Buggy python submission (verdict: Time Limit Exceeded):
#とりあえずKを全探索
def 解():
    N,K=[int(_) for _ in input().split()]
    aA=[int(_) for _ in input().split()]
    aD = [0]*40
    for a in aA:
        a = "{:b}".format(int(a))[::-1]
        for i in range(len(a)):
            if a[i] == "1":
                aD[i] += 1
    dsA = set(aA)
    iL = 0
    for i in range(40):
        if 0 < aD[i]:
            iL = i
    iL +=1
    def calcTotal(iK,aD,N,iL):
        sK = "{:b}".format(int(iK))[::-1]
        iLK = len(sK)
        if iL > iLK:
            sK +="0"*(iL-iLK)
        else:
            iL = iLK
        iC = 0
        for i in range(iL):
            if sK[i] == "0":
                iC += (2 ** i) *  aD[i]
            else:
                iC += (2 ** i) * (N - aD[i])
        return iC
    print(max(calcTotal(i,aD,N,iL) for i in set(range(K+1)) - dsA | {0}))
解()


Fixed submission:
#場合分けがつらい
def 解():
    N,K=[int(_) for _ in input().split()]
    aD =[0]*40
    aA=[int(_) for _ in input().split()]
    iL = 0
    for i in range(40):
        m = 2**i
        aD[i]=sum((_ & m) for _ in aA)
        if aD[i]:
            iL = i
    iLK = len(bin(K))-3
    iML = max(iL,iLK)
    dp = [[0]*2 for _ in range(2)]
    #[x][0] Kに沿う
    #[x][0] Kに沿わない最大
    for i in range(iML,-1,-1):
        ib = i%2 #桁%2 dp
        inb = 0 if ib else 1 #次のdp
        i0 = aD[i] #反転しない
        i1 = (2**i)*N-i0 #反転する
        iM = max(i0,i1) #選べるとき
        bL = i < iLK #Kより長い
        if (K>>i)&1:
            dp[inb][1] = dp[ib][1] + i1
            dp[inb][0] = max(dp[ib][1] + i0 ,dp[ib][0]+(iM if bL else i0))
        else:
            dp[inb][1] = dp[ib][1] + i0
            dp[inb][0] = dp[ib][0] + (iM if bL else i0)
    print(max(dp[inb]))
解()

Buggy python submission (verdict: Time Limit Exceeded):
def muridesu():
    N,K=[int(_) for _ in input().split()]
    aA=sorted(int(_) for _ in input().split())
    dsA = set(aA)
    def f(i,aA):
        return sum(i^a for a in aA)
    def binlen(i):
        return len(bin(i)) - 2
    if 2**binlen(aA[-1]) <= K:
        print(max(f(k,aA) for k in (set(range(aA[-1],K+1)) | {0})))
    else:
        print(max(f(k,aA) for k in (set(range(1,K+1)) - dsA  | {0})))

muridesu()


Fixed submission:
#場合分けがつらい
def 解():
    N,K=[int(_) for _ in input().split()]
    aD =[0]*40
    aA=[int(_) for _ in input().split()]
    iL = 0
    for i in range(40):
        m = 2**i
        aD[i]=sum((_ & m) for _ in aA)
        if aD[i]:
            iL = i
    iLK = len(bin(K))-3
    iML = max(iL,iLK)
    dp = [[0]*2 for _ in range(2)]
    #[x][0] Kに沿う
    #[x][0] Kに沿わない最大
    for i in range(iML,-1,-1):
        ib = i%2 #桁%2 dp
        inb = 0 if ib else 1 #次のdp
        i0 = aD[i] #反転しない
        i1 = (2**i)*N-i0 #反転する
        iM = max(i0,i1) #選べるとき
        bL = i < iLK #Kより長い
        if (K>>i)&1:
            dp[inb][1] = dp[ib][1] + i1
            dp[inb][0] = max(dp[ib][1] + i0 ,dp[ib][0]+(iM if bL else i0))
        else:
            dp[inb][1] = dp[ib][1] + i0
            dp[inb][0] = dp[ib][0] + (iM if bL else i0)
    print(max(dp[inb]))
解()

Buggy python submission (verdict: Time Limit Exceeded):
def muridesu():
    N,K=[int(_) for _ in input().split()]
    aA=[int(_) for _ in input().split()]
    dsA = set(aA)
    def f(i,aA):
        return sum(i^a for a in aA)
    print(max(f(k,aA) for k in (set(range(1,K+1)) - dsA  | {0})))
muridesu()


