Problem:
Score : 400 points

Problem Statement

Let \mathrm{popcount}(n) be the number of 1s in the binary representation of n.
For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.

Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)

For example, when n=7, it becomes 0 after two operations, as follows:

\mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.

\mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.

You are given an integer X with N digits in binary.
For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted.
Find f(X_1), f(X_2), \ldots, f(X_N).

Constraints

1 \leq N \leq 2 \times 10^5

X is an integer with N digits in binary, possibly with leading zeros.

Input

Input is given from Standard Input in the following format:

N
X

Output

Print N lines. The i-th line should contain the value f(X_i).

Sample Input 1

3
011

Sample Output 1

2
1
1

X_1 = 7, which will change as follows: 7 \rightarrow 1 \rightarrow 0. Thus, f(7) = 2.

X_2 = 1, which will change as follows: 1 \rightarrow 0. Thus, f(1) = 1.

X_3 = 2, which will change as follows: 2 \rightarrow 0. Thus, f(2) = 1.

Sample Input 2

23
00110111001011011001110

Sample Output 2

2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3

Solution in PyPy3 (7.3.0) (judge verdict: Wrong Answer)


import sys, io, os, re
from bisect import bisect, bisect_left, bisect_right, insort, insort_left, insort_right
from pprint import pprint
from math import sin, cos, pi, radians, sqrt, floor, ceil
from copy import copy, deepcopy
from collections import deque
from fractions import gcd
from functools import reduce
from itertools import groupby
from heapq import heapify, heappush, heappop

# sys.setrecursionlimit(5000)

n1 = lambda: int(sys.stdin.readline().strip())
nn = lambda: list(map(int, sys.stdin.readline().strip().split()))
f1 = lambda: float(sys.stdin.readline().strip())
fn = lambda: list(map(float, sys.stdin.readline().strip().split()))
s1 = lambda: sys.stdin.readline().strip()
sn = lambda: list(sys.stdin.readline().strip().split())
nl = lambda n: [n1() for _ in range(n)]
fl = lambda n: [f1() for _ in range(n)]
sl = lambda n: [s1() for _ in range(n)]
nm = lambda n: [nn() for _ in range(n)]
fm = lambda n: [fn() for _ in range(n)]
sm = lambda n: [sn() for _ in range(n)]

def array1(n, d=0): return [d] * n
def array2(n, m, d=0): return [[d] * m for x in range(n)]
def array3(n, m, l, d=0): return [[[d] * l for y in xrange(m)] for x in xrange(n)]
def linc(A, d=1): return map(lambda x: x + d, A)
def ldec(A, d=1): return map(lambda x: x - d, A)

N = n1()
X = int(s1(), 2)

def flip(n, i):
    return n ^ (1 << i)

popcount = lambda x: bin(x).count("1")
bits = lambda x, n = 0: bin(x)[2:].zfill(n)

dp = {}

def search(n):
    if n == 0:
        return 0

    if n in dp:
        return dp[n]

    m = n % popcount(n)
    t = search(m) + 1
    dp[n] = t
    return t

#for i in range(N-1, -1, -1):
#    print(search(flip(X, i)))

p = popcount(X)
r1 = X % (p + 1)
r2 = X % (p - 1) if p > 1 else 0

for i in range(N-1, -1, -1):
    Y = flip(X, i)

    if Y == 0:
        print(0)
        continue

    b = bits(X, N)[i]
    if b == 0:
        t = (r1 + 1 << i) % (p + 1)
    else:
        if p == 1:
            print(1)
            continue
        else:
            t = (r2 - 1 << i) % (p - 1)

    print(search(Y))