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