思路:模拟。根据题意,把m和n异或起来,然后统计异或结果中二进制的1数目即可

注意:本题采用的是字符串处理,用python3会比pypy3快

代码:

import sys
input = lambda: sys.stdin.readline().strip()

import math
inf = 10 ** 18

def I():
    return input()

def II():
    return int(input())

def MII():
    return map(int, input().split())

def GMI():
    return map(lambda x: int(x) - 1, input().split())

def LI():
    return input().split()

def LII():
    return list(map(int, input().split()))

def LFI():
    return list(map(float, input().split()))

fmax = lambda x, y: x if x > y else y
fmin = lambda x, y: x if x < y else y
isqrt = lambda x: int(math.sqrt(x))

'''
如果平台的python版本高点,可以写以下代码:
print((m ^ n).bit_count())
'''

def solve():
    m, n = MII()
    print(bin(m ^ n)[2:].count('1'))

t = 1
# t = II()
for _ in range(t):
    solve()