Fixed submission:
#場合分けがつらい
def 解():
    N,K=[int(_) for _ in input().split()]
    aD =[0]*40
    aA=[int(_) for _ in input().split()]
    iL = 0
    for i in range(40):
        m = 2**i
        aD[i]=sum((_ & m) for _ in aA)
        if aD[i]:
            iL = i
    iLK = len(bin(K))-3
    iML = max(iL,iLK)
    dp = [[0]*2 for _ in range(2)]
    #[x][0] Kに沿う
    #[x][0] Kに沿わない最大
    for i in range(iML,-1,-1):
        ib = i%2 #桁%2 dp
        inb = 0 if ib else 1 #次のdp
        i0 = aD[i] #反転しない
        i1 = (2**i)*N-i0 #反転する
        iM = max(i0,i1) #選べるとき
        bL = i < iLK #Kより長い
        if (K>>i)&1:
            dp[inb][1] = dp[ib][1] + i1
            dp[inb][0] = max(dp[ib][1] + i0 ,dp[ib][0]+(iM if bL else i0))
        else:
            dp[inb][1] = dp[ib][1] + i0
            dp[inb][0] = dp[ib][0] + (iM if bL else i0)
    print(max(dp[inb]))
解()

Buggy python submission (verdict: Time Limit Exceeded):

def muridesu():
    N,K=[int(_) for _ in input().split()]
    aA=[int(_) for _ in input().split()]
    dsA = set(aA)
    def f(i,N,aA):
        return sum(i^aA[j] for j in range(N))

    print(max(f(k,N,aA) for k in (set(range(1,K+1)) - dsA  | {0})))
muridesu()


Fixed submission:
#場合分けがつらい
def 解():
    N,K=[int(_) for _ in input().split()]
    aD =[0]*40
    aA=[int(_) for _ in input().split()]
    iL = 0
    for i in range(40):
        m = 2**i
        aD[i]=sum((_ & m) for _ in aA)
        if aD[i]:
            iL = i
    iLK = len(bin(K))-3
    iML = max(iL,iLK)
    dp = [[0]*2 for _ in range(2)]
    #[x][0] Kに沿う
    #[x][0] Kに沿わない最大
    for i in range(iML,-1,-1):
        ib = i%2 #桁%2 dp
        inb = 0 if ib else 1 #次のdp
        i0 = aD[i] #反転しない
        i1 = (2**i)*N-i0 #反転する
        iM = max(i0,i1) #選べるとき
        bL = i < iLK #Kより長い
        if (K>>i)&1:
            dp[inb][1] = dp[ib][1] + i1
            dp[inb][0] = max(dp[ib][1] + i0 ,dp[ib][0]+(iM if bL else i0))
        else:
            dp[inb][1] = dp[ib][1] + i0
            dp[inb][0] = dp[ib][0] + (iM if bL else i0)
    print(max(dp[inb]))
解()

Buggy python submission (verdict: Time Limit Exceeded):
def muridesu():
    N,K=[int(_) for _ in input().split()]
    aA=[int(_) for _ in input().split()]
    def f(i,N,aA):
        return sum(i^aA[j] for j in range(N))
    print(max(f(k,N,aA) for k in range(K+1)))
muridesu()


Fixed submission:
#場合分けがつらい
def 解():
    N,K=[int(_) for _ in input().split()]
    aD =[0]*40
    aA=[int(_) for _ in input().split()]
    iL = 0
    for i in range(40):
        m = 2**i
        aD[i]=sum((_ & m) for _ in aA)
        if aD[i]:
            iL = i
    iLK = len(bin(K))-3
    iML = max(iL,iLK)
    dp = [[0]*2 for _ in range(2)]
    #[x][0] Kに沿う
    #[x][0] Kに沿わない最大
    for i in range(iML,-1,-1):
        ib = i%2 #桁%2 dp
        inb = 0 if ib else 1 #次のdp
        i0 = aD[i] #反転しない
        i1 = (2**i)*N-i0 #反転する
        iM = max(i0,i1) #選べるとき
        bL = i < iLK #Kより長い
        if (K>>i)&1:
            dp[inb][1] = dp[ib][1] + i1
            dp[inb][0] = max(dp[ib][1] + i0 ,dp[ib][0]+(iM if bL else i0))
        else:
            dp[inb][1] = dp[ib][1] + i0
            dp[inb][0] = dp[ib][0] + (iM if bL else i0)
    print(max(dp[inb]))
解()