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 C++ (GCC 9.2.1) (judge verdict: Accepted)


#include<bits/stdc++.h>
 #define ll long long
 #define ii pair<int,int>
 #define endl '\n'
using namespace std;
int inf = 1e9;
void q() { cout << -1 << endl; }

int pc(int n)
{
    int r=0;
    while(n)
    {
        r += n%2;
        n/=2;
    }
    return r;
}
void solve(){

    int dp[300001] = {0,};
    for(int i=1; i<=300000; i++)
        dp[i] = dp[i%pc(i)] + 1;

    int n;
    cin >> n;
    string s;
    cin >> s;

    int rem1 = 0;
    int rem2 = 0;
    int bits = 0;

    int ten1[n];
    int ten2[n];
    for(int i=0; i<s.size(); i++)
        if(s[i] == '1')
            bits++;
        
    int div1 = bits+1;
    int div2 = bits-1;

    for(int i=n-1, r=1; i>=0; i--)
    {
        ten1[i] = r;
        rem1 = (rem1+r*(s[i]=='1'))%div1;
        r = (r*2)%div1;
    }

    if(div2>0)
    {
        for(int i=n-1, r=1; i>=0; i--)
        {
            ten2[i] = r;
            rem2 = (rem2+r*(s[i]=='1'))%div2;
            r = (r*2)%div2;
        }
    }
    for(int i=0; i<n; i++)
    {
        if(s[i]=='1')
        {
            if(div2==0)
                cout << 0 << endl;
            else
            {
            ll t = (rem2 - ten2[i]+div2)%div2;
            cout << dp[t]+1 << endl;
            }
        }
        else
        {
            ll t = (rem1 + ten1[i]+div1)%div1;
            cout << dp[t]+1 << endl;
        }
    }
}
 
int main(){

/* 
    int t;
    cin >> t;
    for(int i=1; i<=t; i++) */
        solve();
}