Problem:


Score : 300 points

Problem Statement

Snuke has N hats. The i-th hat has an integer a_i written on it.

There are N camels standing in a circle.
Snuke will put one of his hats on each of these camels.

If there exists a way to distribute the hats to the camels such that the following condition is satisfied for every camel, print Yes; otherwise, print No.

The bitwise XOR of the numbers written on the hats on both adjacent camels is equal to the number on the hat on itself.

What is XOR?

The bitwise XOR x_1 \oplus x_2 \oplus \ldots \oplus x_n of n non-negative integers x_1, x_2, \ldots, x_n is defined as follows:

- When x_1 \oplus x_2 \oplus \ldots \oplus x_n is written in base two, the digit in the 2^k's place (k \geq 0) is 1 if the number of integers among x_1, x_2, \ldots, x_n whose binary representations have 1 in the 2^k's place is odd, and 0 if that count is even.

For example, 3 \oplus 5 = 6.

Constraints

All values in input are integers.

3 \leq N \leq 10^{5}

0 \leq a_i \leq 10^{9}

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 \ldots a_{N}

Output

Print the answer.

Sample Input 1

3
1 2 3

Sample Output 1

Yes

If we put the hats with 1, 2, and 3 in this order, clockwise, the condition will be satisfied for every camel, so the answer is Yes.

Sample Input 2

4
1 2 4 8

Sample Output 2

No

There is no such way to distribute the hats; the answer is No.



Sample input:
3
1 2 3


Sample output:
Yes

Submission status: Wrong Answer

Solution code:
n = int(input())
a = [int(s) for s in input().split(" ")]
dic_a = {}
for aa in a:
  if aa in dic_a:
    dic_a[aa] += 1
  else:
    dic_a[aa] = 1
a_set = list(dic_a.keys())
n_set = len(a_set)

if (n < 3):
  ans = False
else:
  ans = False
  if 0 in a_set:
    for v in dic_a.values():
      if v >= 2:
        ans = True
        break
  if not ans:
    for i in range(n_set):
      for j in range(i + 1, n_set):
        if a_set[i] ^ a_set[j] in a_set:
          ans = True
          break
      if ans:
        break

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

Submission status: Wrong Answer

Solution code:
import sys,itertools,copy
input = sys.stdin.readline
N=int(input().strip())
an=[int(x) for x in input().strip().split()]

if N==3:
    if an[0]^an[1]==an[2]:
        print("Yes")
    else:
        print("No")
elif N%3!=0:
    if len(set(an))==1:
        print("Yes")
    else:
        print("No")
else:
    bn = list(set(an))
    if len(bn)==1:
        print("Yes")
    elif len(bn)!=3:
        print("No")
    elif bn[0]^bn[1]!=bn[2]:
        print("No")
    else:
        for b in bn:
            if an.count(b)%(N/3)!=0:
                print("No")
                break
        else:
            print("Yes")

Submission status: Wrong Answer

Solution code:
from collections import defaultdict

N = int(input())
a = list(map(int, input().split()))
d = defaultdict(int)
for c in a[1:]:
    d[c] += 1
for i in range(1, N):
    x = a[0]
    y = a[i]
    d[y] -= 1
    for _ in range(N - 2):
        z = x ^ y
        if z in d and d[z] > 0:
            d[z] -= 1
            x = y
            y = z
        else:
            break
    else:
        if a[0] == x ^ y:
            print("Yes")
            exit(0)

print("No")

Submission status: Wrong Answer

Solution code:
n=int(input())
a=list(map(int,input().split()))
aa=set(a)
if len(aa)>3:
    print("No")
    exit()
elif a.count(0)==n:
    print("Yes")
    exit()
elif len(aa)<=1:
    print("No")
    exit()
l=[]
l.append(a[0])
for i in range(1,n):
    if a[i]!=a[0]:
        l.append(a[i])
        break
l.append(l[0]^l[1])

if n==0:
    exit()
else:
    if l[0]==l[1]:
        if n%3==0:
            if a.count(l[0])==(n//3)*2 and a.count(l[2])==n//3:
                print("Yes")
            else:
                print("No")
        else:
            print("No")
    if l[0]==l[2]:
        if n%3==0:
            if a.count(l[0])==(n//3)*2 and a.count(l[1])==n//3:
                print("Yes")
            else:
                print("No")
        else:
            print("No")
    if l[2]==l[1]:
        if n%3==0:
            if a.count(l[1])==(n//3)*2 and a.count(l[0])==n//3:
                print("Yes")
            else:
                print("No")
        else:
            print("No")
    else:
        if n%3==0:
            if a.count(l[0])==n//3 and a.count(l[1])==n//3 and a.count(l[2])==n//3:
                print("Yes")
            else:
                print("No")
        else:
            print("No")

