Problem:
 Score : 400 points    Problem Statement Let f(A, B) be the exclusive OR of A, A+1, ..., B . Find f(A, B) .   What is exclusive OR?  The bitwise exclusive OR of integers c_1, c_2, ..., c_n (let us call it y ) is defined as follows:   When y is written in base two, the digit in the 2^k 's place ( k \geq 0 ) is 1 if, the number of integers among c_1, c_2, ...c_m whose binary representations have 1 in the 2^k 's place, is odd, and 0 if that count is even.   For example, the exclusive OR of 3 and 5 is 6 . (When written in base two: the exclusive OR of 011 and 101 is 110 .)

Solution (Python):
A,B=map(int,input().split())

if A==0:
    n=(B+1)//2
    if n%2==0:
        if B%2==0:
            print(B)
        else:
            print(0)
    else:
        if B%2==0:
            print(1^B)
        else:
            print(1)
else:
    m=A//2
    if m%2==0:
        if (A-1)%2==0:
            a=A-1
        else:
            a=0
    else:
        if (A-1)%2==0:
            a=1^(A-1)
        else:
            a=1
    n=(B+1)//2
    if n%2==0:
        if B%2==0:
            b=B
        else:
            b=0
    else:
        if B%2==0:
            b=1^B
        else:
            b=1
    ans=a^b
    print(ans)

Solution (Python):
A,B=map(int,input().split())
res,B=0,B+1
for j in range(41):
    res+=((((B//(2*(2**j)))*(2**j)+max(0,(B%(2*(2**j)))-2**j))-((A//(2*(2**j)))*(2**j)+max(0,(A%(2*(2**j)))-2**j)))%2)*(2**j)
print(res)

Solution (Python):
A,B=map(int,input().split())
B+=1
Abit=[0]*41
for j in range(41):
    bitp=A//(2*(2**j))
    bitm=A%(2*(2**j))
    Abit[j]+=bitp*(2**j)
    Abit[j]+=max(0,bitm-2**j)
Bbit=[0]*41
for j in range(41):
    bitp=B//(2*(2**j))
    bitm=B%(2*(2**j))
    Bbit[j]+=bitp*(2**j)
    Bbit[j]+=max(0,bitm-2**j)
res=0
for j in range(41):
    Bbit[j]-=Abit[j]
    res+=(Bbit[j]%2)*(2**j)
print(res)

Solution (Python):
A,B=map(int,input().split())
B+=1
Abit=[0]*41
for j in range(41):
    bitp=A//(2*(2**j))
    bitm=A%(2*(2**j))
    Abit[j]+=bitp*(2**j)
    Abit[j]+=max(0,bitm-2**j)
Bbit=[0]*41
for j in range(41):
    bitp=B//(2*(2**j))
    bitm=B%(2*(2**j))
    Bbit[j]+=bitp*(2**j)
    Bbit[j]+=max(0,bitm-2**j)
res=0
for j in range(41):
    Bbit[j]-=Abit[j]
    res+=(Bbit[j]%2)*(2**j)
print(res)

Solution (Python):
a, b = map(int, input().split())
a, b = min(a, b), max(a, b)
def f(a):
    if a % 2 == 0:
        return a ^ (a+1)//2%2
    else:
        return (a+1)//2%2
    
print(f(b) ^ f(a-1))

Solution (Python):


A, B = map(int, input().split())

def rangeXOR(x, y):
    ret = x
    for i in range(x+1, y+1):
        ret ^= i
    return ret

left = rangeXOR(A, A+(3-A%4))
right = rangeXOR(B-(B%4), B)
    
if left==0:
    print(right)
elif right==0:
    print(left)
else:
    print(left^right)

Solution (Python):




import numpy as np
import sys

input = sys.stdin.readline

A,B = map(int,input().split())

def calc(x):
    if (x+1)%4 == 0:
        return 0
    elif (x+1)%4 == 1:
        return x
    elif (x+1)%4 == 2:
        return (x-1)^x
    else:
        return (x-2)^(x-1)^x

if A!=0:A-=1

print(calc(A)^(calc(B)))

Solution (Python):
a, b = map(int, input().split())

def check(n, k):
    if k == 1:
        return ((n + 1) // 2) % 2
    else:
        return max(0, (n + 1) % (k * 2) - k) % 2

k = 1
ans = 0

while k <= b:
    
    ans += abs(check(b, k) - check(a-1, k)) % 2 * k
    k *= 2

print(ans)

Solution (Python):
A, B = map(int,input().split())


def solve(K):
    ans = ((K+1) // 2) % 2
    if K % 2 == 0:
        ans ^= K
    return ans

print(solve(A-1) ^ solve(B))

