N[0]=="1":
    S.reverse()
  elif N[1]="1":
    S.appendleft(N[2])
  else:
    S.append(N[2])
print(S)


rL=list(input())
lL=[]
Q=int(input())
for i in range(Q):
  N=list(input().split())
  if N[0]=="1":
    lL,rL=rL,lL
  elif N[1]=="1":
    lL.append(N[2])
  else:
    rL.append(N[2])
lL.reverse()
print("".join(lL)+"".join(rL))

Verdicts: Runtime Error

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
s = input()
q = int(input())

from collections import deque

ss = deque(s)
flg = True

for i in range(q):
  t = input().split()
  if t[0] == '1':
    flg = !flg
  else:
    if t[1] == '2':
        ss.append(t[2]) if flg else ss.appendleft(t[2])
    else:
        ss.append(t[2]) if flg != True else ss.appendleft(t[2])



print("".join(ss)) if flg else print("".join(ss[::-1]))



s = input()
q = int(input())
ss = []

for i in range(q):
  t = input().split()
  if t[0] == '1':
    ss = ss[::-1]
  else:
    if t[1] == '2':
      ss.append(t[2])
    else:
      ss.insert(0,t[2]

print("".join(ss))



s = input()
q = int(input())

from collections import deque

ss = deque(s)
flg = True

for i in range(q):
  t = input().split()
  if t[0] == '1':
    flg = not flg
  else:
    if t[1] == '2':
        ss.append(t[2]) if flg else ss.appendleft(t[2])
    else:
        ss.append(t[2]) if not flg else ss.appendleft(t[2])

ss = list(ss)

print("".join(ss)) if flg else print("".join(ss[::-1]))



Verdicts: Runtime Error

Runtime Error

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
s = list(str(input()))
q = int(input())
hanten = 0

for i in range(q):
    x = input().split()
    if len(x) == 1:
        t = x[0]
        if t == "1" and hanten == 0:
            hanten = 1
        elif t == "1" and hanten == 1:
            hanten = 0

    else:
        f = int(x[1])
        c = str(x[2])
        if f == 1 and hanten == 0:
            s.insert(0, c)

        elif f == 1 and hanten == 1:
            s.append(c)

        elif f == 2 and hanten == 0:
            s.append(c)

        elif f == 2 and hanten == 1:
            s.insert(0, c)

else:
    d = len(s)
    if hanten == 1:
        for i in range(d):
            print(s[d-1-i], end="")
    else:
        for i in range(d):
            print(s[i], end="")


from collections import deque

s = deque(str(input()))
q = int(input())
hanten = 0

for i in range(q):
    x = input().split()
    if len(x) == 1:
        t = x[0]
        if t == "1" and hanten == 0:
            hanten = 1
        elif t == "1" and hanten == 1:
            hanten = 0

    else:
        f = int(x[1])
        c = str(x[2])
        if f == 1 and hanten == 0:
            s.appendleft(c)

        elif f == 1 and hanten == 1:
            s.append(c)

        elif f == 2 and hanten == 0:
            s.append(c)

        elif f == 2 and hanten == 1:
            s.appendleft(c)

else:
    if hanten == 1:
        s.reverse()

print("".join(s))


Verdicts: Time Limit Exceeded

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
S=input()
Q=int(input())

flip=0
fr=''
ba=''
for i in range(Q):
    Qq=input()
    
    if(Qq.count(' ')>0):
        ti,fi,ci=Qq.split()
        if( flip%2==0   ):
            if(fi=='1'):
                fr=str(ci)+fr
            else:
                ba=ba+str(ci)
        else:
            if(fi=='1'):
                #fr=fr+str(ci)
                ba=ba+str(ci)
            else:
                #ba=str(ci)+ba
                fr=str(ci)+fr
    else:
        flip+=1
        #S=S[::-1]
if(flip&2==0):
    print(fr+S+ba)
else:
    print(ba[::-1]+S[::-1]+fr[::-1])


