Problem:


Score : 200 points

Problem Statement

A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:

S is a palindrome.

Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.

The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.

Determine whether S is a strong palindrome.

Constraints

S consists of lowercase English letters.

The length of S is an odd number between 3 and 99 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

If S is a strong palindrome, print Yes;
otherwise, print No.

Sample Input 1

akasaka

Sample Output 1

Yes

S is akasaka.

The string formed by the 1-st through the 3-rd characters is aka.

The string formed by the 5-th through the 7-th characters is aka.
All of these are palindromes, so S is a strong palindrome.

Sample Input 2

level

Sample Output 2

No

Sample Input 3

atcoder

Sample Output 3

No



Sample input:
akasaka


Sample output:
Yes

Submission status: Accepted

Solution code:

def ispalindrome(s):
    h = len(s)
    if (h % 2) == 0:
        h = int(h/2)
        for a, b in zip(s[:h], reversed(s[h:])):
            if a != b: return 0
    else:
        h = int(h/2)
        for a, b in zip(s[:h], reversed(s[h+1:])):
            if a != b: return 0
    return 1

str_in = input()
N = len(str_in)
result = ispalindrome(str_in)

if result == 0:
    print('No')
else:
    if N == 3:
        print('Yes')
    else:
        str_half = str_in[:int((N-1)/2)]
        result &= ispalindrome(str_half)

        if result == 0:
            print('No')
        else:
            str_half = str_in[int(((N+3)/2)-1):]
            result &= ispalindrome(str_half)

            if result == 0:
                print('No')
            else:
                print('Yes')

Submission status: Accepted

Solution code:
import sys
def fastio():
    from io import StringIO
    from atexit import register
    global input
    sys.stdin = StringIO(sys.stdin.read())
    input = lambda : sys.stdin.readline().rstrip('\r\n')
    sys.stdout = StringIO()
    register(lambda : sys.__stdout__.write(sys.stdout.getvalue()))
fastio()

INF = 10**20
MOD = 10**9 + 7
I = lambda:list(map(int,input().split()))
# from math import gcd
from math import ceil
from collections import defaultdict as dd, Counter
from bisect import bisect_left as bl, bisect_right as br

s = input()
ans = 'No'
n = len(s)
def isPalindrome(s):
    return s == s[::-1]

if isPalindrome(s) and isPalindrome(s[: n // 2]) and isPalindrome(s[(n // 2 + 1): ]):
    ans = "Yes"
print(ans)

Submission status: Accepted

Solution code:
str_inp = list(input())
str_len = len(str_inp)
flag = "ok"
check1 = ""
check2 = ""
dummy1 = (str_len - 1) // 2
for j in range(dummy1):
  check1 = check1 + str_inp[j]
#print(check1)
for i in range(dummy1//2):
  if check1[i] != check1[-i-1]:
    flag = "ng"
    
dummy2 = (str_len + 3) // 2
for k in range(dummy2-1,str_len):
  check2 = check2 + str_inp[k]
#print(check2)
dummy_len = len(check2)
for l in range(dummy_len//2):
  if check2[l] != check2[-l-1]:
    flag = "ng"

if flag == "ok" and check1 == check2:
  print("Yes")
else:
  print("No")

Submission status: Accepted

Solution code:
x = str(input())
check1 = (len(x)-1)/2+1
check2= (len(x)+3)/2
check3 = (len(x)+1)
if x[0:int(check1)-1] == x[int(check2)-1:int(check3)]:
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
org_str = input()
n = len(org_str)
if n%2 == 0:
    N = int(n/2-1)
else:
    N = int((n-1)/2)

org_str1 = org_str[0:(N)]
org_str2 = org_str[(N+1):n]

new_str_list = list(reversed(org_str))
new_str = org_str[::-1]

new_str1_list = list(reversed(org_str1))
new_str1 = org_str1[::-1]

new_str2_list = list(reversed(org_str2))
new_str2 = org_str2[::-1]


if org_str == new_str and org_str1 == new_str1 and org_str2 == new_str2:
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
s = input()
n = len(s)
zen = s[:n // 2]

kou= s[n // 2 + 1:]

s_rev = s[::-1]

zen_rev = zen[::-1]
kou_rev = kou[::-1]

if s == s_rev and zen == zen_rev and kou == kou_rev:
  print('Yes')
else:
  print('No')

