Question

E.g。对于输入5，输出应为7。（bin（1）= 1，bin（2）= 10 ... bin（5）= 101） - ＆gt; 1 + 1 + 2 + 1 + 2 = 7

这是我尝试过的，但它不是一个非常有效的算法，考虑到我为每个整数迭代一次循环。我的代码（Python 3）：

i = int(input())
a = 0
for b in range(i+1):
  a = a + bin(b).count("1")
print(a)

谢谢！

Answer 1

这是基于OEIS的递归关系的解决方案：

def onecount(n):
    if n == 0:
        return 0
    if n % 2 == 0:
        m = n/2
        return onecount(m) + onecount(m-1) + m
    m = (n-1)/2
    return 2*onecount(m)+m+1

>>> [onecount(i) for i in range(30)]
[0, 1, 2, 4, 5, 7, 9, 12, 13, 15, 17, 20, 22, 25, 28, 32, 33, 35, 37, 40, 42, 45, 48, 52, 54, 57, 60, 64, 67, 71]

Answer 2

由于gmpy2 et al ，

Alex Martella似乎表现得更好，至少在我的Win10机器上。

from time import time
import gmpy2

def onecount(n):
    if n == 0:
        return 0
    if n % 2 == 0:
        m = n/2
        return onecount(m) + onecount(m-1) + m
    m = (n-1)/2
    return 2*onecount(m)+m+1

N = 10000

initial = time()
for _ in range(N):
    for i in range(30):
        onecount(i)
print (time()-initial)

initial = time()
for _ in range(N):
    total = 0
    for i in range(30):
        total+=gmpy2.popcount(i)
print (time()-initial)

这是输出：

1.7816979885101318
0.07404899597167969

如果您想要一个列表，并且您正在使用＆gt; Py3.2：

>>> from itertools import accumulate
>>> result = list(accumulate([gmpy2.popcount(_) for _ in range(30)]))
>>> result
[0, 1, 2, 4, 5, 7, 9, 12, 13, 15, 17, 20, 22, 25, 28, 32, 33, 35, 37, 40, 42, 45, 48, 52, 54, 57, 60, 64, 67, 71]

Python：如何有效地计算1到n数字的二进制表示中“1”的数量？

2 个答案: