解题思路:创建一个列表储存临时值。每次大值向后进位。取最大值。空间复杂度=O(1).
while True:
try:
a = int(input())
b = str(bin(a)[2:])
l = [0,0]
for i in b:
if i == '1':
l[0]+=1
else:
if l[0]>l[1]:
l[1] = l[0]
l[0] = 0
print(max(l))
except:
break