Solution (Python):
A, B = map(int,input().split())


def solve(K):
    if K <= 3:
        ans = 0
        for i in range(K+1):
            ans ^= i
        return ans
    ans = ((K+1) // 2) % 2
    if K % 2 == 0:
        ans ^= K
    return ans
print(solve(A-1) ^ solve(B))

Solution (Python):
from sys import stdin
A,B = [int(x) for x in stdin.readline().rstrip().split()]

def f(n):
    res = 0
    tmp = 1
    while tmp <= n:
        tmp *= 2
        num = ((n+1)//tmp)*(tmp//2)
        num += max(0,(n+1)%tmp - tmp//2)
        if num % 2 != 0:
            res += (tmp//2)
    return res

print(f(A-1)^f(B))

Solution (Python):
a,b = map(int,input().split())

def bina(x):
    if x % 2 == 1:
        if (x+1) % 4 == 2:
            return 1
        else:
            return 0
    else:
        return x ^ bina(x-1)
    
ans = bina(a-1) ^ bina(b)

print(ans)

Solution (Python):
import sys
import math  
import bisect  
import queue  


def input():
    return sys.stdin.readline().rstrip()


def count(N):
    cnt = []
    r = 1
    while r < N:
        res = (N // (2 * r)) * r + max(N % (2 * r) - r, 0)
        cnt.append(res)
        r *= 2
    return cnt


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

    cnt_A = count(A)
    cnt_B = count(B + 1)

    cnt = [b for b in cnt_B]
    for i, a in enumerate(cnt_A):
        cnt[i] -= a

    res = 0
    for i, c in enumerate(cnt):
        if c % 2 != 0:
            res += 2 ** i

    print(res)


if __name__ == '__main__':
    main()

Solution (Python):
A,B = map(int,input().split())

def solve(x):
    
    if x < 0:
        return 0
    
    if x % 4 == 0:
        return x
    elif x % 4 == 1:
        return 1
    elif x % 4 == 2:
        return x+1
    elif x % 4 == 3:
        return 0
    
print(solve(A-1)^solve(B))

Solution (Python):
A,B = map(int,input().split())

def solve(x):
    if x % 4 == 0:
        return x
    elif x % 4 == 1:
        return 1
    elif x % 4 == 2:
        return x+1
    elif x % 4 == 3:
        return 0
    
print(solve(A-1)^solve(B))

Solution (Python):
A,B=map(int, input().split())
def fanc(n):
    if n<0:
        return 0
    if n%4==0:
        return n
    elif n%4==1:
        return 1
    elif n%4==2:
        return n+1
    else:
        return 0

print(fanc(A-1)^fanc(B))

Solution (Python):
def main():
    A, B = map(int, input().split())

    def xor_world(k):
        """[0,k]"""
        r = k % 4
        if r == 0:
            return k
        elif r == 1:
            return 1
        elif r == 2:
            return 1 ^ k
        else:
            return 0

    ans = xor_world(B) ^ xor_world(A) ^ A
    print(ans)


if __name__ == '__main__':
    main()

Solution (Python):
def solve(N):
    if N <= 0:
        return 0
    cnt = 0
    while (1 << cnt) <= N:
        cnt += 1

    out = 0 if (N % 4 == 0) or (N % 4 == 3) else 1
    for i in reversed(range(1,cnt)):
        T = 1 << i
        if N < T:
            out += 0
        else:
            N -= T
            if N % 2 == 0:
                out += T
            else:
                out += 0

    return out

A, B = map(int, input().split())
B = solve(B)
A = solve(A-1)

print(A^B)

Solution (Python):
A, B = map(int, input().split())

if A == B:
    print(A)
else:
    A -= 1
    ans = 0
    a = 1
    b = 1
    if ((A+1)//2) % 2 == 1:
        if A % 2 == 0:
            a = 1 ^ A
        else:
            a = 1
    else:
        if A % 2 == 1:
            a = 0
        else:
            a = A

    if ((B+1)//2) % 2 == 1:
        if B % 2 == 0:
            b = 1 ^ B
        else:
            b = 1
    else:
        if B % 2 == 1:
            b = 0
        else:
            b = B

    print(a ^ b)

Solution (Python):
def odd_solve(n):
    return (n + 1) // 2 % 2

def solve(n):
    if n % 2 == 1:
        return odd_solve(n)
    else:
        return odd_solve(n + 1) ^ (n + 1)

a, b = map(int, input().split())
ans = solve(b) ^ solve(a - 1)
print(ans)

Solution (Python):
def f(n):
    if n % 2 == 0:
        
        return n ^ f(n - 1)
    else:
        
        return (n // 2 + 1) % 2

def main():
    a, b = map(int, input().split())
    print(f(a-1) ^ f(b))


if __name__ == '__main__':
    main()