S=input()
Q=int(input())

flip=0
fr=''
ba=''
for i in range(Q):
    Qq=input()
    
    if(Qq.count(' ')>0):
        ti,fi,ci=Qq.split()
        if( flip%2==0   ):
            if(fi=='1'):
                fr=str(ci)+fr
            else:
                ba=ba+str(ci)
        else:
            if(fi=='1'):
                ba=ba+str(ci)
            else:
                fr=str(ci)+fr
    else:
        flip+=1
if(flip%2==0):
    print(fr+S+ba)
else:
    print(ba[::-1]+S[::-1]+fr[::-1])

Verdicts: Wrong Answer

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
from collections import deque
S = list(map(str,input().split()))
Q = int(input())
Query = [list(map(str,input().split())) for _ in range(Q)]
d = deque(S)
for i in range(Q):
  if Query[i][0] == "1":
    d = d.rotate()
  elif Query[i][0] == "2":
    if Query[i][1] == "1":
      d.append(Query[i][2])
    elif Query[i][1] == "2":
      d.appendleft(Query[i][2])
ans = ''.join(d)
print(ans)

from collections import deque
S = list(map(str,input().split()))
Q = int(input())
Query = [list(map(str,input().split())) for _ in range(Q)]
d = deque(S)
for i in range(Q):
  if Query[i][0] == "1":
    d = ''.join(list(reversed(d)))
  elif Query[i][0] == "2":
    if Query[i][1] == "1":
      d.append(Query[i][2])
    elif Query[i][1] == "2":
      d.appendleft(Query[i][2])
ans = ''.join(d)
print(ans)

from collections import deque
S = str(input())
Q = int(input())
d = deque()
reverse = 1
for i in S:
    d.append(i)
for j in range(Q):
  Query = input().split()
  if len(Query) == 1:
    reverse *= -1
  else:
    F = int(Query[1])
    C = str(Query[2])
    if reverse == 1:
      if F == 1:
        d.appendleft(C)
      elif F == 2:
        d.append(C)
    elif reverse == -1:
      if F == 1:
        d.append(C)
      elif F == 2:
        d.appendleft(C)
if reverse == -1:
    d.reverse()
ans = ''.join(d)
print(ans)

Verdicts: Runtime Error

Runtime Error

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
import copy

S=input()
Q=int(input())
Query=[]
temp=[]

revflag=0
sbef=[]
saft=[]

for i in range(Q):
    temp=input().split()
    if len(temp) ==1:
        revflag += 1
    else:
        if (revflag + int(temp[1])-1 )%2 == 0:
            sbef.append(temp[2])
        else:
            saft.append(temp[2])

ans=""

for i in range(len(sbef)):
    ans+=sbef[len(sbef)-i-1]

ans += S
for i in range(len(saft)):
    ans+=saft[i]
print(ans)

tmp=""
tmp2 = copy.copy(ans)


if revflag%2 ==1:
    for i in range(len(ans)):
        tmp+=tmp2[len(ans)-i-1]
    ans=tmp


print(ans)


import copy

S=input()
Q=int(input())
Query=[]
temp=[]

revflag=0
sbef=[]
saft=[]

for i in range(Q):
    temp=input().split()
    if len(temp) ==1:
        revflag += 1
    else:
        if (revflag + int(temp[1])-1 )%2 == 0:
            sbef.append(temp[2])
        else:
            saft.append(temp[2])

ans=""

for i in range(len(sbef)):
    ans+=sbef[len(sbef)-i-1]

ans += S
for i in range(len(saft)):
    ans+=saft[i]

tmp=""
tmp2 = copy.copy(ans)


if revflag%2 ==1:
    for i in range(len(ans)):
        tmp+=tmp2[len(ans)-i-1]
    ans=tmp


print(ans)


Verdicts: Wrong Answer

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
s=input()
n=int(input())
isreverse=False
top=''

for i in range(n):
    ss=input().split()
    if len(ss)==1:
        isreverse=not isreverse
    else:
        if ss[1]=="1":
            if isreverse: s+=ss[2]
            else: top+ss[2]
        else:
            if isreverse: top+=ss[2]
            else: s+=ss[2]
if isreverse:
    s=s[::-1]+top
else:
    s=top[::-1]+s


print(s)

s=input()
n=int(input())
isreverse=False
top=''

for i in range(n):
    ss=input().split()
    if len(ss)==1:
        isreverse=not isreverse
    else:
        if ss[1]=="1":
            if isreverse: s+=ss[2]
            else: top+=ss[2]
        else:
            if isreverse: top+=ss[2]
            else: s+=ss[2]
if isreverse:
    s=s[::-1]+top
else:
    s=top[::-1]+s


print(s)