Submission status: Wrong Answer

Solution code:
N=int(input())
A=list(map(int,input().split()))
def binbin(a):
    return int(bin(a).split("b")[1])
arr=[A[-1]]
arr.extend(A)
arr.append(A[0])
ans=[]
for i in range(1,N+1):
    ans.append(int(binbin(arr[i])) == int("".join(list(map(lambda x:str(int(x)%2),str(binbin(arr[i-1])+binbin(arr[i+1])))))))

if all(ans):
    if any(ans)==False:
        print("Yes")
else:
    if any(ans)==False:
        print("No")

Submission status: Wrong Answer

Solution code:
import collections
n=int(input())
a=list(map(int,input().split()))
f=0
ct=0
C=collections.Counter(a)
D=C.most_common()
for i in range(n):
  if a[i]==0:
    ct=1
    f=1
for i in range(n):
  a[i]=format(a[i],'020b')
l=[0 for i in range(20)]
for i in range(n):
  for j in range(20):
    if a[i][j]=='1':
      l[j]+=1
for j in range(20):
  l[j]=l[j]%2

if f==0:
  if 1 in l:
    print('No')
  elif D[0][1]>=2:
    print('No')
  elif n%3!=0:
    print('No')
  else:
    print('Yes')
elif f==1:
  if len(D)==1:
    print('Yes')
  elif len(D)==2 and D[0][0]!=0 and D[0][1]==D[1][1]*2:
    print('Yes')
  else:
    print('No')

Submission status: Wrong Answer

Solution code:
N = int(input())
a = [int(i) for i in input().split()]

a.sort()
for i in range(N-1):
    if a[i-1] ^ a[i+1] == a[i]:
        continue
    else:
        print("No")
        exit()

if a[N-2] ^ a[0] != a[N-1]:
    print("No")
    exit()

print("Yes")

Submission status: Wrong Answer

Solution code:
N = int(input())
a = list(map(int,input().split()))

res = "No"
a_set = list(set(a))
if len(a_set) == 3:
    res = "Yes"
print(res)

Submission status: Wrong Answer

Solution code:
n = input()
a = list(map(int, input().split()))

print("No")

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import random
input = sys.stdin.readline

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

def main():

	N = int(input())
	#S = list(input())
	#L,R = map(int, input().split(" "))
	#L = list(map(int, input().split(" ")))

	A = input().split()
	A = [int(s) for s in A]
	A.sort()

	a_max = A[-1]
	#A[-1] = -1

	a = 2
	i = 1
	ll = [1]
	while a < 100000+1:
		a = 2 ** i
		ll.append(a)
		i += 1
		if a_max == a:
			print("No")
			sys.exit()

	tmp = 0
	flg = False
	for i in ll:
		tmp += i
		if a_max == tmp:
			flg = True
			break
	
	if flg == False:
			print("No")
			sys.exit()

	bk = -1
	next1 = -1
	for i, val in enumerate(A):
		#A[i] = -1
		if i == 0:
			tmp = a_max - val
			if tmp in A:
				bk = val
				next1 = tmp
				continue
			else:
				print("No")
				sys.exit()
		
		if i == N-1:
			if next1 == a_max and bk ^ A[0] == a_max:
				print("Yes")
				sys.exit()
		
		if next1 == a_max:
			tmp = next1 - bk
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				print("No")
				sys.exit()

		elif next1 == A[0]:
			tmp = a_max - next1
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				#print(next1)
				print("No")
				sys.exit()

		else:
			tmp = next1 + bk
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				print("No")
				sys.exit()
		
	print("No")


if __name__ == "__main__":
	main()