Submission status: Accepted

Solution code:
S = input()
T = S[0:int((len(S)-1)/2)]
U = S[int((len(S)+3)/2):int(len(S)-1)]
if S == S[::-1] and T == T[::-1] and U == U[::-1]:
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
s = input()
n = len(s)

def j(x:str):
    for i in range(len(x)//2):
        if x[i] != x[-(i+1)]:
            return False
    return True

if not j(s):
    print ("No")
    exit ()

if not j(s[0:((n-1)//2)]):
    print ("No")
    exit ()

if not j(s[((n+3)//2)-1::]):
    print ("No")
    exit ()

print ("Yes")

Submission status: Accepted

Solution code:
A = input()
E = int((len(A)-1)/2)
B = A[0:E]
C = A[int((len(A)+3)/2)-1:]
if A == A[::-1] and B == B[::-1] and C == C[::-1]:
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
def check_str( ss ):
  sslen = len(ss)
  
  if( sslen %2 == 0 ): 
    return (ss[:sslen//2] == ss[sslen//2:][::-1])
  else:
  	return (ss[:sslen//2] == ss[sslen//2+1:][::-1])
  
s = input()

slen = len(s)

if( not check_str( s ) ):
  print( "No" )
else:
#print(s[:slen//2])
#print(s[slen//2+1:][::-1])

  s1 =  s[:slen//2]
  s2 = s[slen//2+1:]
  
  
  if( (not check_str( s1 )) or (not check_str( s2 )) ):
    print( "No" )
  else:
    print( "Yes" )

#s1 = s[:len(s)-1]

#s1

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
s = list(input())
n = len(s)
ans = "Yes"
x = (n-1)//2-1
y = (n+3)//2-1
n-=1
for i in range(x+1):
    if s[i] != s[n-i]:
        ans = "No"
        break
for j in range(x+1):
    if s[j] != s[x-j]:
        ans = "No"
        break
    if s[y+j] != s[n-j]:
        ans = "No"
        break
print(ans)

Submission status: Accepted

Solution code:
line = input().rstrip()
line_len = len(line)
line_reverse = list(reversed(line))
line_reverse = ''.join(line_reverse)

start = int((line_len -1) /2)
end = int((line_len + 3) / 2 - 1)
# print(line_len - 1)
start_len = line[0:start]
end_len = line[end:(line_len)]


if line == line_reverse and start_len == end_len:
    print('Yes')
else:
    print('No')

Submission status: Accepted

Solution code:
s = str(input())
forward = int((len(s)-1)/2)
back = int((len(s)+3)/2 -1)
if s == ''.join(list(reversed(s))):
    if s[:forward] == ''.join(list(reversed(s[:forward]))):
        if s[back:] == ''.join(list(reversed(s[back:]))):
            print("Yes")
            exit()
print("No")

Submission status: Accepted

Solution code:
s = input()
n = len(s)

sub_1 = s[0:int((n-1)/2)]
sub_1 == sub_1[::-1]

sub_2 = s[int(n+3/2):]
sub_2 == sub_2[::-1]

if(s==s[::-1] and sub_1 == sub_1[::-1] and sub_2 == sub_2[::-1]):
    print('Yes')
else:
    print('No')