Verdicts: Wrong Answer

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
string=list(input())
Q=int(input())
def func1(s):
    s.reverse()
    
def func2(s,F,C):
    if F=="1":
        s.insert(0,C)
        #return s
    if F=="2":
        s.append(C)
        #return s

            
for i in range(Q):
    Input=list(map(str,input().split()))
    if Input[0]=='1':
        func1(string)
        
    elif Input[0]=='2':
        func2(string,Input[1],Input[2])
for i in string:
    print(i,end="")

string=list(input())
Q=int(input())
reverse=0
def func1(s):
    s.reverse()
    
def func2(s,F,C):
    if F=="1":
        s.insert(0,C)
        #return s
    if F=="2":
        s.append(C)
        #return s

            
for i in range(Q):
    Input=list(map(str,input().split()))
    if Input[0]=='1':
        func1(string)
        
    elif Input[0]=='2':
        func2(string,Input[1],Input[2])
for i in string:
    print(i,end="")

string=list(input())
Q=int(input())
def func1(s):
    s.reverse()
    return s
def func2(s,F,C):
    if F=="1":
        s.insert(0,C)
        #return s
    if F=="2":
        s.append(C)
        #return s

            
for i in range(Q):
    Input=list(map(str,input().split()))
    if Input[0]=='1':
        func1(string)
        
    elif Input[0]=='2':
        func2(string,Input[1],Input[2])
for i in string:
    print(i,end="")

S = str(input())
Q = int(input())
 
hanten = 0
front = ""
end = ""
 
for i in range(Q):
    query=list(map(str,input().split()))
    if query[0]=="1":
        hanten+=1
        hanten=hanten%2
    else:
        if query[1]=="1":
            if hanten==0:
                front=query[2]+front
            else:
                end=end+query[2]
        else:
            if hanten==0:
                end=end+query[2]
            else:
                front=query[2]+front
ans=front+S+end
if hanten==1:
    ans=ans[::-1]
print(ans)

Verdicts: Time Limit Exceeded

Time Limit Exceeded

Time Limit Exceeded

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
from collections import deque

s=list(input())
q=int(input())
d = deque(s)
reverse=False

for i in range(q):
  qs=list(input().split())
  if qs[0]=='1':
    reverse=not reverse
  else:
    if qs[1]=='1':
      d.appendleft(qs[2]) if reverse else d.append(qs[2])
    else:
      d.append(qs[2]) if reverse else d.appendleft(qs[2])
#  print(d)
print("".join(d))


from collections import deque

s=list(input())
q=int(input())
d = deque(s)
reverse=False

for i in range(q):
  qs=list(input().split())
  if qs[0]=='1':
    reverse=not reverse
  else:
    if qs[1]=='1':
      d.appendleft(qs[2]) if not reverse else d.append(qs[2])
    else:
      d.append(qs[2]) if not reverse else d.appendleft(qs[2])
#  print(d)
if reverse:
  d.reverse()
print("".join(d))


Verdicts: Wrong Answer

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
import numpy as np
s=input()

ss=[]
for i in range(len(s)):
    ss.append(s[i])
arr=np.array(ss)

q=int(input())
for i in range(q):
    l=list(map(str,input().split()))
    if l[0]=="1":
        arr=arr[::-1]
        
    else:
        if l[1]=="1":
            arr=arr[::-1]
            arr=np.hstack((arr,l[2]))
            arr=arr[::-1]
        if l[1]=="2":
            arr=np.hstack((arr,[l[2]]))
        

a=arr.tolist()
for i in a:
    print(i,end="")

import numpy as np
s=input()

ss=[]
for i in range(len(s)):
    ss.append(s[i])
arr=np.array(ss)

q=int(input())
for i in range(q):
    l=list(map(str,input().split()))
    if l[0]=="1":
        arr=arr[::-1]
        
