Problem:


Score : 200 points

Problem Statement

We will say that two integer sequences of length N, x_1, x_2, ..., x_N and y_1, y_2, ..., y_N, are similar when |x_i - y_i| \leq 1 holds for all i (1 \leq i \leq N).

In particular, any integer sequence is similar to itself.

You are given an integer N and an integer sequence of length N, A_1, A_2, ..., A_N.

How many integer sequences b_1, b_2, ..., b_N are there such that b_1, b_2, ..., b_N is similar to A and the product of all elements, b_1 b_2 ... b_N, is even?

Constraints

1 \leq N \leq 10

1 \leq A_i \leq 100

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the number of integer sequences that satisfy the condition.

Sample Input 1

2
2 3

Sample Output 1

7

There are seven integer sequences that satisfy the condition:

1, 2

1, 4

2, 2

2, 3

2, 4

3, 2

3, 4

Sample Input 2

3
3 3 3

Sample Output 2

26

Sample Input 3

1
100

Sample Output 3

1

Sample Input 4

10
90 52 56 71 44 8 13 30 57 84

Sample Output 4

58921



Sample input:
2
2 3


Sample output:
7

Submission status: Accepted

Solution code:
import sys
import math
def I():return int(sys.stdin.readline().replace("\n",""))
def I2():return map(int,sys.stdin.readline().replace("\n","").split())
def S():return str(sys.stdin.readline().replace("\n",""))
def L():return list(sys.stdin.readline().replace("\n",""))
def Intl():return [int(k) for k in sys.stdin.readline().replace("\n","").split()]
def Lx(k):return list(map(lambda x:int(x)*-k,sys.stdin.readline().replace("\n","").split()))

if __name__ == "__main__":
    n = I()
    a = Intl()
    all = 1
    cnt = 1
    for i in range(n):
        all *= 3
        if a[i]%2 == 0:cnt *= 2
    print(all-cnt)

Submission status: Accepted

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

B = [a-1 for a in A]
C = [a+1 for a in A]


def Base_10_to_n(X, n):
    if (int(X/n)):
        return Base_10_to_n(int(X/n), n)+str(X%n)
    return str(X%n)

count = 0
for i in range(3**N):
    i3 = Base_10_to_n(i, 3).zfill(N)
    Pick = []
    for j in range(N):
        if i3[j] == '0':
            Pick.append(A[j])
        elif i3[j] == '1':
            Pick.append(B[j])
        elif i3[j] == '2':
            Pick.append(C[j])
#    print(i3, ':', Pick)
    if any([p%2 == 0 for p in Pick]):
        count += 1
print(count)

Submission status: Accepted

Solution code:
n=int(input())
a=list(map(int,input().split()))
cnt=1
for i in a:
    if i%2==0:
        cnt*=2
    
print(3**(len(a))-cnt)

Submission status: Accepted

Solution code:
n=int(input())
a=list(map(int,input().split()))
ans=3**n
tmp=1
for i in range(n):
    if a[i]%2==0:
        tmp*=2
ans-=tmp
print(ans)

Submission status: Accepted

Solution code:
N = int(input())
A = list(map(int, input().split()))
x = 1
y = 1
for a in A:
    x *= 3
    if a%2 == 0:
        y *= 2
print(x-y)

Submission status: Accepted

Solution code:
import sys

def dfs(x,m,nex):
    if m==n:
        #print(x)
        if x%2==0:
            global ans
            ans += 1
    else:
        ai = a[nex]
        m += 1
        nex += 1
        for i in (ai,ai-1,ai+1):
            dfs(x*i,m,nex)

sys.setrecursionlimit(10**6)

n = int(input())
a = list(map(int, input().split()))
ans = 0
for i in (a[0],a[0]-1,a[0]+1):
    dfs(i,1,1)
print(ans)

Submission status: Accepted

Solution code:
#-*-coding:utf-8-*-

def main():
    base_array=[]
    similar_array=[]
    odd_num_array=[]
    counter=0
    ans=1
    array_length = int(input())
    base_array=list(map(int,input().split()))
    for i in base_array:
        data = [i-1,i,i+1]
        similar_array.append(data)
    for array in similar_array:
        for i in array:
            if i % 2 !=0:
                counter+=1
        odd_num_array.append(counter)
        counter=0
    for number in odd_num_array:
        ans*=number
    print(3**array_length-ans)

if __name__=="__main__":
    main()

Submission status: Accepted

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

ans = 3**N
cnt = 1
for i in range(N):
    if A[i]%2==0:
        cnt *= 2

print(ans-cnt)