Solution (Python):
a,b = map(int,input().split())
kazu = [a+i for i in range(4 - a%4)]
kazu.extend([b-i for i in range(b%4+1)])
ans = kazu[0]
for i in range(1,len(kazu)):
    ans = ans ^ kazu[i]
print(ans)

Solution (Python):
from math import ceil
def f(x):
  xor_cnt=ceil(x/2)
  res=0
  if x%2==1:
    if xor_cnt%2==0:
      res=0
    else:
      res=1
  else:
    if xor_cnt%2==0:
      res=0^x
    else:
      res=1^x
  return res

a,b=map(int,input().split())
print(f(b)^f(a-1))

Solution (Python):
import sys
input=sys.stdin.readline
def F(N):

    if not N & 1:
        x = N >> 1
        return x & 1
    return (N - 1) ^ F(N - 1)
A, B = map(int, input().split())
answer = F(B + 1) ^ F(A)
print(answer)

Solution (Python):
a,b = map(int, input().split())
if a%2:
    if (b-a)%2:
        print((( (b-a-1) // 2) % 2) *1 ^ a ^ b)
    else:
        print((( (b-a) // 2) % 2) *1 ^ a)
else:
    if (b-a+1)%2:
        print((( (b-a) // 2) % 2) *1 ^ b)
    else:
        print((( (b-a+1) // 2) % 2) *1)

Solution (Python):
def resolve():
    import math
    a, b = map(int, input().split())
    if a == 1:
        bina = 0
    elif a == 2:
        bina = 1
    elif a == 3:
        bina = 3
    elif a%4 ==3:
        bina = a
    else:
        lowera = math.floor((a - 3) / 4) * 4 + 3
        bina = 0
        for i in range(lowera + 1, a):
            bina = bina ^ i

    if b ==1:
        binb = 1
    elif b ==2:
        binb = 3
    elif b==3:
        binb =0
    else:
        if b%4 == 3:
            binb = 0
        else:
            lowerb = math.floor((b-3)/4)*4+3
            binb = 0
            for j in range(lowerb+1, b+1):
                binb = binb^j

    print(bina^binb)
resolve()

Solution (Python):
def split(word): 
    return [char for char in word]
def convert(s): 
  
    
    new = "" 
  
    
    for x in s: 
        new += x  
  
    
    return new 
a,b=map(int,input().split())
ans=[]
abin="{0:b}".format(a)
bbin="{0:b}".format(b)
abin=split(abin)
bbin=split(bbin)
abin.reverse()
bbin.reverse()
for i in range(len(bbin)-len(abin)):
  abin.append('0')
if (abin[0]=='1' and ((b-a)%4==0 or (b-a)%4==1)) or (abin[0]=='0' and ((b-a)%4==1 or (b-a)%4==2)) :
  ans.append('1')
else:
  ans.append('0')
for i in range(1,len(bbin)):
  flag=0
  if abin[i]=='1' and a%2==1:
    flag+=1
  if bbin[i]=='1' and b%2==0:
    flag+=1
  if flag==1:
    ans.append('1')
  else:
    ans.append('0')
ans.reverse()
print(int(convert(ans),2))

Solution (Python):


import sys
readline = sys.stdin.buffer.readline
sys.setrecursionlimit(10 ** 7)

a, b = map(int, readline().split())

def f(x):
    s = 0
    for i in range(x // 4 * 4, x + 1):
        s ^= i
    return s

print(f(a - 1) ^ f(b))

Solution (Python):
a, b = map(int, input().split())

def cal_xor(n):
    if n % 2 == 0:
        if (n // 2) % 2 == 0:
            return 0 ^ n
        else:
            return 1 ^ n
    else:
        if ((n+1)//2) % 2 == 0:
            return 0
        else:
            return 1

print(cal_xor(b)^cal_xor(a-1))

Solution (Python):
a,b = map(int, input().split())

def f(x):
    if x % 2 == 1:
        return x // 2 % 2
    return x ^ f(x - 1)

print(f(a-1) ^ f(b))

Solution (Python):
import sys, re, os
from collections import deque, defaultdict, Counter
from math import ceil, sqrt, hypot, factorial, pi, sin, cos, radians
from itertools import permutations, combinations, product, accumulate
from operator import itemgetter, mul
from copy import deepcopy
from string import ascii_lowercase, ascii_uppercase, digits
from heapq import heapify, heappop, heappush
 
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def S_MAP(): return map(str, input().split())
def LIST(): return list(map(int, input().split()))
def S_LIST(): return list(map(str, input().split()))
 
sys.setrecursionlimit(10 ** 9)
INF = float('inf')
mod = 10 ** 9 + 7

A, B = MAP()

def f(X):
    if X % 2 == 1:
        one_num = (X + 1) // 2
        if one_num % 2 == 0:
            return 0
        else:
            return 1
    else:
        one_num = X // 2
        if one_num % 2 == 0:
            return 0 ^ X
        else:
            return 1 ^ X
print(f(B) ^ f(A-1))

