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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define N 200005

char a[N];
int n,num,x,y,modx[N],mody[N],ans[N],pre[N],suf[N];


int calc(int x){//把x除到0需要的步数 
	int res=0;
	while(x){
		res++;
		int tot=0;
		for(int i=0;(1<<i)<=x;i++)
			tot+=(x>>i & 1);
		x%=tot;
	} 
	return res;
}

int main(){
	cin>>n;
	cin>>(a+1);
	
	for(int i=1;i<=n;i++)
		if(a[i]=='1')num++;
	if(num==0){//没有1的时候 
		for(int i=1;i<=n;i++)cout<<1<<'\n';
		return 0;
	}
	if(num==1){//只有一个1的时候 
		if(a[n]=='1')
			for(int i=1;i<=n;i++)ans[i]=(i==n?0:2); 
		else 
			for(int i=1;i<=n;i++)ans[i]=(i==n?2:1);
		for(int i=1;i<=n;i++)cout<<ans[i]<<'\n';
		return 0;
	}	
		
	
	x=num+1;y=num-1;//y>=1
	// 先把操作一次得到的结果求出来
	 
	long long base=1;
	for(int i=n;i>=1;i--){
		base%=x;
		modx[i]=base;
		base*=2;
	}
	for(int i=1;i<=n;i++){
		if(a[i]=='1')pre[i]=modx[i]+pre[i-1];
		else pre[i]=pre[i-1];
		pre[i]%=x;
	}
	for(int i=n;i>=1;i--){
		if(a[i]=='1')suf[i]=modx[i]+suf[i+1];
		else suf[i]=suf[i+1];
		suf[i]%=x;
	}
	for(int i=1;i<=n;i++)if(a[i]=='0'){
		ans[i]=pre[i-1]+suf[i+1]+modx[i];
		ans[i]%=x;
	}
	
	base=1;
	for(int i=n;i>=1;i--){
		base%=y;
		mody[i]=base;
		base*=2;
	}
	for(int i=1;i<=n;i++){
		if(a[i]=='1')pre[i]=mody[i]+pre[i-1];
		else pre[i]=pre[i-1];
		pre[i]%=y;
	}
	for(int i=n;i>=1;i--){
		if(a[i]=='1')suf[i]=mody[i]+suf[i+1];
		else suf[i]=suf[i+1];
		suf[i]%=y;
	}
	for(int i=1;i<=n;i++)if(a[i]=='1'){
		ans[i]=pre[i-1]+suf[i+1];
		ans[i]%=y;
	}
	 
	for(int i=1;i<=n;i++)
		cout<<calc(ans[i])+1<<'\n';
}