Buggy python submission (verdict: Time Limit Exceeded):
def binary(x):
    '''40桁の2進数表記の作成'''

    k = 40
    s = ''
    for i in reversed(range(k)):
        s += str(x // 2 ** i)
        x %= 2 ** i

    return s

# 未知のxkを既知のkkに置き換える
# k += 1 として、x < k を満たす
# この時、xの未知の40桁のいくつかを、既知のkで置き換える
n, k = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]

ck = [0 for i in range(40)]
for i in a:
    ab = binary(i)
    for j in range(40):
        ck[j] += int(ab[j])


k1b = binary(k + 1)

for i in reversed(range(len(k1b))):
    if k1b[-i-1] == '1':
        lenk = i
        break

fx = []
for i in reversed(range(lenk + 1)):
    x = ''
    for j in reversed(range(40)):
        if j > i:
            x += str(int(k1b[-j-1]))
        elif j == i:
            if k1b[-j-1] == '1':
                x += str(0)
            else:
                break
        else:
            if ck[-j-1] * 2 > n:
                x += str(0)
            else:
                x += str(1)

    if len(x) == 40:
        fx_temp = 0
        for k in range(40):
            if x[-k-1] == '1':
                fx_temp += 2 ** k * (n - ck[-k-1])
            else:
                fx_temp += 2 ** k * ck[-k-1]
        fx.append(fx_temp)

print(max(fx))


Fixed submission:
# bit演算という方法がある、ビット演算子
# a & b 1,1で1
# a | b 1,1, 1,0, 0,1で1
# a ^ b xor
# ~a ビット反転
# << 左シフト
# >> 右シフト
# 未知のxkを既知のkkに置き換える
# k += 1 として、x < k を満たす
# この時、xの未知の40桁のいくつかを、既知のkで置き換える
# 注意として、xはkより大きくなってはいけない

n, k = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]

ck = [0 for i in range(40)]
for i in a:
    for j in range(40):
        if i>>j & 1:
            ck[j] += 1

fx = []
for i in range(40):

    if (k+1)>>i & 1 == 0:
        continue

    ans = 0
    for j in range(40):
        if j > i:
            if ((k+1)>>j) & 1:
                ans += (1<<j) * (n-ck[j])
            else:
                ans += (1<<j) * ck[j]
        elif j == i:
            ans += (1<<j) * ck[j]
        else:
            if ck[j] * 2 > n:
                ans += (1<<j) * ck[j]
            else:
                ans += (1<<j) * (n-ck[j])

    fx.append(ans)

print(max(fx))

Buggy python submission (verdict: Time Limit Exceeded):
def binary(x):
    '''40桁の2進数表記の作成'''

    k = 40
    s = ''
    for i in reversed(range(k)):
        s += str(x // 2 ** i)
        x %= 2 ** i

    return s

def decimal(s):
    '''40桁の2進数を10進数に変換'''

    k = 40
    x = 0
    for i in range(k):
        x += int(s[i]) * 2 ** (k - 1 - i) 

    return x

def xor(x, y):
    '''xor calcuration'''

    bx = binary(x)
    by = binary(y)
    s = ''

    for i in range(len(bx)):
        if abs(int(bx[i]) - int(by[i])):
            s += '1'
        else:
            s += '0'
    return decimal(s)

def f(x, a):
    '''f(x) calcuration'''

    fx = 0
    for i in a:
        fx += xor(x, i)

    return fx

# 未知のxkを既知のkkに置き換える
# k += 1 として、x < k を満たす
# この時、xの未知の40桁のいくつかを、既知のkで置き換える
n, k = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]

ck = [0 for i in range(40)]
for i in a:
    ab = binary(i)
    for j in range(40):
        ck[j] += int(ab[j])


k1b = binary(k + 1)

for i in reversed(range(len(k1b))):
    if k1b[-i-1] == '1':
        lenk = i
        break

fx = []
for i in reversed(range(lenk + 1)):
    x = ''
    for j in reversed(range(40)):
        if j > i:
            x += str(int(k1b[-j-1]))
        elif j == i:
            if k1b[-j-1] == '1':
                x += str(0)
            else:
                break
        else:
            if ck[-j-1] * 2 > n:
                x += str(0)
            else:
                x += str(1)

    if len(x) == 40:
        x = decimal(x)
        fx.append(f(x, a))

print(max(fx))


Fixed submission:
# bit演算という方法がある、ビット演算子
# a & b 1,1で1
# a | b 1,1, 1,0, 0,1で1
# a ^ b xor
# ~a ビット反転
# << 左シフト
# >> 右シフト
# 未知のxkを既知のkkに置き換える
# k += 1 として、x < k を満たす
# この時、xの未知の40桁のいくつかを、既知のkで置き換える
# 注意として、xはkより大きくなってはいけない

n, k = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]

ck = [0 for i in range(40)]
for i in a:
    for j in range(40):
        if i>>j & 1:
            ck[j] += 1

fx = []
for i in range(40):

    if (k+1)>>i & 1 == 0:
        continue

    ans = 0
    for j in range(40):
        if j > i:
            if ((k+1)>>j) & 1:
                ans += (1<<j) * (n-ck[j])
            else:
                ans += (1<<j) * ck[j]
        elif j == i:
            ans += (1<<j) * ck[j]
        else:
            if ck[j] * 2 > n:
                ans += (1<<j) * ck[j]
            else:
                ans += (1<<j) * (n-ck[j])

    fx.append(ans)

print(max(fx))

Buggy python submission (verdict: Time Limit Exceeded):
def binary(x):
    '''40桁の2進数表記の作成'''

    k = 40
    s = ''
    for i in reversed(range(k)):
        s += str(x // 2 ** i)
        x %= 2 ** i

    return s

def decimal(s):
    '''40桁の2進数を10進数に変換'''

    k = 40
    x = 0
    for i in range(k):
        x += int(s[i]) * 2 ** (k - 1 - i) 

    return x

def xor(x, y):
    '''xor calcuration'''

    bx = binary(x)
    by = binary(y)
    s = ''

    for i in range(len(bx)):
        if abs(int(bx[i]) - int(by[i])):
            s += '1'
        else:
            s += '0'
    return decimal(s)

def f(x, a):
    '''f(x) calcuration'''

    fx = 0
    for i in a:
        fx += xor(x, i)

    return fx


n, k = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]


fx = []
for x in range(k + 1):
    fx.append(f(x, a))

print(max(fx))


Fixed submission:
# bit演算という方法がある、ビット演算子
# a & b 1,1で1
# a | b 1,1, 1,0, 0,1で1
# a ^ b xor
# ~a ビット反転
# << 左シフト
# >> 右シフト
# 未知のxkを既知のkkに置き換える
# k += 1 として、x < k を満たす
# この時、xの未知の40桁のいくつかを、既知のkで置き換える
# 注意として、xはkより大きくなってはいけない

n, k = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]

ck = [0 for i in range(40)]
for i in a:
    for j in range(40):
        if i>>j & 1:
            ck[j] += 1

fx = []
for i in range(40):

    if (k+1)>>i & 1 == 0:
        continue

    ans = 0
    for j in range(40):
        if j > i:
            if ((k+1)>>j) & 1:
                ans += (1<<j) * (n-ck[j])
            else:
                ans += (1<<j) * ck[j]
        elif j == i:
            ans += (1<<j) * ck[j]
        else:
            if ck[j] * 2 > n:
                ans += (1<<j) * ck[j]
            else:
                ans += (1<<j) * (n-ck[j])

    fx.append(ans)

print(max(fx))