    else:
        if l[1]=="1":
            arr=np.hstack((l[2],arr))
        if l[1]=="2":
            arr=np.hstack((arr,[l[2]]))
        

a=arr.tolist()
for i in a:
    print(i,end="")

import numpy as np
s=input()
ss=[]
for i in range(len(s)):
    ss.append(s[i])

    
arr=np.array(ss)

q=int(input())
for i in range(q):
    l=list(map(str,input().split()))
    if l[0]=="1":
        arr=arr[::-1]
        
    else:
        if l[1]=="1":
            arr=arr[::-1]
            arr=np.append(arr,l[2])
            arr=arr[::-1]
        if l[1]=="2":
            arr=np.append(arr,[l[2]])
        

a=arr.tolist()
for i in a:
    print(i,end="")


s=input()
q=int(input())
p=0
a=""
b=""
for i in range(q):
    l=list(map(str,input().split()))
    if l[0]=="1":
        p=(p+1)%2
    else:
        if p==0:
            if l[1]=="1":
                a+=l[2]
            else:
                b+=l[2]
        else:
            if l[1]=="1":
                b+=l[2]
            else:
                a+=l[2]
if p==1:
    print(b[::-1]+s[::-1]+a)
else:
    print(a[::-1]+s+b)


Verdicts: Time Limit Exceeded

Time Limit Exceeded

Time Limit Exceeded

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
from collections import deque
d = deque(input())
flag = 0
for T_F_C in (input().split() for _ in range(int(input()))):
    if T_F_C[0] == '1':
        flag ^= 1
        continue
    _, f, c = T_F_C
    if (flag+int(f)) % 2:
	d.appendleft(c)
    else:
        d.append(c)
if flag:
    d = reversed(d)
print("".join(d))

from collections import deque
d = deque(input())
flag = 0
for T_F_C in (input().split() for _ in range(int(input()))):
    print("".join(d) if flag else "".join(reversed(d)))
    if T_F_C[0] == '1':
        flag ^= 1
        continue
    _, f, c = T_F_C
    if (flag+int(f)) % 2:
        d.append(c)
    else:
        d.appendleft(c)
if flag:
    print("".join(d))
else:
    print("".join(reversed(d)))

#!/usr/bin/env python3

def main():
    from itertools import chain
    S = input()
    Q = int(input())
    Queries = [input().split() for _ in range(Q)]
    H_T = H, T = [], []
    flag = 0
    for query in Queries:
        if len(query) == 1:
            flag ^= 1
            continue
        _, f, c = query
        H_T[(int(f) + flag - 1) % 2].append(c)
    t = ''.join(chain(reversed(H), S, T))
    if flag:
        t = reversed(t)
    print(''.join(t))


if __name__ == '__main__':
    main()


Verdicts: Runtime Error

Runtime Error

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
from sys import stdin
s = str(stdin.readline().rstrip())
q = int(stdin.readline().rstrip())
for i in range(q):
    query = list(map(int, stdin.readline().rstrip().split()))
    if int(query[0])==1:
        s = s[::-1]
    else:
        if int(query[1])==2:
            s += query[2]
        else:
            s = query[2] + s

print(s)


from sys import stdin
s = str(stdin.readline().rstrip())
#A, B, C = map(int, stdin.readline().rstrip().split())

q= int(stdin.readline().rstrip())
hanten = 0
t=""
u=""

for i in range(q):
    l = list(map(str, stdin.readline().rstrip().split()))
    if int(l[0]) == 1:
        t,u = u,t
        hanten ^= 1
    else:
        if int(l[1])==1:
            t+=l[2]
        else:
            u+=l[2]


if hanten==0:
    print(t[::-1]+s+u)
else:
    print(t[::-1]+s[::-1]+u)