Submission status: Accepted

Solution code:
n=int(input())
a=list(map(int,input().split()))
ans=3**n
cnt=0
for i in range(n):
  if a[i]%2==0:
    cnt=cnt+1
print(ans-2**cnt)

Submission status: Accepted

Solution code:
import sys
input = sys.stdin.readline
# sys.setrecursionlimit(100000)


def main():
    N = int(input().strip())
    A = [int(i) for i in input().strip().split()]
    odds = 1
    for a in A:
        if a % 2 == 0:
            odds *= 2
        else:
            continue

    return 3**N - odds


if __name__ == "__main__":
    print(main())

Submission status: Accepted

Solution code:
n=int(input())
li=list(map(int,input().split()))
ev=0
for i in li:
  if i%2==0:
    ev+=1
print(3**n-2**ev)

Submission status: Accepted

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

tmp = 1
for aa in A:
	if aa%2 == 0:
		tmp *= 2
	else:
		tmp *= 1

#if len(A) == 1:

print(3**N-tmp)

Submission status: Accepted

Solution code:
n = int(input())
li = list(map(int,input().split()))
ans = 3**n
x = 1

for i in li:
    if i%2 == 0:
        x *= 2

print(ans-x)

Submission status: Accepted

Solution code:
n=int(input())
a=list(map(int,input().split()))
total=3**len(a)
kisu=1
for x in a:
    b=1
    if x%2==0:
        b=2
    kisu*=b
print(total-kisu)

Submission status: Accepted

Solution code:
n = int(input())
a = list(map(int, input().split()))
ans = 3 ** n
x = 1
for i in range(n):
    if a[i] % 2 == 0:
        x *= 2
print(ans - x)

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-

#############
# Libraries #
#############

import sys
input = sys.stdin.readline

import math
#from math import gcd
import bisect
from collections import defaultdict
from collections import deque
from collections import Counter
from functools import lru_cache


#############
# Constants #
#############

MOD = 10**9+7
INF = float('inf')

#############
# Functions #
#############

######INPUT######
def I(): return int(input().strip())
def S(): return input().strip()
def IL(): return list(map(int,input().split()))
def SL(): return list(map(str,input().split()))
def ILs(n): return list(int(input()) for _ in range(n))
def SLs(n): return list(input().strip() for _ in range(n))
def ILL(n): return [list(map(int, input().split())) for _ in range(n)]
def SLL(n): return [list(map(str, input().split())) for _ in range(n)]

######OUTPUT######
def P(arg): print(arg); return
def Y(): print("Yes"); return
def N(): print("No"); return
def E(): exit()
def PE(arg): print(arg); exit()
def YE(): print("Yes"); exit()
def NE(): print("No"); exit()

#####Shorten#####
def DD(arg): return defaultdict(arg)


#####Inverse#####
def inv(n): return pow(n, MOD-2, MOD)

######Combination######
kaijo_memo = []
def kaijo(n):
  if(len(kaijo_memo) > n):
    return kaijo_memo[n]
  if(len(kaijo_memo) == 0):
    kaijo_memo.append(1)
  while(len(kaijo_memo) <= n):
    kaijo_memo.append(kaijo_memo[-1] * len(kaijo_memo) % MOD)
  return kaijo_memo[n]

gyaku_kaijo_memo = []
def gyaku_kaijo(n):
  if(len(gyaku_kaijo_memo) > n):
    return gyaku_kaijo_memo[n]
  if(len(gyaku_kaijo_memo) == 0):
    gyaku_kaijo_memo.append(1)
  while(len(gyaku_kaijo_memo) <= n):
    gyaku_kaijo_memo.append(gyaku_kaijo_memo[-1] * pow(len(gyaku_kaijo_memo),MOD-2,MOD) % MOD)
  return gyaku_kaijo_memo[n]

def nCr(n,r):
  if(n == r):
    return 1
  if(n < r or r < 0):
    return 0
  ret = 1
  ret = ret * kaijo(n) % MOD
  ret = ret * gyaku_kaijo(r) % MOD
  ret = ret * gyaku_kaijo(n-r) % MOD
  return ret

######Factorization######
def factorization(n):
  arr = []
  temp = n
  for i in range(2, int(-(-n**0.5//1))+1):
    if temp%i==0:
      cnt=0
      while temp%i==0: 
        cnt+=1 
        temp //= i
      arr.append([i, cnt])
  if temp!=1:
    arr.append([temp, 1])
  if arr==[]:
    arr.append([n, 1])
  return arr

#####MakeDivisors######
def make_divisors(n):
  divisors = []
  for i in range(1, int(n**0.5)+1):
    if n % i == 0:
      divisors.append(i)
      if i != n // i: 
        divisors.append(n//i)
  return divisors

