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 Java (OpenJDK 1.8.0) (judge verdict: Wrong Answer)


import java.util.Arrays;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        char[] c = sc.next().toCharArray();

        long popcount = 0;       //xのpopcount 実際使うのはこれの+1か-1
        for(int i=0;i<n;i++){
            if(c[i] == '1'){
                popcount++;
            }
        }

        long[] num = new long[200001];
        for(int i=0;i<200001;i++){
            num[i] = i;
        }
        long[] popcalc = new long[200001];    //200000までのpopcount計算の結果
        for(int i=0;i<18;i++){
            for(int j=(int) Math.pow(2,i); j<200001; j++){
                if(num[j]%2==1){
                    popcalc[j]++;
                }
                num[j] /= 2;
            }
        }
        for(int i=1;i<200001;i++){
            popcalc[i] = i % popcalc[i];
        }

        //System.out.println(popcalc[26] + ", " + popcalc[27] + ", " + popcalc[28] + ", " + popcalc[29] + ", " + popcalc[30]);

        long[] popplus = new long[n];
        long[] popminus = new long[n];
        int flag = -1;
        for(int i=0;i<n;i++){
            if(flag != -1){
                popplus[n-1-i] = popplus[n-1-i + flag];
            }else {
                popplus[n - 1 - i] = (long) (Math.pow(2, i) % (popcount + 1));
                if(i != 0 && popplus[n-1-i] == popplus[n-1]){
                    flag = i;
                }
            }
        }
        flag = -1;
        if(popcount <= 1){
            Arrays.fill(popminus, 0);
        }else {
            for (int i = 0; i < n; i++) {
                if (flag != -1) {
                    popminus[n - 1 - i] = popminus[n - 1 - i + flag];
                } else {
                    popminus[n - 1 - i] = (int) (Math.pow(2, i) % (popcount - 1));
                    if (i != 0 && popminus[n - 1 - i] == popminus[n - 1]) {
                        flag = i;
                    }
                }
            }
        }
        long popxplus = 0;
        long popxminus = 0;
        for(int i=0;i<n;i++){
            if(c[n-1-i] == '1') {
                popxplus += popplus[n-1-i];
                popxplus %= popcount+1;
                popxminus += popminus[n-1-i];
                if(popcount > 1) {
                    popxminus %= popcount - 1;
                }
            }
        }

        for(int i=0;i<n;i++){
            long count = 1;
            if(c[i] == '0'){
                long search = popxplus + popplus[i];
                search %= popcount+1;
                while(search != 0){
                    count++;
                    search = popcalc[(int) search];
                }
            }else{
                if(popcount > 1){
                    long search = popxminus - popminus[i] + popcount-1;
                    search %= popcount-1;
                    while(search != 0){
                        count++;
                        search = popcalc[(int) search];
                    }
                }
            }
            System.out.println(count);
        }
    }
}