Submission status: Wrong Answer

Solution code:
N = int(input())
a = list(map(int, input().split()))

def main():
    if a[0] != a[-1] ^ a[1]:
        return "No"
    for i in range(1, N - 1):
        if a[i] != a[i + 1] ^ a[i - 1]:
            return "No"
    if a[-1] != a[0] ^ a[-2]:
        return "No"
    return "Yes"

print(main())

Submission status: Wrong Answer

Solution code:
a = int(input())
b = list(map(int, input().split()))
c = {}

for i in range(a):
    if b[i] not in c:
        c.setdefault(b[i], 1)
    else:
        c[b[i]] += 1

if len(c) == 3:
    f = list(c.values())
    for i in range(1, len(f)):
        if f[i] != f[i-1]:
            print("No")
            exit()
    g = list(c.keys())
    if g[0] ^ g[1] == g[2]:
        print("Yes")
    else:
        print("No")
elif len(c) == 2:
    f = list(c.values())
    if f[0] != f[1]:
        print("No")
        exit()
    g = list(c.keys())
    if g[0] == 0 or g[1] == 0:
        print("Yes")
    else:
        print("No")
elif len(c) == 1:
    print("Yes")
else:
    print("No")
    exit()

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import random
input = sys.stdin.readline

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

def main():

	N = int(input())
	#S = list(input())
	#L,R = map(int, input().split(" "))
	#L = list(map(int, input().split(" ")))

	A = input().split()
	A = [int(s) for s in A]
	A.sort()

	a_max = A[-1]
	#A[-1] = -1

	a = 2
	i = 1
	ll = [1]
	while a < 100000+1:
		a = 2 ** i
		ll.append(a)
		i += 1
		if a_max == a:
			print("Yes")
			sys.exit()

	tmp = 0
	flg = False
	for i in ll:
		tmp += i
		if a_max == tmp:
			flg = True
			break
	
	if flg == False:
			print("No")
			sys.exit()

	bk = -1
	next1 = -1
	for i, val in enumerate(A):
		#A[i] = -1
		if i == 0:
			tmp = a_max - val
			if tmp in A:
				bk = val
				next1 = tmp
				continue
			else:
				print("No")
				sys.exit()
		
		if i == N-1:
			if next1 == a_max and bk ^ A[0] == a_max:
				print("Yes")
				sys.exit()
		
		if next1 == a_max:
			tmp = next1 - bk
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				print("No")
				sys.exit()

		elif next1 == A[0]:
			tmp = a_max - next1
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				#print(next1)
				print("No")
				sys.exit()

		else:
			tmp = next1 + bk
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				print("No")
				sys.exit()
		
	print("No")


if __name__ == "__main__":
	main()

Submission status: Wrong Answer

Solution code:
def solve(N, _as):
    if N % 3 != 0:
        return False

    max_bit_len = 0
    max_bit_len_ints = []
    other_ints = []
    for a in _as:
        bl = int.bit_length(a)
        if bl > max_bit_len:
            max_bit_len = bl
            other_ints.extend(max_bit_len_ints)
            max_bit_len_ints.clear()
            max_bit_len_ints.append(a)
        elif bl == max_bit_len:
            max_bit_len_ints.append(a)
        else:
            other_ints.append(a)

    if N / 3 != len(other_ints):
        return False

    first = max_bit_len_ints.pop()
    for second in other_ints:
        return _solve(first, second, max_bit_len_ints, other_ints)


def _solve(first, second, _max_bit_len_ints, _other_ints):
    max_bit_len_ints = _max_bit_len_ints[:]
    other_ints = _other_ints[:]
    other_ints.remove(second)

    pre_latest = second
    latest = first^second
    try:
        max_bit_len_ints.remove(latest)
    except ValueError:
        return False

    count = 0
    while len(max_bit_len_ints) > 0 and len(other_ints) > 0:
        _latest = latest
        latest = pre_latest^latest
        pre_latest = _latest
        if count == 0 or count == 2:
            try:
                max_bit_len_ints.remove(latest)
            except ValueError:
                return False
        else:
            try:
                other_ints.remove(latest)
            except ValueError:
                return False
        count = (count + 1) % 3

    return pre_latest^latest == first and latest^first == second


if __name__ == '__main__':
    N = int(input())
    _as = [int(x) for x in input().split()]
    print('Yes' if solve(N, _as) else 'No')

