Problem:
Score : 300 points

Problem Statement

Given an integer N, find the base -2 representation of N.

Here, S is the base -2 representation of N when the following are all satisfied:

S is a string consisting of 0 and 1.

Unless S = 0, the initial character of S is 1.

Let S = S_k S_{k-1} ... S_0, then S_0 \times (-2)^0 + S_1 \times (-2)^1 + ... + S_k \times (-2)^k = N.

It can be proved that, for any integer M, the base -2 representation of M is uniquely determined.

Constraints

Every value in input is integer.

-10^9 \leq N \leq 10^9

Input

Input is given from Standard Input in the following format:

N

Output

Print the base -2 representation of N.

Sample Input 1

-9

Sample Output 1

1011

As (-2)^0 + (-2)^1 + (-2)^3 = 1 + (-2) + (-8) = -9, 1011 is the base -2 representation of -9.

Sample Input 2

123456789

Sample Output 2

11000101011001101110100010101

Sample Input 3

0

Sample Output 3

0

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <string>
#include <vector>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <queue>
#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <cstring>

using namespace std;

int neg(long long int x){
	int cnt=0;
	if(x==1){
		return 0;
	}
	while(x!=0){
		x/=-2;
		cnt++;
	}
	return cnt;
}

int main(){
	long long int N;
	cin>>N;
	if(N==0){
		cout<<"0"<<endl;
		return 0;
	}
	vector<char> s(64);
	for(int i=0;i<s.size();i++){
		s[i]='0';
	}
	int offset;
	if(N<0){
		offset=1;
	}else{
		offset=-1;
	}
	while(N!=0){
		int idx=neg(N);
		long long int tmp1=pow(-2,idx);
		long long int tmp2=pow(-2,idx-1);
		long long int tmp3=pow(-2,idx+1);

		if(N==tmp1){
			offset++;
		}else if(N==tmp2){
			idx--;
		}else if(N==tmp3){
			idx++;
		}else if(N<0 &&abs(N-tmp1)>=abs(N-tmp2)){
			idx--;
		}else if(N>0 && abs(N-tmp1)>=abs(N-tmp3)){
			idx++;
		}
		s[idx]='1';
		N-=pow(-2,idx);
		//cout<<N<<" ("<<pow(-2,idx)<<endl;

	}
	bool flg=false;
	for(int i=s.size()-1;i>=0;i--){
		if(s[i]=='1'){
			flg=true;
		}
		if(flg){
			cout<<s[i];
		}
	}
	cout<<endl;
	return 0;
}