#####MakePrimes######
def make_primes(N):
  max = int(math.sqrt(N))
  seachList = [i for i in range(2,N+1)]
  primeNum = []
  while seachList[0] <= max:
    primeNum.append(seachList[0])
    tmp = seachList[0]
    seachList = [i for i in seachList if i % tmp != 0]
  primeNum.extend(seachList)
  return primeNum

#####GCD#####
def gcd(a, b):
    while b:
        a, b = b, a % b
    return a

#####LCM#####
def lcm(a, b):
    return a * b // gcd (a, b)

#####BitCount#####
def count_bit(n):
  count = 0
  while n:
    n &= n -1
    count += 1
  return count

#####ChangeBase#####
def base_10_to_n(X, n):
  if X//n:
    return base_10_to_n(X//n, n)+[X%n]
  return [X%n]

def base_n_to_10(X, n):
  return sum(int(str(X)[-i-1])*n**i for i in range(len(str(X))))

#####IntLog#####
def int_log(n, a):
  count = 0
  while n>=a:
    n //= a
    count += 1
  return count

#############
# Main Code #
#############

N = I()
A = IL()

ans = 1
for a in A:
  if a%2==0:
    ans *= 2
print(pow(3,N)-ans)

Submission status: Accepted

Solution code:
#! /usr/bin/env python3

import sys
import numpy as np
int1 = lambda x: int(x) - 1
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(500000)

N, *A = map(int, read().split())

res = pow(3, N)
odd = 1

for a in A:
    if a % 2 == 0:
        odd *= 2

print(res - odd)

Submission status: Accepted

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

cnt = 1
for i in A:
    if not i%2:
        cnt *= 2

print(3**N-cnt)

Submission status: Accepted

Solution code:
def main():
    N = int(input())
    A = list(map(int,(input().split())))

    al = 3 ** N
    minus = 1
    for a in A:
        if a % 2 ==0:
            minus *= 2
    print(al-minus)


if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
n,*a=map(int,open(0).read().split())
b=1
for i in a:
    b*=2-i%2
print(pow(3,len(a))-b)

Submission status: Accepted

Solution code:
N=int(input())
*A,=map(int,input().split())

count=0
for a in A:
    count+=1*(a%2==0)

ans=3**N-2**count
print(ans)

Submission status: Accepted

Solution code:
n=int(input())
a=list(map(int,input().split()))
is_even=[]
c=1
for x in a:
    c*=2 if x%2==0 else 1
print(3**n-c)

Submission status: Accepted

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

al = 1
bad = 1

for a in A:
    al *= 3
    if a % 2 == 0:
        bad *= 2

print(al - bad)

Submission status: Accepted

Solution code:
n=int(input())
a=list(map(int,input().split()))
ptn=1
for i in range(len(a)):
    if a[i]%2==0:
        ptn*=2
print(3**n-ptn)

Submission status: Accepted

Solution code:
N=int(input())
A=list(map(int,input().split()))
count=0
for a in A:
    if a%2==0:
        count+=1
print(3**N-(2**count))

Submission status: Accepted

Solution code:
n=int(input())
a=list(map(int,input().split()))
x=0
for i in a:
  if i%2==0:
   	x+=1
print(3**n-2**x)

Submission status: Accepted

Solution code:
N = int(input())
A = list(map(int, input().split()))
res = 3**N
odd_cnt = 0
for i in A:
    if i%2 == 0:
        odd_cnt += 1
print(res - 2**odd_cnt)

Submission status: Accepted

Solution code:
N = int(input())
An = list(map(int, input().split()))
even = sum([1 for a in An if a % 2 == 0])
print(3**N - 2**even)

Submission status: Accepted

Solution code:
import sys

input = sys.stdin.readline


def main():
    N = int(input())
    A = list(map(int, input().split()))

    odd_product = 1
    for a in A:
        if a % 2 == 0:
            count = 2
        else:
            count = 1
        odd_product *= count

    ans = 3 ** N - odd_product
    print(ans)


if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
n = int(input())
a = [int(i) for i in input().split()]
t = 1
for i in a:
    if i % 2 == 0: t *= 2
print(3 ** n - t)

Submission status: Accepted

Solution code:
N = int(input())
A = list(map(int, input().split()))
ans = 0
for i in range(3 ** N):
    mul = 1
    for j in range(N):
        mod = i // (3 ** j) % 3
        if mod == 0:
            mul *= A[j] - 1
        elif mod == 1:
            mul *= A[j]
        else:
            mul *= A[j] + 1
    if mul % 2 == 0:
        ans += 1
print(ans)