Submission status: Wrong Answer

Solution code:
n = int(input())
a = [int(x) for x in input().split(' ')]

b = a
flg = 'No'
flgg = 'no'

for k in range(n-1) :
        # print(k)
        flgg = 'ok'

        for b in range(k+1,n-1) :
                flgg = 'no'

                # print(str(a[b]) + " " + str(a[k-1]) + " " + str(a[k]))

                if(a[b] ^ a[k-1] == a[k]) :
                        tmp = a[k+1]
                        a[k+1] = a[b]
                        a[b] = tmp
                        # print(a[k+1])
                        flgg = 'ok'
                        break
        else :
                if (flgg == 'no') :
                        break
else :
        if (flgg == 'ok') :
                flg = 'Yes'



print(flg)

Submission status: Wrong Answer

Solution code:
n = int(input())
figure = list(map(int, input().split()))
figure1 = []
p=0
for j in range(1,n):
  figure1.append(figure[0]^figure[j])
for k in range(2,n-1):
  if figure1[0]==figure[k]:
    print("Yes")
    #number1 = figure[0]
    #number2 = figure[1]
    #number3 = figure[k]
    p+=1
    break
if p==0:
  print("No")

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import random
input = sys.stdin.readline

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

def main():

	N = int(input())
	#S = list(input())
	#L,R = map(int, input().split(" "))
	#L = list(map(int, input().split(" ")))

	A = input().split()
	A = [int(s) for s in A]
	A.sort()

	a_max = A[-1]
	#A[-1] = -1

	a = 2
	i = 1
	ll = [1]
	while a < 100000+1:
		a = 2 ** i
		ll.append(a)
		i += 1
		if a_max == a:
			print("Yes")
			sys.exit()

	tmp = 0
	flg = False
	for i in ll:
		tmp += i
		if a_max == tmp:
			flg = True
			break
	
	if flg == False:
			print("Yes")
			sys.exit()

	bk = -1
	next1 = -1
	for i, val in enumerate(A):
		#A[i] = -1
		if i == 0:
			tmp = a_max - val
			if tmp in A:
				bk = val
				next1 = tmp
				continue
			else:
				print("No")
				sys.exit()
		
		if i == N-1:
			if next1 == a_max and bk ^ A[0] == a_max:
				print("Yes")
				sys.exit()
		
		if next1 == a_max:
			tmp = next1 - bk
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				print("No")
				sys.exit()

		elif next1 == A[0]:
			tmp = a_max - next1
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				#print(next1)
				print("No")
				sys.exit()

		else:
			tmp = next1 + bk
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				print("No")
				sys.exit()
		
	print("No")


if __name__ == "__main__":
	main()

Submission status: Wrong Answer

Solution code:
n = int(input())
a_lists = list(input().split())

lists = []
index = [0] * n

cnt = 0

def search(a,b,c):

    if int(a_lists[a]) == int(a_lists[b]) ^ int((a_lists[c])):
        index[c] = 1
        index[a] = 1
        index[b] = 1

        return 1
    else:
        return -1

for i in range(0,int(n)):
    if index[i] == 1:
        next

    cnt = 0
    for j in range(0,int(n)):
        if index[j] == 1:
            next
        if i == j:
            next

        cnt_j = 0
        for k in range(0,int(n)):
            if index[k] == 1:
                next
            if i == k:
                next
            if j == k:
                next
            ret = search(i,j,k)
            if ret > 0:
                cnt_j = 1
                break
        if cnt_j == 1:
            cnt = 1
            break
    if cnt == 0:
        print("No")
        exit()

print("Yes")