Verdicts: Runtime Error

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
temp = input()
s = ["tikan"]
flag = "off"
q = int(input())
dummy = ""
for i in range(q):
  q_list = input().split()
  if q_list[0] == "1":
    if flag == "on":
      flag = "off"
    else:
      flag = "on"
  else:
    if q_list[1] == "1":
      if flag == "off":
        s.insert(0,q_list[2])
      else:
        s.append(q_list[2])
    else:
      if flag == "off":
        s.append(q_list[2])
      else:
        s.insert(0,q_list[2])
#  print(s)
ans = ""
for i in s:
  if i == "tikan":
    i = temp
    if flag == "on":
      i = list(i)
      i.reverse()
      i = "".join(i)
  ans += i
if flag == "on":
  ans = list(ans)
  ans.reverse()
  ans = "".join(ans)
print(ans)v

s = input()
q = int(input())
turn = 1
start = []
end = []
for i in range(q):
    q_list = list(input().split())
    if q_list[0] == "1":
        turn *= -1
    else:
        if q_list[1] == "1":
            if turn == 1:
                start.append(q_list[2])
            else:
                end.append(q_list[2])
        else:
            if turn == -1:
                start.append(q_list[2])
            else:
                end.append(q_list[2])

dum1 = "".join(start)
dum2 = "".join(end)
if turn == 1:
    print(dum1[::-1]+s+dum2)
else:
    print(dum2[::-1] + s[::-1] + dum1)

Verdicts: Runtime Error

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
#D

S=input()
Q=int(input())
#Query=[]

#    Query.append(list(input().split()))
    
    

t=0
b=""
a=""
for j in range(Q):
    i=list(input().split())
    if i[0]=="1":
        t+=1
    elif i[0]=="2":
        if t%2==0:
            if i[1]=="1":
                b=i[2]+b
            elif i[1]=="2":
                a=a+i[2]
        elif t%2==1:
            if i[1]=="1":
                a=i[2]+a
            elif i[1]=="2":
                b=b+i[2]
if t%2==0:
    print(b+S+a)
    
else:
    print(a+S[::-1]+b)

#D

S=input()
Q=int(input())
#Query=[]

#    Query.append(list(input().split()))
    
    

t=0
b=""
a=""
for j in range(Q):
    i=list(input().split())
    if i[0]=="1":
        t+=1
    elif i[0]=="2":
        if t%2==0:
            if i[1]=="1":
                b=i[2]+b
            elif i[1]=="2":
                a=a+i[2]
        elif t%2==1:
            if i[1]=="1":
                a=a+i[2]
            elif i[1]=="2":
                b=i[2]+b
if t%2==0:
    print(b+S+a)
    
else:
    print(a[::-1]+S[::-1]+b[::-1])

Verdicts: Wrong Answer

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
#D
s=input()
q=int(input())
l=[]
ans=[s]
for i in range(q):
    l.append(list(input().split()))
for i in range(q):
    if int(l[i][0])==1:
        ans.reverse()
    elif int(l[i][1])==1:#2 1
        ans.reverse()
        ans.append(l[i][2])
        ans.reverse()
    else:#2 2
        ans.append(l[i][2])
for i in ans:
    print(i,end="")

#D
s=input()
q=int(input())
l=[]
left=[]
right=[]
ans=[s]
for i in range(q):
    l.append(list(input().split()))
    if int(l[i][0])==1:
        left,right=right,left
    elif int(l[i][1])==1:#2 1
        left.append(l[i][2])
    else:#2 2
        right.append(l[i][2])
ans.extend(right)
ans.reverse()
ans.extend(left)
ans.reverse()
for i in ans:
    print(i,end="")

#D
from collections import deque
s=input()
q=int(input())
l=[]
turn=0
ans=deque()
ans.append(s)
for i in range(q):
    query=input().split()
    if query[0]=="1":
        turn+=1
    if query[0]=="2":
        if query[1]=="1":
            if turn%2==0:
                ans.appendleft(query[2])
            else:
                ans.append(query[2])
        elif query[1]=="2":
            if turn%2==0:
                ans.append(query[2])
            else:
                ans.appendleft(query[2])
ans="".join(ans)
if turn%2==1:
    ans=ans[::-1]
print(ans)

