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 Perl (5.26.1) (judge verdict: Accepted)

#!/usr/bin/perl
use Data::Dumper;
my ($n) = map { $_ - 0 } split(/\s+/,<STDIN>);
( my $x = scalar(<STDIN>) ) =~ s/[\x00-\x20]+$//ios;

my $bit_cnt = 0;
for(my $i=0;$i<$n;$i++){
  my $bit = substr($x,$n-1-$i,1) - 0;
  $bit_cnt++ if $bit == 1;
}

if( $n == 1 ){
  my $r = &f( oct('0b' . $x) ^ 1 );
  print "$r\n"; exit(0);
}
if( $n == 2 ){
  my $r1 = &f( oct('0b' . $x) ^ 2 );
  my $r2 = &f( oct('0b' . $x) ^ 1 );
  print "$r1\n$r2\n"; exit(0);
}
if( $n == 3 ){
  my $r1 = &f( oct('0b' . $x) ^ 4 );
  my $r2 = &f( oct('0b' . $x) ^ 2 );
  my $r3 = &f( oct('0b' . $x) ^ 1 );
  print "$r1\n$r2\n$r3\n"; exit(0);
}

if( $n == 4 and 1 == 0 ){
  my $r1 = &f( oct('0b' . $x) ^ 8 );
  my $r2 = &f( oct('0b' . $x) ^ 4 );
  my $r3 = &f( oct('0b' . $x) ^ 2 );
  my $r4 = &f( oct('0b' . $x) ^ 1 );
  print "$r1\n$r2\n$r3\n$r4\n"; exit(0);
}

my @d1 = ();
$#d1 = 200000;
my @d3 = ();
$#d3 = 200000;

my $m_1 = ( $bit_cnt > 1 ? ( 2 % ($bit_cnt-1) ) : 0 );
my $m_3 = ( 2 % ($bit_cnt+1) );

my $b_1 = 1;
my $b_3 = 1;

my $t_1 = 0;
my $t_3 = 0;

for(my $i=0;$i<$n;$i++){
  my $bit = substr($x,$n-1-$i,1) - 0;
  if( $bit_cnt > 1 ){
    my $v1 = $b_1 * $bit;
    $d1[$i] = $b_1;
    $t_1 = ($t_1 + $v1) % ($bit_cnt-1);
    $b_1 = ( $b_1 * $m_1 ) % ($bit_cnt-1);
  }
  my $v3 = $b_3 * $bit;
  $d3[$i] = $b_3;
  $t_3 = ($t_3 + $v3) % ($bit_cnt+1);
  $b_3 = ( $b_3 * $m_3 ) % ($bit_cnt+1);
}

#print &Dumper(@d1[0..22]);
#print &Dumper(@d3[0..22]);

#print "t1 =$t_1 , t3=$t_3\n";

my @out = ();
$#out = $n-1;

for(my $i=0;$i<$n;$i++){
  my $bit = substr($x,$n-1-$i,1) - 0;
  if( $bit == 1 ){
    if( $bit_cnt == 1 ){
      $out[$i] = 0;
    } else {
      my $tv1 = ( $t_1 - $d1[$i] + ($bit_cnt - 1) ) % ($bit_cnt - 1);
      my $v1 = 1 + &f($tv1);
      $out[$i] = $v1;
    }
  } else {
    my $tv3 = ( $t_3 + $d3[$i] + ($bit_cnt + 1) ) % ($bit_cnt + 1);
    #print "i=$i , tv3=$tv3\n";
    my $v3 = 1 + &f($tv3);
    $out[$i] = $v3;
  }
}

for(my $i=0;$i<$n;$i++){
  print "$out[$n-1-$i]\n";
}

exit(0);

sub f {
  my $n1 = shift;
  my $c = 0;
  return 0 if $n1 == 0;
  return 1 if $n1 == 1;
  while($n1>0){
    my $bit_c = 0;
    my $n2 = $n1;
    while($n2>0){
      $bit_c ++ if ( $n2 & 1 ) != 0;
      $n2 >>= 1;
    }
    $n1 = $n1 % $bit_c;
    $c++;
    if( $n1 == 1 ){
      $c++;
      last;
    }
  }
  return $c;
}