Submission status: Wrong Answer

Solution code:
n=int(input())
a=list(map(int,input().split()))
aa=set(a)
if len(aa)>3:
    print("No")
    exit()
elif a.count(0)==n:
    print("Yes")
    exit()
elif len(aa)<=1:
    print("No")
    exit()
l=[]
l.append(a[0])
for i in range(1,n):
    if a[i]!=a[0]:
        l.append(a[i])
        break
l.append(l[0]^l[1])

if n==0:
    exit()
else:
    if l[0]==l[1]:
        if n%3==0:
            if a.count(l[0])==(n//3)*2 and a.count(l[2])==n//3:
                print("Yes")
            else:
                print("No")
        elif n%3==1:
            if a.count(l[0])==(n//3)*2+1 and a.count(l[2])==n//3:
                print("Yes")
            else:
                print("No")
        else:
            if a.count(l[0])==(n//3)*2+2 and a.count(l[2])==n//3:
                print("Yes")
            else:
                print("No")
    if l[0]==l[2]:
        if n%3==0:
            if a.count(l[0])==(n//3)*2 and a.count(l[1])==n//3:
                print("Yes")
            else:
                print("No")
        elif n%3==1:
            if a.count(l[0])==(n//3)*2+1 and a.count(l[1])==n//3:
                print("Yes")
            else:
                print("No")
        else:
            if a.count(l[0])==(n//3)*2+1 and a.count(l[1])==n//3+1:
                print("Yes")
            else:
                print("No")
    if l[2]==l[1]:
        if n%3==0:
            if a.count(l[1])==(n//3)*2 and a.count(l[0])==n//3:
                print("Yes")
            else:
                print("No")
        elif n%3==1:
            if a.count(l[1])==(n//3)*2 and a.count(l[0])==n//3+1:
                print("Yes")
            else:
                print("No")
        else:
            if a.count(l[1])==(n//3)*2+1 and a.count(l[0])==n//3+1:
                print("Yes")
            else:
                print("No")
    else:
        if n%3==0:
            if a.count(l[0])==n//3 and a.count(l[1])==n//3 and a.count(l[2])==n//3:
                print("Yes")
            else:
                print("No")

        else:
            print("No")

Submission status: Wrong Answer

Solution code:
a=int(input())
b=list(map(int, input().split()))
c=list(set(b))
if len(c)!=3:
  print("No")
else:  
  d=[0 for i in range(3)]
  for i in range(len(c)):
    e=c[i]
    d[i]=b.count(e)
  if d[0]==d[1]and d[0]==d[2]:
    print("Yes")
  else:
    print("No")

Submission status: Wrong Answer

Solution code:
n=int(input())
a=list(map(int,input().split()))
a.sort()
b=int(n/3)
if(n%3!=0):
    print('No')
else:
    if(a[0] == a[b-1] and a[b-1] != a[b]):
        if(a[b]==a[2*b-1] and a[2*b-1]!=a[2*b]):
            if(a[2*b]==a[n-1]):
                if(a[0]^a[b]^a[n-1]==0):
                    print('Yes')
                else:
                    print('No')
            else:
                print('No')
        else:
            print('No')
    else:
        print('No')

Submission status: Wrong Answer

Solution code:
import sys

def solve(n, a):  
    set_a = set(a)
    a_0 = a[0]
    for i in range(n):
        if i == 0:
            continue

        prev_a = a[i]
        next_a = a_0 ^ a[i]
        if next_a not in set_a:
            continue
        count = 0
        while count < n :
            next_next_a = prev_a ^ next_a
            if next_next_a not in set_a:
               break
            count += 1
            prev_a = next_a
            next_a = next_next_a
        
        if count == n:
            return 'Yes'

    return 'No'


if __name__ == '__main__':
    n = int(sys.stdin.readline().strip())
    a = map(int, sys.stdin.readline().strip().split(" "))
    print solve(n, a)

Submission status: Wrong Answer

Solution code:
n = input()
a = list(map(int, input().split()))

print("Yes")

Submission status: Wrong Answer

Solution code:
a = int(input())
b = list(map(int, input().split()))
c = {}

for i in range(a):
    if b[i] not in c:
        c.setdefault(b[i], 1)
    else:
        c[b[i]] += 1

if len(c) == 3:
    f = list(c.values())
    for i in range(1, len(f)):
        if f[i] != f[i-1]:
            print("No")
            exit()
    g = list(c.keys())
    if g[0] ^ g[1] == g[2]:
        print("Yes")
    else:
        print("No")
elif len(c) == 2:
    f = list(c.values())
    if f[0] != f[1]:
        print("No")
        exit()
    g = list(c.keys())
    if g[0] == 0 or g[1] == 0:
        print("Yes")
    else:
        print("No")
else:
    print("No")
    exit()