Submission status: Accepted

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

odd_pair = 1
all_pair = 1
for a in A:
    all_pair*=3
    if a%2==0:
        odd_pair*=2
print(all_pair-odd_pair)

Submission status: Accepted

Solution code:
import sys, re
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import accumulate, permutations, combinations, product, groupby
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from bisect import bisect, bisect_left
from fractions import gcd
from heapq import heappush, heappop
from functools import reduce
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))
def ZIP(n): return zip(*(MAP() for _ in range(n)))
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7

N = INT()
A = LIST()

ans = 3**N
tmp = 1
for a in A:
	if a % 2 == 0:
		tmp *= 2
print(ans-tmp)

Submission status: Accepted

Solution code:
N = int(input())
A = list(map(int, input().split()))
print(3 ** N - 2 ** sum([1 for i in A if not i % 2]))

Submission status: Accepted

Solution code:
n = int(input())
a = list(map(int,input().split()))
k = 1
for d in a:
  if d%2 == 0:
    k *= 2
print(3**n-k)

Submission status: Accepted

Solution code:
n=int(input())
A=list(map(int,input().split()))
#全パターンは3**n
x=1
for i in range(n):
    if A[i]%2==0:
        x*=2
    else:
        x*=1
print(3**n-x)

Submission status: Accepted

Solution code:
n = int(input())
cnt = 0
for i in map(int,input().split()):
    if i % 2 == 0:
        cnt += 1
print(3**n-2**cnt)

Submission status: Accepted

Solution code:
N=int(input())
A=list(map(int,input().split()))
e=[a%2 for a in A].count(0)
print(3**N-2**e)

Submission status: Accepted

Solution code:
N=int(input())
A=list(map(int,input().split()))
odd=[a%2 for a in A].count(0)
print(3**N-2**odd)

Submission status: Accepted

Solution code:
n = int(input())
A = list(map(int, input().split()))
o = 1
for a in A:
	if a % 2 == 0:
		o *= 2
print(3**n - o)

Submission status: Accepted

Solution code:
# https://atcoder.jp/contests/code-festival-2017-qualc/tasks/code_festival_2017_qualc_b


n = int(input())
nums = [int(i) for i in input().split()]

odd = 0
even = 0
for num in nums:
    if num % 2 == 0:
        even += 1
    else:
        odd += 1

ans = 3 ** n - 2 ** even
print(ans)

Submission status: Accepted

Solution code:
from operator import mul
from functools import reduce

n = int(input())
a = list(map(int, input().split()))
k = [2 if i % 2 == 0 else 1 for i in a]

count = 3**len(a) - reduce(mul, k)

print(count)

Submission status: Accepted

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

dis_cnt = 1
for a in A:
  if(a%2 == 0):
    dis_cnt = dis_cnt*2
  else:
    dis_cnt = dis_cnt*1
    
ans = 3**N - dis_cnt 
print(ans)

Submission status: Accepted

Solution code:
import sys

N = int(input())
A = list(map(int, sys.stdin.readline().rsplit()))

e = 0
for a in A:
    if a % 2 == 0:
        e += 1 

print(3 ** N - 2 ** e)

Submission status: Accepted

Solution code:
n = int(input())
a = [int(x) for x in input().split()]
odds = len([1 for x in a if x % 2 == 0])

ans = 3**n - 2**odds
print(ans)

Submission status: Accepted

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


ans = 0


def solve(i, l):
    global ans

    if len(l) == n:
        total = 1
        for k in l:
            total *= k
        if total % 2 == 0:
            ans += 1
        return

    for j in range(-1, 2):
        l.append(a[i] + j)
        solve(i + 1, l)
        l.pop()


solve(0, [])
print(ans)

Submission status: Accepted

Solution code:
N = int(input())
A = list(map(int,input().split()))
ans = 3**N
odd = 1
for i in range(N):
  if A[i] % 2 == 0:
    odd *= 2
print(ans - odd)

Submission status: Accepted

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

e = 0
for i in range(n):
    if a[i] % 2 == 0:
        e += 1
    
print(3**n - 2**e)

Submission status: Accepted

Solution code:
def main():
    N = int(input())
    A = list(map(int, input().split()))

    res = 1
    for a in A:
        if a % 2 == 0:
            res *= 2

    print(3**N - res)


if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
N = int(input())
src = list(map(int,input().split()))
 
whole = 3**N
even = 0
for i in range(N):
    if src[i]%2 == 0: even += 1
 
print(whole - 2**even)

Submission status: Accepted

Solution code:

N = int(input())