Verdicts: Wrong Answer

Wrong Answer

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
import sys
input = sys.stdin.readline

qs = [list(input().split()) for _ in range(q)]

left = ''
right = ''


front = False
for i in qs:
    if(i[0] == '1'):
        front = not front
        continue

    f = int(i[1])
    c = i[2]
    if(front + f == 2):
        right = right + c
    else:
        left = left + c

if(not front):
    ans = left[::-1] + s + right
else:
    ans = right[::-1] + s[::-1] + left

print(ans)

import sys
input = sys.stdin.readline

qs = [list(input().split()) for _ in range(q)]

left = ''
right = ''


front = False
for i in qs:
    if(i[0] == '1'):
        front = not front
        continue

    f = int(i[1])
    c = i[2]
    if(front + f == 2):
        right = right + c
    else:
        left = left + c

if(not front):
    ans = left[::-1] + s + right
else:
    ans = right[::-1] + s[::-1] + left

print(ans)

s = input()
q = int(input())

import sys
input = sys.stdin.readline

qs = [list(input().split()) for _ in range(q)]

left = ''
right = ''


front = False
for i in qs:
    if(i[0] == '1'):
        front = not front
        continue

    f = int(i[1])
    c = i[2]
    if(front + f == 2):
        right = right + c
    else:
        left = left + c

if(not front):
    ans = left[::-1] + s + right
else:
    ans = right[::-1] + s[::-1] + left

print(ans)


Verdicts: Runtime Error

Runtime Error

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
S = list(input())
Q = int(input())
for i in range(Q):
    a = list(input().split())
    if len(a) == 1:##T_i=1
        S.reverse()
    else:#T_i == 2
        if a[1] == "1":
            S.insert(0,a[2])
        else:
            S.append(a[2])

for i in range(len(S)):
    print(S[i],end="")


S = input()
Q = int(input())
Atama = []
Osiri = []
rev = [0]*Q
for i in range(Q):
    a = list(input().split())
    if len(a) == 1:##T_i=1
        rev[i] = 1
    else:#T_i == 2
        if a[1] == "1":
            Atama.append([i,a[2],1])
        else:
            Osiri.append([i,a[2],-1])


Rev = []
tmp = 0
r = sum(rev)
for i in range(len(rev)):
    r -= rev[i]
    Rev.append(r)
##print(Rev)
for i in range(len(Atama)):
    Atama[i][2] *= (-1)**Rev[Atama[i][0]]
for i in range(len(Osiri)):
    Osiri[i][2] *= (-1)**Rev[Osiri[i][0]]

Ans = Atama+Osiri
Ans.sort()
##print(Ans)
fw = ""
aw = ""
N = len(Ans)
for i in range(N):
    if Ans[i][2] == 1:
        fw += Ans[i][1]
    else:
        aw += Ans[i][1]

if sum(rev)%2 == 1:
    S = S[::-1]
fw = fw[::-1]

print(fw+S+aw)

Verdicts: Time Limit Exceeded

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
S = input().split()
N = int(input())
Q = [[] for _ in range(N)]
for i in range(N):
    line = input().split()
    Q[i] = line
reverse = 0
r_b = []
r_f = []
for i in range(N):
    if Q[i][0] == '1' :
        reverse = (reverse + 1) % 2
    else :
        if reverse == 0 :
            if Q[i][1] == '1' :
                r_f = [Q[i][2]] + r_f
            else :
                r_b.append(Q[i][2])
        else:
            if Q[i][1] == '1':
                r_b.append(Q[i][2])
            else:
                r_f = [Q[i][2]] + r_f

print(1)

S = input().split()
N = int(input())
reverse = 0
r_b = []
r_f = []
for i in range(N):
    line = input().split()
    if line[i][0] == '1' :
        reverse = (reverse + 1) % 2
    else :
        if reverse == 0 :
            if line[i][1] == '1' :
                r_f = [line[i][2]] + r_f
            else :
                r_b.append(line[i][2])
        else:
            if line[i][1] == '1':
                r_b.append(line[i][2])
            else:
                r_f = [line[i][2]] + r_f

if reverse == 0 :
    S = r_f + S + r_b
    print("".join(S))
else:
    S.reverse()
    r_b.reverse()
    r_f.reverse()
    S = r_b + S + r_f
    print("".join(S))