Submission status: Wrong Answer

Solution code:
import numpy as np

N = int(input())
a = input().split(" ")
A = [int(i) for i in a]

bit = [A[0],A[1]]
flag = 0
num = 2
while True:
    k = 0
    flag2 = 0
    for i in range(len(bit)):
        if i == 0:
            b = bin(bit[0]^bit[len(bit)-1])[2::]
            if b == np.binary_repr(A[num]):
                k = i
                flag2 = 1
                break
        else:
            b = bin(bit[i-1]^bit[i])[2::]
            if b == np.binary_repr(A[num]):
                k = i
                flag2 = 1
                break
    if flag2 == 1:
        bit.insert(k, A[num])
        num += 1
        if num == N:
            flag = 1
            break
    if flag2 == 0:
        print("No")
        break
if flag == 1:
    print("Yes")

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import random
input = sys.stdin.readline

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

def main():

	N = int(input())
	#S = list(input())
	#L,R = map(int, input().split(" "))
	#L = list(map(int, input().split(" ")))

	A = input().split()
	A = [int(s) for s in A]
	A.sort()

	a_max = A[-1]
	#A[-1] = -1

	a = 2
	i = 1
	ll = [1]
	while a < 100000+1:
		a = 2 ** i
		ll.append(a)
		i += 1
		if a_max == a:
			print("Yes")
			sys.exit()

	tmp = 0
	flg = False
	for i in ll:
		tmp += i
		if a_max == tmp:
			flg = True
			break
	
	if flg == False:
			print("Yes")
			sys.exit()

	bk = -1
	next1 = -1
	for i, val in enumerate(A):
		#A[i] = -1
		if i == 0:
			tmp = a_max - val
			if tmp in A:
				bk = val
				next1 = tmp
				continue
			else:
				print("Yes")
				sys.exit()
		
		if i == N-1:
			if next1 == a_max and bk ^ A[0] == a_max:
				print("Yes")
				sys.exit()
		
		if next1 == a_max:
			tmp = next1 - bk
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				print("No")
				sys.exit()

		elif next1 == A[0]:
			tmp = a_max - next1
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				#print(next1)
				print("No")
				sys.exit()

		else:
			tmp = next1 + bk
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				print("No")
				sys.exit()
		
	print("No")


if __name__ == "__main__":
	main()

Submission status: Wrong Answer

Solution code:
a=int(input())
b= list(map(int, input().split()))
b.append(b[0])
b.append(b[1])
d=0

for i in range(a):
    c= int(b[i]^b[i+2])
    if c!=b[i+1]:
        d += 1
        break

if d==0:
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
N = int(input())
n = list(map(int, input().split()))
t = 0
m = list(range(1, N+1))
if set(n) != set(m):
    print("No")
else:
    for i in range(N):
      if i == 0:
        if n[N-1] ^ n[1] == n[0]:
            t += 1
      elif i == N-1:
        if n[N-2] ^ n[0] == n[N-1]:
          t += 1
      else:
        if n[i-1] ^ n[i+1] == n[i]:
          t += 1
    if t == N:
      print("Yes")
    else:
      print("No")

Submission status: Wrong Answer

Solution code:
# -*- coding: utf-8 -*-
import sys
import math
from bisect import bisect_left
from bisect import bisect_right
import collections
import copy
import heapq
from collections import defaultdict
from heapq import heappop, heappush
import random
input = sys.stdin.readline

##### リストの 二分木検索 #####
# bisect_left(lists, 3)
# bisect_right(lists, 3)

##### プライオリティキュー #####
# heapq.heapify(a) #リストaのheap化
# heapq.heappush(a,x) #heap化されたリストaに要素xを追加
# heapq.heappop(a) #heap化されたリストaから最小値を削除＆その最小値を出力

# heapq.heappush(a, -x) #最大値を取り出す時は、pushする時にマイナスにして入れよう
# heapq.heappop(a) * (-1) #取り出す時は、-1を掛けて取り出すこと

##### タプルリストのソート #####
# sorted(ans) #(a, b) -> 1st : aの昇順, 2nd : bの昇順
# sorted(SP, key=lambda x:(x[0],-x[1])) #(a, b) -> 1st : aの昇順, 2nd : bの降順
# sorted(SP, key=lambda x:(-x[0],x[1])) #(a, b) -> 1st : aの降順, 2nd : bの昇順
# sorted(SP, key=lambda x:(-x[0],-x[1])) #(a, b) -> 1st : aの降順, 2nd : bの降順