Submission status: Accepted

Solution code:
s = input()

res = 0
s
s[::-1]
if s == s[::-1]:
    res += 1

t = int((len(s)-1)/2)
tm = s[:t]
if tm == tm[::-1]:
    res += 1

u = int(((len(s) +3) / 2 )-1)
tm = s[u:]
if tm == tm[::-1]:
    res +=1

if res == 3:
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
S=input().strip()
r_S = S[::-1]
a = S[:int(((len(S)-1)/2)):]
b = S[int(((len(S)+1)/2))::]
r_a = S[int(((len(S)-3)/2))::-1]
r_b = S[:int(((len(S)-1)/2)):-1]

if a == r_a and b == r_b and S == r_S:
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
S = input()
N = len(S)

left_end = (N - 1)//2
right_start = (N + 3) //2 - 1

left_S = S[:left_end]
right_S = S[right_start:]

rev_left_S = left_S[::-1]
rev_right_S = right_S[::-1]

check_1 = False
if (left_S == rev_right_S):
  check_1 = True

check_2 = False
if (left_S == rev_left_S):
  check_2 = True

check_3 = False
if (right_S == rev_right_S):
  check_3 = True

result = 'Yes' if (check_1 and check_2 and check_3) else 'No'

print(result)

Submission status: Accepted

Solution code:
s = input()

n = int(len(s))

n1 = int ( (n - 1 ) / 2 )
n2 = int ( (n + 1 ) / 2 )

s1 = s[0:n1]
s2 = s[n2:n]

def if_kai( st ):
  ret = True
  nn = len(st)
  nn1 = int ( (nn - 1 ) / 2 )
  for i in range(nn1+1):
    if st[i] != st[nn-i-1] :
       ret = False
       break
  return ret      

if if_kai( s ) and if_kai( s1 ) and if_kai( s2 ) :
  print('Yes')
else:
  print('No')

Submission status: Accepted

Solution code:
S = input()

def check_if_kaibun(string):
  # odd
  if len(string) % 2 == 1:
    middle = int((len(string) - 1)  / 2)
    # print(string[0:middle])
    # print(string[middle + 1:])
    if (string[0:middle]) == (string[middle + 1:][::-1]):
      return True
    else:
      return False
  else:
    middle = int(len(string) / 2)
    if (string[0:middle]) == (string[middle:][::-1]):
      return True
    else:
      return False
    
a = int((len(S) - 1) / 2)
b = int((len(S) + 3) / 2)
# print(a, b)
# print(check_if_kaibun("asabasa"))
# print(check_if_kaibun("aabbaa"))

if check_if_kaibun(S):
  # print(S[0:a])
  if check_if_kaibun(S[0:a]):
    # print(S[b-1:])
    if check_if_kaibun(S[b-1:]):
      print("Yes")
    else:
      print("No")
  else:
    print("No")
else:
  print("No")

Submission status: Accepted

Solution code:
from copy import copy

S = input()
Slist = []
par = []
rap = []

for s in S:
    Slist.append(s)

if len(Slist)==3:
    if Slist==list(reversed(Slist)):
        print("Yes")
    else:
        print("No")
else:
    par = copy(Slist)
    rap = copy(Slist)
    rap = list(reversed(rap))

    if par==rap:
        rap = []
        par = copy(Slist)
        par = par[:(len(par)-1)//2]
        rap = copy(par)
        rap = list(reversed(rap))

        if par==rap:
            rap = []
            par = copy(Slist)
            par = par[(len(par)+3)//2-1:]
            rap = copy(par)
            rap = list(reversed(rap))

            if par==rap:
                print("Yes")
            else:
                print("No")
        else:
            print("No")
    else:
        print("No")

Submission status: Accepted

Solution code:
#Input
s = input()
n = len(s)
#Process
#Progress1
o1 = True
for i in range(n):
    if(i != int(n/2) + 1):
        #print(s[i],s[n - i - 1])
        o1 = o1 and s[i] == s[n - i -1]

#Progress2
o2 = True
n2 = int((n - 1) / 2)
for i in range(n2):
    #print(s[i],s[n2 - i - 1])
    o2 = o2 and s[i] == s[n2 - i -1]
    
#Progress3
o3 = True
n3 = int((n + 3) / 2)
for i in range(n - n3):
    #print(s[i + n3 -1],s[n - i -1])
    o3 = o3 and s[i + n3 -1] == s[n - i -1]


#EndProgress
if(o1 and o2 and o3): 
    Ans = "Yes"
else:
    Ans = "No"