Solution (Python):
A, B = map(int, input().split())

A -= 1
ans = 0

def calc(x):
    x += 1
    lst = [0] * 65
    count = 2
    for i in range(64):
        
        tmp = x // count
        a = x % count
        a = min(a, count // 2)
        lst[i] = (x - (a + tmp * count // 2)) % 2
        count *= 2
    return lst

lstA = calc(A)
lstB = calc(B)


tmp = 1
for i in range(65):
    if abs(lstA[i] - lstB[i]) == 1:
        ans += tmp
    tmp *= 2

print (ans)

Solution (Python):
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 7)

a, b = map(int, read().split())
cnt_a = []
ans = 0
check = 1
while check <= a - 1:
    x, y = divmod(a, check * 2)
    cnt_a.append(x * check + max(0, y - check))
    check *= 2
check = 1
while check <= b:
    if len(cnt_a) != 0:
        aa = cnt_a.pop(0)
    else:
        aa = 0
    x, y = divmod(b + 1, check * 2)
    cnt = x * check + max(0, y - check) - aa
    if cnt % 2 == 1:
        ans += check
    check *= 2
print(ans)

Solution (Python):
a, b = map(int, input().split())
b = b + 1
d = 1
res = 0
t1 = a % 4
t2 = b % 4
ls = [0, 1, 0, 1]
res += (sum(ls[:t2]) - sum(ls[:t1])) % 2
while 2 ** d < max(a, b):
    tmp1 = a % (2 ** (d + 1))
    tmp2 = b % (2 ** (d + 1))
    tmp3 = max(tmp1, tmp2)
    tmp4 = min(tmp1, tmp2)
    if tmp3 >= 2 ** d:
        tmp3 = tmp3 % 2
    else:
        tmp3 = 0
    if tmp4 >= 2 ** d:
        tmp4 = tmp4 % 2
    else:
        tmp4 = 0
    tmp = (tmp4 - tmp3) % 2
    res += tmp * (2 ** d)
    d += 1

print(res)

Solution (Python):
import sys
sys.setrecursionlimit(1 << 25)
read = sys.stdin.readline
ra = range
enu = enumerate


def read_ints():
    return list(map(int, read().split()))


def read_a_int():
    return int(read())


def read_tuple(H):
    '''
    H is number of rows
    '''
    ret = []
    for _ in range(H):
        ret.append(tuple(map(int, read().split())))
    return ret


def read_col(H):
    '''
    H is number of rows
    A列、B列が与えられるようなとき
    ex1)A,B=read_col(H)    ex2) A,=read_col(H) 
    '''
    ret = []
    for _ in range(H):
        ret.append(list(map(int, read().split())))
    return tuple(map(list, zip(*ret)))


def read_matrix(H):
    '''
    H is number of rows
    '''
    ret = []
    for _ in range(H):
        ret.append(list(map(int, read().split())))
    return ret
    


MOD = 10**9 + 7
INF = 2**31  

from collections import defaultdict, Counter, deque
from operator import itemgetter
from itertools import product, permutations, combinations
from bisect import bisect_left, bisect_right  











def f(x):
    start = x // 4 * 4
    ret = 0
    for i in ra(start, x + 1):
        ret ^= i
    return ret


A, B = read_ints()

print(f(B) ^ f(A - 1))

Solution (Python):
A, B = map(int, input().split())

def g(x):
    if x % 2 == 1:
        return x // 2 % 2
    return x ^ g(x - 1)

print(g(A - 1) ^ g(B))

Solution (Python):
from math import ceil


def main():
    a, b = map(int, input().split())
    answer = 0
    if a < b:
        for i in range(a, ceil(a / 4) * 4):
            answer ^= i
        for i in range((b // 4) * 4, b + 1):
            answer ^= i
    else:
        answer = a
    print(answer)


if __name__ == '__main__':
    main()

Solution (Python):
def main():
    A, B = (int(i) for i in input().split())

    def f(X):
        X += 1
        bi = []
        d = 2
        for i in range(40):
            v = (X//d) * (d//2) + max((X % d) - (d//2), 0)
            bi.append(v)
            d *= 2
        return bi

    ans = 0
    d = 0
    fA = f(A-1)
    fB = f(B)
    
    for a, b in zip(fA, fB):
        v = max(b - a, 0)
        if v % 2 == 1:
            ans += 2**d
        d += 1
    print(ans)


if __name__ == '__main__':
    main()

Solution (Python):
a,b=map(int,input().split())
if a==b:
  print(a)
  exit(0)
ans=0
if a&1:
  ans=a
  a+=1
if ((b-a+1)//2)&1:ans^=1
if not(b&1):ans^=b
print(ans)