A = list(map(int, input().split()))
c = 1
for a in A:
    if a % 2 == 0:
        c *= 2
print(3**N-c)

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-
import sys

import fractions
import copy
import bisect
import math
import numpy as np
import itertools
from itertools import combinations_with_replacement
#import math#数学的計算はこれでいける。普通に0.5乗しても計算可能
#w=input()
from operator import itemgetter
from sys import stdin
#input = sys.stdin.readline#こっちの方が入力が早いが使える時に使っていこう
from operator import mul
from functools import reduce
from collections import Counter
#from collections import deque

#input = stdin.readline


j=0
k=0
n=3
r=1
a=[0]
#n=int(input())
#r=int(input())
#print(M)
#A=int(input())
#B=int(input())
#print(N)
"1行1つの整数を入力を取得し、整数と取得する"


#number_list=list(map(int, input().split(" ")))#数字の時
#print(number_list)
"12 21 332 とか入力する時に使う"
"1行に複数の整数の入力を取得し、整数として扱う"

'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
#メモ
for i in number_list:#こっちの方がrage使うより早いらしい
    print(number_list[i-1])#

'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''

'''
x=[]
y=[]
for i in range(N):
    x1, y1=[int(i) for i in input().split()]
    x.append(x1)
    y.append(y1)
print(x)
print(y)
"複数行に2数値を入力する形式 x座標とy座標を入力するイメージ"
'''

'''
mixlist=[]
for i in range(N):
    a,b=input().split()
    mixlist.append((int(a),b))
print(mixlist)
"複数行にintとstrを複合して入力するやつ，今回はリスト一つで処理している"
'''

'''
#array=[input().split()for i in range(N)]
#print(type(array[0][0]))
#print(array)
"正方行列にstr型の値を入力"
'''

#brray=[list(map(int, input().split(" ")))for i in range(N)]
#print(type(brray[0][0]))
#print(brray)
'''
入力
1 2
4 5
7 8
出力結果
[[1, 2], [4, 5], [7, 8]]
'''
"列数に関して自由度の高いint型の値を入力するタイプの行列"
#以下に別解を記載
#N, M = [int(i) for i in input().split()]
'''
table = [[int(i) for i in input().split()] for m in range(m)]

print(type(N))
print(N)
print(type(M))
print(M)

print(type(table))
print(table)
'''




#s=input()
#a=[int(i) for i in s]
#print(a[0])
#print([a])
#単数値．桁ごとに分割したい．入力と出力は以下の通り
#イメージとして1桁ごとにリストに値を入れているかんじ
#intを取ると文字列分解に使える
'''
入力
1234
出力
1
[[1, 2, 3, 4]]
'''




'''
word_list= input().split(" ")
print(word_list[0])
"連続文字列の入力"
"qw er ty とかの入力に使う"
"入力すると空白で区切ったところでlistの番号が与えられる"
'''



'''
A, B, C=stdin.readline().rstrip().split()#str style 何個でもいけることが多い
print(A)
"リストではなく独立したstr型を入れるなら以下のやり方でOK"
'''


#a= stdin.readline().rstrip()
#print(a.upper())
"aという変数に入っているものを大文字にして出力"


#a,b=map(int, input().split())  #int style 複数数値入力　「A B」みたいなスペース空いた入力のとき

#なんかうまく入力されるけど
#a=[[int(i) for i in 1.strip()]for 1 in sys.stdin]
#a = [[int(c) for c in l.strip()] for l in sys.stdin]]
#print(a)
#複数行の数値を入力して正方行列を作成


##############################################################################################
##############################################################################################
#under this line explains example calculation



'''
コンビネーションの組み合わせの中身を出力する形式
for i in itertools.combinations(brray, 2)
combinationsをpermutationsにすれば順列になる
今回なら(abc133B)
入力
1 2
5 5
-2 8
出力
[[1, 2], [5, 5], [-2, 8]]
もちろん一次元リストでも使えるし
何よりiもリストのように使えるので便利
'''

#nCr combination
'''
def cmb(n,r):
    #When n < r , this function isn't valid
    r= min(n-r,r)
    #print(n,r)
    if r == 0: return 1
    over = reduce(mul, range(n, n-r, -1))
    #flochart mul(n,n-1)=x
    #next     mul(x,n-2)........(n-r+1,n-r)
    #mul read a,b and returns a*b

    under = reduce(mul, range(1, r+1))
    #print(over, under)
    #reduce is applied mul(1,2)=2
    #next   mul(2,3)=6
    #next   mul(6,4)=4.........last(r!,r+1)=r+1!

    return over // under
    #// is integer divide
    #calc example 5C2
    #over=5*4*3
    #under=3*2*1
a = cmb(n, r)
#print(a)
'''

