题解 | #查找输入整数二进制中1的个数#

'''十进制转二进制：
5=====101  1*2^0 + 0*2^1 + 1*2^2 = 1+0+4=5
5/2=2---1
2/2=1---0
1/2=0---1

7=====111  1*2^0 + 1*2^1 + 1*2^2 + 1+2+4=7
7/2=3---1
3/2=1---1
1/2=0---1

2/2=1---0  10
1/2=0---1

4/2=2---0  100
2/2=1---0
1/2=0---1


print('10101'.count('1'))
print([1,2,3].count(1))
'''

while True:
    try:
        n=int(input())
        binary=''
        while n//2>=1:
            binary =  binary+str(n%2)
            n = n//2
        binary = binary+str(n)
        binary = binary[::-1]
        print(binary.count('1'))
    except:
        break