# sorted(SP, key=lambda x:(x[1])) #(a, b) -> 1st : bの昇順
# sorted(SP, key=lambda x:(-x[1])) #(a, b) -> 1st : bの降順

def main():

	N = int(input())
	#S = list(input())
	#L,R = map(int, input().split(" "))
	#L = list(map(int, input().split(" ")))

	A = input().split()
	A = [int(s) for s in A]
	A.sort()

	a_max = A[-1]
	#A[-1] = -1

	a = 2
	i = 1
	ll = [1]
	while a < 100000+1:
		a = 2 ** i
		ll.append(a)
		i += 1
		if a_max == a:
			print("Yes")
			sys.exit()

	tmp = 0
	flg = False
	for i in ll:
		tmp += i
		if a_max == tmp:
			flg = True
			break
	
	if flg == False:
			print("Yes")
			sys.exit()

	bk = -1
	next1 = -1
	for i, val in enumerate(A):
		#A[i] = -1
		if i == 0:
			tmp = a_max - val
			if tmp in A:
				bk = val
				next1 = tmp
				continue
			else:
				print("Yes")
				sys.exit()
		
		if i == N-1:
			if next1 == a_max and bk ^ A[0] == a_max:
				print("No")
				sys.exit()
		
		if next1 == a_max:
			tmp = next1 - bk
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				print("Yes")
				sys.exit()

		elif next1 == A[0]:
			tmp = a_max - next1
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				#print(next1)
				print("No")
				sys.exit()

		else:
			tmp = next1 + bk
			if tmp in A:
				bk = next1
				next1 = tmp
				continue
			else:
				print("No")
				sys.exit()
		
	print("No")


if __name__ == "__main__":
	main()

Submission status: Wrong Answer

Solution code:

import numpy as np 

def make_bit(A):
    arr = np.zeros(31)
    i = 30
    while A:
        arr[i] = A % 2
        A = A // 2
        i -= 1

    return arr

N = int(input())

a_lst = list(map(int, input().split()))



if N == 3:
    a_lst.sort(reverse=True)

    a1 = make_bit(a_lst[0])
    a2 = make_bit(a_lst[1])
    a3 = make_bit(a_lst[2])
    bits_arr = np.array([a1, a2, a3])

    judge = True
    for i in range(31):
        if bits_arr[0][i] == bits_arr[1][i] and bits_arr[2][i] == 1:
            judge = False
            break

        elif bits_arr[0][i] != bits_arr[1][i] and bits_arr[2][i] == 0:
            judge = False
            break
            
    if judge:
        print("Yes")
    else:
        print("No")


else:
    print("No")

Submission status: Wrong Answer

Solution code:
import sys

N = int(input())
a = list(map(int, input().split()))

l = a[0]
m = -1
n = -1
o = -1

for i in range(1,N):
    if a[i] != l and m == -1:
        m = a[i]
    elif a[i] != l and a[i] != m and n == -1 :
        n = a[i]
    elif a[i] != l and a[i] != m and a[i] != n and o == -1:
        print("No")
        sys.exit()

if (m == -1 and n == -1) and l^l == l and a.count(l) == N:
    print("Yes")
elif n == -1 and a.count(l)+a.count(m) == N and ( (a.count(l)*2==a.count(m) and l^m==m) or (a.count(l)==a.count(m)*2 and l^m == l)):
    print("Yes")
elif a.count(l)+a.count(m)+a.count(n) == N and l^m==n:
    print("Yes")
else:
    print("No")

Submission status: Wrong Answer

Solution code:
n = int(input())
figure = list(map(int, input().split()))
figure1 = []
p=0
for j in range(1,n):
  figure1.append(figure[0]^figure[j])
for k in range(n-1):
  if figure1[0]==figure[k]:
    print("Yes")
    #number1 = figure[0]
    #number2 = figure[1]
    #number3 = figure[k]
    p+=1
    break
if p==0:
  print("No")