S = input()
N = int(input())
Q = [[] for _ in range(N)]
for i in range(N):
    line = input().split()
    Q[i] = line
reverse = 0
r_b = ""
r_f = ""
for i in range(N):
    if Q[i][0] == '1' :
        reverse = (reverse + 1) % 2
    else :
        if reverse == 0 :
            if Q[i][1] == '1' :
                r_f = Q[i][2] + r_f
            else :
                r_b = r_b +Q[i][2]
        else:
            if Q[i][1] == '1':
                r_b = r_b +Q[i][2]
            else:
                r_f = Q[i][2] + r_f
if reverse == 0 :
    print(r_f + S + r_b)
else:
    r_f = r_f[::-1]
    r_b = r_b[::-1]
    S = S[::-1]
    print(r_b + S + r_f)

Verdicts: Wrong Answer

Runtime Error

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
from collections import deque

S = list(input())
Q = int(input())
d = deque(S)
flag = -1
for _ in range(Q):
    q = list(input().split())
    if q[0] == '1':
        flag *= -1
    else:
        f = int(q[1])
        c = q[2]
        if flag == 1 and f == 1 or flag == -1 and f == 2:
            d.appendleft(c)
        else:
            d.append(c)

if flag == -1:
    d = d[::-1]

print(''.join(d))


from collections import deque

S = list(input())
Q = int(input())
d = deque(S)
flag = -1
for _ in range(Q):
    q = list(input().split())
    if q[0] == '1':
        flag *= -1
    else:
        f = int(q[1])
        c = q[2]
        if flag == 1 and f == 1 or flag == -1 and f == 2:
            d.appendleft(c)
        else:
            d.append(c)
d = list(d)
if flag == 1:
    d = d[::-1]

print(''.join(d))


from collections import deque

S = list(input())
Q = int(input())
d = deque(S)
flag = -1
for _ in range(Q):
    q = list(input().split())
    if q[0] == '1':
        flag *= -1
    else:
        f = int(q[1])
        c = q[2]
        if flag == -1 and f == 1 or flag == 1 and f == 2:
            d.appendleft(c)
        else:
            d.append(c)
d = list(d)
if flag == 1:
    d = d[::-1]

print(''.join(d))


Verdicts: Runtime Error

Wrong Answer

Accepted
Problem: Takahashi has a string S consisting of lowercase English letters.  Starting with this string, he will produce a new one in the procedure given as follows.  The procedure consists of Q operations. In Operation i (1 \leq i \leq Q), an integer T_i is provided, which means the following:    * If T_i = 1: reverse the string S.    * If T_i = 2: An integer F_i and a lowercase English letter C_i are additionally provided.      * If F_i = 1 : Add C_i to the beginning of the string S.     * If F_i = 2 : Add C_i to the end of the string S.  Help Takahashi by finding the final string that results from the procedure.

Submission attempts (in order):
from collections import deque

S = str(input())
Q = int(input())

tmp = deque(S)
houkou = 1 

for i in range(Q):
    inp = input().split()
    if inp[0] == '1':
        if houkou == 1:
            houkou = 2
        else:
            houkou = 1
    else:
        if houkou == 1 and inp[1] == '1':
            tmp.appendleft(inp[2])
        elif houkou == 1 and inp[1] == '2':
            tmp.append(inp[2])
        elif houkou == 2 and inp[1] == '1':
            tmp.append(inp[2])
        else:
            tmp.appendleft(inp[2])
    
    print(houkou, tmp)

output = ''.join(list(tmp))
if houkou == 2:
    output = output[::-1]

print(output)


from collections import deque

S = str(input())
Q = int(input())

tmp = deque(S)
houkou = 1 

for i in range(Q):
    inp = input().split()
    if inp[0] == '1':
        if houkou == 1:
            houkou = 2
        else:
            houkou = 1
    else:
        if houkou == 1 and inp[1] == '1':
            tmp.appendleft(inp[2])
        elif houkou == 1 and inp[1] == '2':
            tmp.append(inp[2])
        elif houkou == 2 and inp[1] == '1':
            tmp.append(inp[2])
        else:
            tmp.appendleft(inp[2])
    
#    print(houkou, tmp)

output = ''.join(list(tmp))
if houkou == 2:
    output = output[::-1]

print(output)


Verdicts: Runtime Error

Accepted