'''
import itertools
from itertools import combinations_with_replacement

combinationについて
以下の違いを意識しよう
combinations() p, r 長さrのタプル列、ソートされた順で重複なし

combinations_with_replacement() p, r 長さrのタプル列、ソートされた順で重複あり


使用例 　　　　　　　　　　　                  出力　　
combinations('ABCD', 2)　　　　　　　　　　　　AB AC AD BC BD CD

combinations_with_replacement('ABCD', 2)　　 AA AB AC AD BB BC BD CC CD DD
'''



'''
#集計
#example
#a=[2,2,2,3,4,3,1,2,1,3,1,2,1,2,2,1,2,1]
#a=Counter(a)
for i in a.most_common(n):print(i)
#most_common()メソッドは、出現回数が多い要素順にCounterオブジェクトを並び替えます。
#引数にint型の数字nを設定した場合は、出現回数が高い上位ｎ個の要素を返します。
#何も設定しなければ、コンテナ型にあるすべての要素を出現回数の順番に並び替えたタプル型オブジェクトを返します。
#out put
#(2, 8)
 #(1, 6)
#(3, 3)
#(4, 1)
'''


#二部探索(binary search)
#A = [1, 2, 3, 3, 3, 4, 4, 6, 6, 6, 6]
#print(A)
#index = bisect.bisect_left(A, 5) # 7 最も左(前)の挿入箇所が返ってきている
#A.insert(index, 5)
#print(index)
#print(A)
'''
bisect.bisect_left(a, x, lo=0, h=len(a))
引数
a: ソート済みリスト
x: 挿入したい値
lo: 探索範囲の下限
hi: 探索範囲の上限
(lo, hiはスライスと同様の指定方法)
bisect_leftはソートされたリストaに対して順序を保ったままxを挿入できる箇所を探索します。leftが示す通り、aにすでにxが存在している場合は、挿入箇所は既存のxよりも左側になります。また、lo, hiを指定することで探索範囲を絞り込むことも可能です。デフォルトはaの全体が探索対象です。
'''

'''
素数の判定
'''


def is_prime(n):
    if n == 1: return False

    for k in range(2, int(np.sqrt(n)) + 1):
        #sqrt(n)+1以上は考えて約数はないので却下
        if n % k == 0:
            return False
            #割り切れたらFalse
    return True


'''
npのmaxとmaximumの違い
xs = np.array([1, -2, 3])
np.max(xs, 0)
この出力は3となります．[1, -2, 3]と0の4つの数字のうち，最も大きい値を出力します．

一方で，[max(1, 0), max(-2, 0), max(3, 0)]を出力したい時があります．
その時は，numpyのmaximum関数を用います．

xs = np.array([1, -2, 3])
np.maximum(xs, 0)  # [1, 0, 3]

'''

def factorial(x, mod=10**9+7):
    # x!
    # 階乗
    # ex) factorial(5) = 120
    tmp = 1
    for i in range(1, x+1):
        tmp = (tmp * i) % mod
    return tmp
    '''
    Basically, when you do a**d % n, you actually have to calculate a**d, which could be quite large.
    But there are ways of computing a**d % n without having to compute a**d itself, and that is what pow does.
    '''


########################################################################
########################################################################
#b2=a[:]  #1次元のときはコピーはこれで良い
#print(b2)

#a= [[0]*3 for i in range(5)] #2次元配列はこう準備、[[0]*3]*5だとだめ
#b3=copy.deepcopy(a)  #2次元配列はこうコピーする
#print(b3)

'''
ascii_letters = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
ascii_lowercase = 'abcdefghijklmnopqrstuvwxyz'
ascii_uppercase = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
digits = '0123456789'
hexdigits = '0123456789abcdefABCDEF'
letters = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuv...\xaf\xb0...
lowercase = 'abcdefghijklmnopqrstuvwxyz'
octdigits = '01234567'
printable = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTU...
punctuation = '!"#$%&\'()*+,-./:;<=>?@[\\]^_`{|}~'
uppercase = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
whitespace = '\t\n\x0b\x0c\r '
'''
'''
引数endに任意の文字列を指定すると、その文字列が改行の代わりに最後に挿入される。例えば連続するprint()の出力を改行なしで単純に連結したい場合は空文字列''を指定すればよい。

print('abc', end='---')
print('xyz')
# abc---xyz

print('abc', end='')
print('xyz')
# abcxyz
source: print_basic.py
なお、デフォルトで改行されるのは引数endのデフォルト値が改行文字'\n'に設定されているから。
'''

def main():
    n=3
    steps=0
    state=0
    w=1
    j=0
    k=0
    #lenを変数に置くと事故るから注意
    n=int(input())
    #a,b=input().split(" ")#a,bがともに文字列のとき
    #a,b=map(int, input().split(" "))
    #mapでとってもstrで変換すればappend可能
    #s=input()
    #a=[i for i in s]#int(i)にすれば数字も行ける
    nl=list(map(int, input().split(" ")))#数字の時
    times=1
    for i in range(n):
        if nl[i]%2==1:
            times=times*1
        else:
            times=times*2
    print(3**n-times)