Solution (Python):
if __name__ == '__main__':
    def judge(x):
        if x % 4 == 0:
            return x
        elif x % 4 == 1:
            return 1
        elif x % 4 == 2:
            return x + 1
        else:
            return 0

    a, b = map(int, input().split())
    fa = judge(a - 1)
    fb = judge(b)
    print(fa ^ fb)

Solution (Python):











import sys
sys.setrecursionlimit(10**6)
input=sys.stdin.readline
from math import floor,sqrt,factorial,hypot,log 
from heapq import heappop, heappush, heappushpop
from collections import Counter,defaultdict,deque
from itertools import accumulate,permutations,combinations,product,combinations_with_replacement
from bisect import bisect_left,bisect_right
from copy import deepcopy
from fractions import gcd
from random import randint
def ceil(a,b): return (a+b-1)//b
inf=float('inf')
mod = 10**9+7
def pprint(*A): 
    for a in A:     print(*a,sep='\n')
def INT_(n): return int(n)-1
def MI(): return map(int,input().split())
def MF(): return map(float, input().split())
def MI_(): return map(INT_,input().split())
def LI(): return list(MI())
def LI_(): return [int(x) - 1 for x in input().split()]
def LF(): return list(MF())
def LIN(n:int): return [I() for _ in range(n)]
def LLIN(n: int): return [LI() for _ in range(n)]
def LLIN_(n: int): return [LI_() for _ in range(n)]
def LLI(): return [list(map(int, l.split() )) for l in input()]
def I(): return int(input())
def F(): return float(input())
def ST(): return input().replace('\n', '')
def f(x):
    res = 0
    while(((x+1)%4)!=0):
        res ^= x
        x-=1

    return res
def main():
    A,B=MI()
    print(f(B)^f(A-1))

if __name__ == '__main__':
    main()

Solution (Python):
def F(X):
    if X % 4 == 1:
        return 1
    if X % 4 == 2:
        return X + 1
    if X % 4 == 3:
        return 0
    if X % 4 == 0:
        return X


def test():
    x = 0
    for i in range(1, 50):
        x ^= i
        print(i, ":", x)


if __name__ == "__main__":
    A, B = map(int, input().split())
    if A == 0:
        print(F(B))
    else:
        print(F(B) ^ F(A-1))

Solution (Python):

from collections import defaultdict, deque
from heapq import heappush, heappop
from itertools import permutations, accumulate
import sys
import math
import bisect
def LI(): return [int(x) for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS():return [list(x) for x in sys.stdin.readline().split()]
def S():
    res = list(sys.stdin.readline())
    if res[-1] == "\n":
        return res[:-1]
    return res
def IR(n):
    return [I() for i in range(n)]
def LIR(n):
    return [LI() for i in range(n)]
def SR(n):
    return [S() for i in range(n)]
def LSR(n):
    return [LS() for i in range(n)]

sys.setrecursionlimit(1000000)
mod = 1000000007

def solve():
    def s(n):
        if n < 0:
            return 0
        m = n >> 2
        res = 0
        for i in range(m << 2,n+1):
            res ^= i
        return res
    a,b = LI()
    print(s(a-1)^s(b))
    return


if __name__ == "__main__":
    solve()

Solution (Python):
import sys
read = sys.stdin.buffer.read
readline = sys.stdin.buffer.readline
readlines = sys.stdin.buffer.readlines
sys.setrecursionlimit(10 ** 8)
INF = float('inf')
MOD = 10 ** 9 + 7

def main():
    A, B = map(int, readline().split())
    if A%2==0 and B%2==0:
        count_1 = (B-A)//2
        ans = (count_1%2)^B
    elif A%2==0 and B%2==1:
        count_1 = (B-A+1)//2
        ans = count_1%2
    elif A%2==1 and B%2==0:
        count_1 = (B-A-1)//2
        ans = A^B^(count_1%2)
    else:
        count_1 = (B-A)//2
        ans = A^(count_1%2)
    print(ans)

if __name__ == '__main__':
    main()

Solution (Python):
import sys
sys.setrecursionlimit(10**6)
from math import floor,ceil,sqrt,factorial,log
from heapq import heappop, heappush, heappushpop
from collections import Counter,defaultdict,deque
from itertools import accumulate,permutations,combinations,product,combinations_with_replacement
from bisect import bisect_left,bisect_right
from copy import deepcopy
from operator import itemgetter
from fractions import gcd
mod = 10 ** 9 + 7
inf = float('inf')
ninf = -float('inf')
 

def ii(): return int(sys.stdin.readline().rstrip()) 
def mii(): return map(int,sys.stdin.readline().rstrip().split())
def limii(): return list(mii()) 
def lin(n:int): return [ii() for _ in range(n)]
def llint(n: int): return [limii() for _ in range(n)]