#Output
print(Ans)

Submission status: Accepted

Solution code:
s = input()
n = len(s)
s1 = s[0:int(((n-1)/2))]
s2 = s[int(((n+3)/2)-1)::]
if (s != s[::-1]) or (s1 != s1[::-1]) or (s2 != s2[::-1]):
    print('No')
else:
    print('Yes')

Submission status: Accepted

Solution code:
s=input()
s=list(s)
n=len(s)
a=s[:(n-1)//2]
b=s[(n+2)//2:]
m=len(a)
yes=0
for i in range((n-1)//2):
    if s[i]==s[-(i+1)]:
        yes+=1
if n==3:
    if yes==1:
        print('Yes')
    else:
        print('No')
else:    
    if m%2==1:
        for i in range((m-1)//2):
            if a[i]==a[-(i+1)]:
                yes+=1
            if b[i]==b[-(i+1)]:
                yes+=1
    else:
       for i in range(m//2):
            if a[i]==a[-(i+1)]:
                yes+=1
            if b[i]==b[-(i+1)]:
                yes+=1
    if yes>=(m-1)+(n-1)//2:
        print('Yes')
    else:
        print('No')

Submission status: Accepted

Solution code:
s = input()
n = len(s)
count = 0

for i in range((n-1)//2):
  if s[i] != s[n-1-i]:
    count = 1
    break

if count == 0:
  for i in range((n-1) // 2 // 2):
    if s[i] != s[(n-3) // 2 - i]:
      count = 1
      break
    elif s[(n+1)//2 + i] != s[n-1-i]:
      count = 1
      break
print("Yes" if count==0 else "No")

Submission status: Accepted

Solution code:
s=str(input())

cnt=0

num=len(s)//2

a=s[:num]
b=s[num+1:]

num2=len(a)//2
if len(a)%2 ==1:
    c=a[:num2]
    d=a[num2+1:]
else:
    c=a[:num2]
    d=a[num2:]


num3=int((len(s)+3)/2-1)
s1=s[num3:]
num4=len(s1)//2
if len(s1)%2==1:
    e=s1[:num4]
    f=s1[num4+1:]
else:
    e=s1[:num4]
    f=s1[num4:]


if a==b[::-1]:
    cnt +=1

if c==d[::-1]:
    cnt +=1

if e==f[::-1]:
    cnt +=1

if cnt==3:
    print("Yes")
else:
    print("No")

Submission status: Accepted

Solution code:
S=str(input())

N=len(S)
i=0
r="No"

s=int((N-1)/2)
t=int((N+3)/2)

a=S[0:s]
b=S[t-1:N]

if(N==3 and S==S[::-1]):
  print("Yes")
elif(S==S[::-1] and a==(a[::-1]) and b==(b[::-1])):
  print("Yes")
else:
  print("No")

Submission status: Accepted

Solution code:
def judge(s):
    ok = True
    n = len(s)
    m = (n - 1) // 2
    if (n == 1):
        ok = True
    else:
        if (m == 0):
            ok = (s[0] == s[1])
        else:
            for i in range(m):
                if s[i] != s[(n-1)-i]:
                    ok = False
                    break
    return ok

def strong_judge(s):
    ok = judge(s)
    n = len(s)
    m = (n - 1) // 2
    kf = (m - 1) // 2
    kb = m + kf + 1
    sf = s[0:m]
    sb = s[(m + 1):n]
    if ok:
        ok = judge(sf) and judge(sb)
    return ok

s = str(input())

if ((len(s) % 2 == 0) or (len(s) > 99) or (len(s) < 3)):
    print("No")
else:
    if strong_judge(s):
        print("Yes")
    else:
        print("No")

Submission status: Accepted

Solution code:
S = input()

N = len(S)

def func(S,ans,N):
    for i in range(N):
        if S[i] != S[N - i-1]:
            ans ='No'
    return (ans)

ans = func(S, 'Yes',(len(S)-1//2))
S1 = S[(N+3)//2-1:N]
ans = func(S1, ans,len(S1))

print(ans)