    #print(a)
    #brray=[list(map(int, input().split(" ")))for i in range(N)]
    #print(brray)


    for i in range(1,n+1,2):#これで1個飛ばしループになる
     pass
     #print(i)

    l = ['aaa', 'bbb', 'ccc']
    x=l.index('aaa')#ある要素のインデックス番号を取得
    #s = ''.join(l)
    #s = " ".join(map(str,a))#スペースを空ける時はこう
    #print(s)
    # aaabbbccc

    #if state in known_states and not flag: # state => 2 steps => 5
        #こうやってある要素がリストの中にあるか探すのか
    #print(f'steps: {steps}, state: {state}')
    #こう書いてあるとデバックしやすい

    #dp= [[0]*3 for i in range(5)]#列　行
    #dp用の0の入ったやつ
    #dp= [[0]*(w+1) for i in range(n+1)]#0からwまでのw+1回計算するから
    #print(dp)＃初期条件が入る分計算回数+1列分必要(この場合は判断すべきものの数)
    DP = np.zeros(w+1, dtype=int)#これでも一次元リストが作れる
    exdp=np.zeros((3,4)) # ３×４の２次元配列を生成。2次元ならこう
    #dtypeは指定しないとfloatになる
    #for i in range(n):#ちょっとした入力に便利
    #    a, b = map(int, input().split())
        #dp[i][0] += [a]
        #これだとintとlistをつなぐことになって不適
    #    dp[i] += [a]
    #    dp[i] += [b]
        #これはうまくいく
        #やり方はいろいろあるということ
    #print(dp)


    #brray=[list(map(int, input().split(" ")))for i in range(N)
    #print(brray)



    '''
    入力
    1234
    出力
    [1, 2, 3, 4]
    '''
    pin_l=["x" for i in range(10)]#内包表記に慣れろ
    #print(pin_l)

    ls = ["a", "b", "c", "d", "e"]


    #print(ls[2:5])
    #スライスでの取得

    #print(ls[:-3])
    #一番左端から右から3番目より左まで取得

    #print(ls[:4:2])
    #スライスで1個飛ばしで取得
    #ないときは左端スタート
    #始点のインデックス番号 : 終点のインデックス番号 : スキップする数+1
    #print(ls[::2])
    '''
    lsというリストの場合に、1つ飛びの値を取得したい場合には
    ls[::2]
    のようにします。こうすると、
    ["a", "c", "d"]と出力される
    '''

if __name__ == "__main__":
    main()

Submission status: Accepted

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

cnt=1
for num in a:
  if num%2==0:
    cnt*=2

print((3**n)-cnt)

Submission status: Accepted

Solution code:
N = int(input())
A = list(map(int,input().split()))
ALL = 3**N
even_count = 0
for a in A:
    if a%2 == 0:
        even_count += 1
print(ALL-2**even_count)

Submission status: Accepted

Solution code:
N = int(input())
A = list(map(int, input().split()))
kisuu = 1
for i in range(N):
  if A[i] % 2 == 0:
    kisuu *= 2
print(3**N-kisuu)

Submission status: Accepted

Solution code:
import sys
import heapq
import re
from itertools import permutations
from bisect import bisect_left, bisect_right
from collections import Counter, deque
from fractions import gcd
from math import factorial, sqrt, ceil
from functools import lru_cache, reduce
INF = 1 << 60
MOD = 1000000007
sys.setrecursionlimit(10 ** 7)

# UnionFind
class UnionFind():
    def __init__(self, n):
        self.n = n
        self.parents = [-1] * n

    def find(self, x):
        if self.parents[x] < 0:
            return x
        else:
            self.parents[x] = self.find(self.parents[x])
            return self.parents[x]

    def union(self, x, y):
        x = self.find(x)
        y = self.find(y)

        if x == y:
            return

        if self.parents[x] > self.parents[y]:
            x, y = y, x

        self.parents[x] += self.parents[y]
        self.parents[y] = x

    def size(self, x):
        return -self.parents[self.find(x)]

    def same(self, x, y):
        return self.find(x) == self.find(y)

    def members(self, x):
        root = self.find(x)
        return [i for i in range(self.n) if self.find(i) == root]

    def roots(self):
        return [i for i, x in enumerate(self.parents) if x < 0]

    def group_count(self):
        return len(self.roots())

    def all_group_members(self):
        return {r: self.members(r) for r in self.roots()}

    def __str__(self):
        return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots())

def is_prime(n):
    if n == 1:
        return False
    for i in range(2,int(n**0.5)+1):
        if n % i == 0:
            return False
    return True