def ss(): return sys.stdin.readline().rstrip() 
def mss(): return sys.stdin.readline().rstrip().split()
def limss(): return list(mss()) 
def lst(n:int): return [ss() for _ in range(n)]
def llstr(n: int): return [limss() for _ in range(n)]





a,b=mii()

if a%2==0 and b%2==1:
    if ((b-a+1)//2)%2==0:
        print(0)
    else:
        print(1)

elif a%2==1 and b%2==1:
    if ((b-a)//2)%2==0:
        print(a^0)
    else:
        print(a^1)

elif a%2==1 and b%2==0:
    if ((b-a-1)//2)%2==0:
        print(a^b^0)
    else:
        print(a^b^1)

else:
    if ((b-a)//2)%2==0:
        print(b^0)
    else:
        print(b^1)

Solution (Python):
a,b=map(int,input().split())
f=lambda x:x//2%2^x if x%2==0 else (x//2+1)%2
print(f(a-1)^f(b))

Solution (Python):
import string
import sys

sys.setrecursionlimit(10 ** 5 + 10)


def input(): return sys.stdin.readline().strip()


def resolve():
    A, B = map(int, input().split())
    def xor(B):
        if B >= 1:
            if B % 2 == 1:
                if (B - 1) % 4 == 0:
                    return 1
                else:
                    return 0

            else:
                if (B - 2) % 4 == 0:
                    return 1 ^ B
                else:
                    return 0 ^ B
        else:
            return 0
    print(xor(B)^xor(A-1))


resolve()

Solution (Python):
a, b = map(int, input().split())


def f(keta, x):
    val = 2**keta
    if x < (2 ** (keta)):
        return False
    else:
        x -= (val-1)
        ans = x // (2*val) * val
        ans += min(val,x-val*2*(x//(2*val)))
        return ans
if b == 0:
    print(0)
    exit()
ans = []
i = 0
while f(i,b) != False:
    
    val = f(i,b)
    if f(i,a-1) != False:
        val -= f(i,a-1)
    
    ans += str(val % 2)
    i += 1

ans.reverse()
print(int(''.join(ans),2))

Solution (Python):
a, b = map(int, input().split())
def o(x):
    return(x, 1, x+1, 0)[x%4]
print(o(~-a) ^ o(b))

Solution (Python):
import math
a,b = map(int, input().split())

def xorcalc(x):
	if x %2 == 0:
		y = x//2
		if y %2 == 0:
			return x
		else:
			return x+1
	else:
		y = math.ceil(x/2)
		if y%2 == 0:
			return 0
		else:
			return 1
ans = xorcalc(b) ^ xorcalc(a-1)
print (ans)

Solution (Python):


"""
・c1~cnについて全体のxorをとるとき、それぞれのcについて各bitで１が奇数個立っているとそのbitが１、偶数個立っているとそのbitが０になる
　→　xorは順番を入れ替えても答えは変わらない

・0 xor N = N、A xor~B = (0 xor ~A-1) xor (0 xor ~B)

・それぞれの偶数nについて、n xor (n+1) = 1 (下位１ビットだけが異なるから)
"""


A, B = map(int, input().split())

def solve(n):

    if n == 0:
        return 0

    nums = (n + 1) % 4

    if nums == 0:
        return 0

    if nums == 1:
        return n

    if nums == 2:
        return 1 

    if nums == 3:
        return 1 ^ n


print( solve(A-1) ^ solve(B))

Solution (Python):
A,B=map(int,input().split())
if A%2==0:
    d=B-A
    if d%4==0:
        print(B)
    elif d%4==1:
        print(1)
    elif d%4==2:
        print(1^B)
    else:
        print(0)
else:
    d=B-A-1
    if d%4==0:
        print(A^B)
    elif d%4==1:
        print(A^1)
    elif d%4==2:
        print(A^1^B)
    else:
        print(A^0)

Solution (Python):
A, B = map(int, input().split())







if A == B:
    print(A)
    exit()

if A % 2 == 0 and B % 2 == 1:
    if (A & 2) == (B & 2):
        print(1)
    else:
        print(0)

    exit()

if A % 2 == 0 and B % 2 == 0:
    temp = 0
    if (A & 2) != (B & 2):
        temp = 1

    print(B + temp)
    exit()

if A % 2 == 1 and B % 2 == 1:
    temp = 0
    if (A & 2) != (B & 2):
        temp = 1

    print(A - temp)
    exit()

temp = 0
if (A & 2) == (B & 2):
    temp = 1

xo = A ^ B
print(xo - temp)

Solution (Python):
def f(x):
    if x % 4 == 0:
        return x
    elif x % 4 == 1:
        return 1
    elif x % 4 == 2:
        return 1^x
    else:
        return 0

def main():
    import sys
    input = sys.stdin.readline
    a, b = map(int, input().split())
    print(f(a-1)^f(b))
    

if __name__ == '__main__':
    main()