Submission status: Accepted

Solution code:
s = str(input())

tmp_p = 0

tmp = ''
for i in range(len(s)//2):
    tmp += s[-(i+1)]
#print(tmp)
if tmp != s[0:len(s)//2]:
    tmp_p += 1

tmp_2 = s[0:(len(s)-1)//2]
tmp = ''
if len(tmp_2) != 1:
    for i in range(len(tmp_2)//2):
        tmp += tmp_2[-(i+1)]

    #print(tmp)
    #print(tmp_2[0:len(tmp_2)//2])
    if tmp != tmp_2[0:len(tmp_2)//2]:
        tmp_p += 1


tmp_3 = s[(len(s)+3)//2-1:]
tmp = ''
if len(tmp_3) != 1:
    for i in range(len(tmp_3)//2):
        tmp += tmp_3[-(i+1)]
    #print(tmp)
    #print(tmp_3[0:len(tmp_3)//2])
    if tmp != tmp_3[0:len(tmp_3)//2]:
        tmp_p += 1

if tmp_p != 0:
    print('No')
else:
    print('Yes')

Submission status: Accepted

Solution code:
S = input()
N = len(S)
a = N % 4
ans = "Yes"
if N == 3:
    if S[0] != S[2]:
        ans = "No"
elif a == 1:
    x = N // 4
    for i in range(x):
        if (S[i] != S[2 * x - i - 1]) or (S[2 * x + 1 + i] != S[4 * x - i]):
            ans = "No"
    for i in range (2 * x):
        if S[i] != S[N - 1 - i]:
            ans = "No"
elif a == 3:
    x = N // 4
    for i in range(x):
        if (S[i] != S[2 * x - i]) or (S[2 * x + 2 + i] != S[4 * x + 2 - i]):
            ans = "No"
    for i in range (2 * x + 1):
        if S[i] != S[N - 1 - i]:
            ans = "No"
print(ans)

Submission status: Accepted

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
    AtoZ = list('abcdefghijklmnopqrsruvwxyz')
    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_float = lambda: float(r())
    R = lambda: list(map(int, r().split()))
    R_str = lambda: r().split()
    R_map = lambda: map(int, r().split())
    R_float = lambda: list(map(float, r().split()))
    R_tuple = 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()
    N = len(S) - 1
    def uragaeshi(u):
        return u[::-1]
    
    if S == uragaeshi(S) and S[:(N-1)//2+1] == uragaeshi(S[:(N-1)//2+1]) and S[(N+3)//2:] == uragaeshi(S[(N+3)//2:]):
        print('Yes')
    else:
        print('No')
    
    #%%
    # 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: Accepted

Solution code:
S = list(input())
N = len(S)
if N%4 == 3 or N%4 == 1:
    if S[0:int((N-1)/2)] == S[int((N+1)/2):N]:
        flag = True
        s = S[0:int((N-1)/2)]
        for i in range(N//4):
            if s[i] != s[-i-1]:
                flag=False
        if flag:
            print('Yes')
        else:
            print('No')
    else:
        print('No')    
else:
    print('No')

Submission status: Accepted

Solution code:
s = input()

l = len(s)
cent = int((l - 1) / 2)

# 1
for i in range(cent):
    if s[i] != s[l-i-1]:
        print('No')
        exit(0)

# 2
for i in range(cent//2):
    if s[i] != s[cent - i - 1] or s[cent + 1 + i] != s[l-i-1]:
        print('No')
        exit(0)

print('Yes')

Submission status: Accepted

Solution code:
s = input()
n = len(s)
#print(s)
rev = ''.join(list(reversed(s)))

#print(rev)
a = int((n-1)/2)
s1 = s[:a]
s_1 = ''.join(list(reversed(s1)))
#print(a, s1, s_1)
b = int((n+3)/2)
s2 = s[b-1:n+1]
s_2 = ''.join(list(reversed(s2)))
#print(b, s2, s_2)

if (s==rev) and (s1 == s_1) and (s2==s_2):
  print('Yes')
else:
  print('No')