# ワーシャルフロイド (任意の2頂点の対に対して最短経路を求める)
# 計算量n^3 (nは頂点の数)
def warshall_floyd(d, n):
    #d[i][j]: iからjへの最短距離
    for k in range(n):
        for i in range(n):
            for j in range(n):
                d[i][j] = min(d[i][j],d[i][k] + d[k][j])
    return d


# ダイクストラ
def dijkstra_heap(s, edge, n):
    #始点sから各頂点への最短距離
    d = [10**20] * n
    used = [True] * n #True:未確定
    d[s] = 0
    used[s] = False
    edgelist = []
    for a,b in edge[s]:
        heapq.heappush(edgelist,a*(10**6)+b)
    while len(edgelist):
        minedge = heapq.heappop(edgelist)
        #まだ使われてない頂点の中から最小の距離のものを探す
        if not used[minedge%(10**6)]:
            continue
        v = minedge%(10**6)
        d[v] = minedge//(10**6)
        used[v] = False
        for e in edge[v]:
            if used[e[1]]:
                heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1])
    return d

# 素因数分解
def factorization(n):
    arr = []
    temp = n
    for i in range(2, int(-(-n**0.5//1))+1):
        if temp%i==0:
            cnt=0
            while temp%i==0:
                cnt+=1
                temp //= i
            arr.append([i, cnt])

    if temp!=1:
        arr.append([temp, 1])

    if arr==[]:
        arr.append([n, 1])

    return arr

# 2数の最小公倍数
def lcm(x, y):
    return (x * y) // gcd(x, y)

# リストの要素の最小公倍数
def lcm_list(numbers):
    return reduce(lcm, numbers, 1)

# リストの要素の最大公約数
def gcd_list(numbers):
    return reduce(gcd, numbers)

# 素数判定



# limit以下の素数を列挙
def eratosthenes(limit):
    A = [i for i in range(2, limit+1)]
    P = []

    while True:
        prime = min(A)
        
        if prime > sqrt(limit):
            break
            
        P.append(prime)
            
        i = 0
        while i < len(A):
            if A[i] % prime == 0:
                A.pop(i)
                continue
            i += 1
            
    for a in A:
        P.append(a)
            
    return P

# 同じものを含む順列
def permutation_with_duplicates(L):

    if L == []:
        return [[]]

    else:
        ret = []

        # set（集合）型で重複を削除、ソート
        S = sorted(set(L))

        for i in S:

            data = L[:]
            data.remove(i)

            for j in permutation_with_duplicates(data):
                ret.append([i] + j)

        return ret


# ここから書き始める
def func(a, n, i, p):
    if i == n:
        if p % 2 == 0:
            return 1
        return 0
    return func(a, n, i + 1, p * (a[i] - 1)) + func(a, n, i + 1, p * a[i]) + func(a, n, i + 1, p * (a[i] + 1))



n = int(input())    
a = list(map(int, input().split()))        
ans = func(a, n, 0, 1)
print(ans)

Submission status: Accepted

Solution code:
n=int(input())
a=list(map(int,input().split()))
even=0
for i in range(n):
  if a[i]%2 == 0:
    even += 1
print(3**n - 2**even)

Submission status: Accepted

Solution code:
#!/usr/bin/env python
# coding: utf-8

# In[11]:


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


# In[12]:


even_list = [x for x in A if x%2 == 0]
ans = 3**len(A) - 2**(len(even_list))
print(ans)


# In[ ]:

Submission status: Accepted

Solution code:
n = int(input())
a = list(map(int, input().split()))
en = 1
for i in a:
    if i%2==0:
        en*=2
print((3**n)-en)

Submission status: Accepted

Solution code:
n = int(input())
a = list(map(int,input().split()))
ans = 3**n
cou = 1
for i in range(n):
    if a[i] % 2 == 0:
        cou *= 2
print(ans-cou)

Submission status: Accepted

Solution code:
from itertools import product

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

s = sum(any(map(lambda x: x % 2 == 0, (u + v for u, v in zip(A, list(x)))))
        for x in product((-1, 0, 1), repeat=N))
print(s)