Solution (Python):
a,b=map(int,input().split())
def f(x):
  if x%4==0:
    return x
  elif x%4==1:
    return 1
  elif x%4==2:
    return 1^x
  else:
    return 0
print(f(a-1)^f(b))

Solution (Python):

a,b = map(int,input().split())

A = format(max(0,a-1), 'b')
B = format(b, 'b')


AA = []
BB = []

if(len(A) >= 2):
    for i in range(len(A)):
        if(i == len(A)-1):
            if(((a-1)%4 == 1) or ((a-1)%4 == 2)):
                AA.append("1")
            else:
                AA.append("0")
        else:
            if((A[i] == "1") and (A[-1] == "0")):
                AA.append("1")
            else:
                AA.append("0")
else:
    AA.append(A[0])

if(len(B) >= 2):
    for i in range(len(B)):
        if(i == len(B)-1):
            if((b%4 == 1) or (b%4 == 2)):
                BB.append("1")
            else:
                BB.append("0")
        else:
            if((B[i] == "1") and (B[-1] == "0")):
                BB.append("1")
            else:
                BB.append("0")
else:
    BB.append(B[0])


AA = "".join(AA)
BB = "".join(BB)

AA = int('0b'+AA,0)
BB = int('0b'+BB,0)

ans = AA^BB

print(ans)

Solution (Python):
a,b=map(int,input().split())
c,d=0,0
if a!=0:
  a-=1
  m=(-(-a//2))%2
  c=m if a%2 else m^a
m=m=(-(-b//2))%2
d=m if b%2 else m^b
print(c^d)

Solution (Python):
a,b=map(int,input().split())
print(((((a+1)&2)>>1)^(a*(~a&1)))^a^(((b+1)&2)>>1)^(b*(~b&1)))

Solution (Python):
a,b=map(int,input().split())

def solve(x):
    if x <= 0 : return 0
    
    return (x+1)//2%2^(x * ((x+1)%2))

print(solve(b)^solve(a-1))

Solution (Python):
a,b=map(int,input().split())

def solve(x):
    if x <= 0 : return 0
    res = 0
    keta = 0
    limit =x
    while limit:
        if keta == 0:
            tmp=x%4
            if tmp == 1 or tmp == 2:
                res += 1
        else:
            tmp = x%(1<<(keta+1))
            tmp -= 1<<keta
            if tmp >= 0:
                if tmp %2 == 0:
                    res += 1<<keta
        keta += 1
        limit>>=1
    return res

print(solve(b)^solve(a-1))

Solution (Python):
def g(x):
 p=1
 while p*2<=x:p*=2
 return[0,1,3,0][x]if x<4else(p*((x-p+1)&1))|g(x-p)
a,b=map(int,input().split());print(g(b)^g(a-1))

Solution (Python):
a, b = map(int, input().split())


def cum_xor(x):
    if x < 0:
        return 0
    if x < 4:
        t = [0, 1, 3, 0]
        return t[x]
    p = 1
    while (p << 1) <= x:
        p <<= 1
    mask = p * ((x - p + 1) & 1)
    return mask | cum_xor(x - p)


print(cum_xor(b) ^ cum_xor(a - 1))

Solution (Python):


def f(data):
    if data<=0: return 0
    if data%2==1:
        ara = (data+1)//2
        if ara%2==0: return 0
        else: return 1
    else:
        ara = data//2
        if ara%2==0: return data
        else: return data+1
nim,mike = map(int,input().split())
print(f(nim-1)^f(mike))

Solution (Python):
def xor(v):
    result = 0
    for i in range(v//4 * 4, v+1):
        result ^= i
    return result


a, b = map(int, input().split())

print(xor(b) ^ xor(a-1))

Solution (Python):
a,b = map(int, input().split())

def xor_solve(x):
    if x % 4 == 0:
        return x
    elif x % 4 == 1:
        return 1
    elif x % 4 == 2:
        return x+1
    else:
        return 0
        
print(xor_solve(b) ^ xor_solve(a-1))

Solution (Python):

import sys

sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))

A, B = lr()

def F(x):
    '''x-1までの排他的論理和の合計を返す'''
    if x%2 == 0:
        y = x >> 1
        return y & 1
    else:
        return (x-1)^F(x-1)

answer = F(A) ^ F(B+1)
print(answer)

Solution (Python):
A,B=map(int,input().split())
def F(x):
  if x%2==0:
    return ((x//2)%2)^x
  else:
    return ((x+1)//2)%2
if A==0:
  print(F(B))
else:
  print(F(B)^F(A-1))

Solution (Python):
a,b = map(int,input().split())

if a == b:
    print(a)
    exit()
ansl = []
if a % 2 == 1:
    ansl.append(a)
    a += 1
if b % 2 == 0:
    ansl.append(b)
    b -= 1

k = b - a + 1
k = k // 2
if k % 2 == 1:
    ansl.append(1)
if len(ansl) == 0:
    print(0)
elif len(ansl) == 1:
    print(ansl[0])
elif len(ansl) == 2:
    print(ansl[0] ^ ansl[1])
else:
    print(ansl[0] ^ ansl[1] ^ ansl[2])

Solution (Python):
A, B = map(int, input().split())

if A % 2 == 0:
    ans = 0
else:
    ans = A
    A += 1

if B % 2 == 0:
    ans ^= B
    B -= 1

if B - A + 1 <= 0:
    pass
elif (B - A + 1) // 2 % 2 == 0:
    pass
else:
    ans ^= 1

print(ans)

Solution (Python):
a,b=map(int,input().split())
def func(x):
    cnt=(x+1)//2
    if x%2==1:
        if cnt%2==0:
            return 0
        else: return 1
    else:
        if cnt%2==0:
            return 0^x
        else: return x^1

print(func(b)^func(a-1))

Solution (Python):
A, B = map(int,input().split())
def f(N):
    if N%4 == 0:
        return N
    elif N%4 == 1:
        return 1
    elif N%4 == 2:
        return N+1
    else:
        return 0
print(f(A-1)^f(B))

Solution (Python):
def main():
    A, B = (int(i) for i in input().split())

    def f(C):
        if C <= 0:
            return [0]*40
        arr = []
        for digit in range(1, 41):
            cur = (C+1)//(2**digit)
            cur *= 2**(digit-1)
            y = (C+1) % (2**digit)
            y -= 2**(digit-1)
            if y > 0:
                cur += y
            arr.append(cur)
            
        return arr

    ans = 0
    i = 0
    fA = f(A-1)
    fB = f(B)
    for a, b in zip(fA, fB):
        ans += ((b-a) % 2)*(2**i)
        i += 1
    print(ans)


if __name__ == '__main__':
    main()

Solution (Python):
A, B = map(int, input().split())

def f(c, i):
    if c == 0: return 0
    if i == 1:
        tmp = (c + 1) // 2
    else:
        tmp = max(0, (c + 1) % (2 ** i) - 2 ** i // 2)
    
    return tmp % 2

ret = ""

for i in range(1, 45 + 1):
    
    val = (f(B, i) - f(max(0, A - 1), i)) % 2
    ret += str(val)

ret = list(map(str, ret))
ret.reverse()
ret = "".join(ret)

print(int(ret, 2))

Solution (Python):
def find_xor_from_zero(n):
    
    return [n, 1, n + 1, 0][n % 4]


def main():
    a, b = map(int, input().split())
    if a != b:
        print(find_xor_from_zero(a - 1) ^ find_xor_from_zero(b))
    else:
        print(a)


if __name__ == '__main__':
    main()

Solution (Python):
A, B = map(int, input().split())


def F(N):
    """
    0,1,2,..,Nの排他的論理和
    """
    if N < 2:
        return N
    if N % 4 == 3:
        return 0
    if N % 4 == 1:
        return 1
    if N % 4 == 0:
        return N
    if N % 4 == 2:
        return N + 1








if A == 0:
    print(F(B))
else:
    print(F(B) ^ F(A-1))

Solution (Python):
def f(x):
  xor_cnt,r=divmod(x,2)
  if r==1:xor_cnt+=1
  res=0
  if x%2==1:
    if xor_cnt%2==0:
      res=0
    else:
      res=1
  else:
    if xor_cnt%2==0:
      res=0^x
    else:
      res=1^x
  return res  

a,b=map(int,input().split())
print(f(b)^f(a-1))

Solution (Python):
a,b=map(int,input().split())
one_num=0
if(a%2==0 and b%2==0):
    one_num=(((b-1)-a)+1)//2
    one_num=0 if one_num%2==0 else 1
    print(one_num^b)
elif(a%2==0):
    one_num=((b-a)+1)//2
    one_num=0 if one_num%2==0 else 1
    print(one_num)
elif(b%2==0):
    one_num=(((b-1)-(a+1))+1)//2
    one_num=0 if one_num%2==0 else 1
    print(one_num^a^b)
else:
    one_num=((b-(a+1))+1)//2
    one_num=0 if one_num%2==0 else 1
    print(one_num^a)

Solution (Python):
A,B=map(int, input().split())

def f(x):
    D=1
    res=0
    while True:
        d,m=divmod(x+1,D<<1)
        if (d*D + max(0, m-D))&1: res+=D
        if d==0: break
        D<<=1
    return res

ans=f(B)
if A: ans^=f(A-1)
print(ans)